Polynomials

Introduction

{{FORMULA: expr=ax² + bx + c | symbols=a: coefficient of x², b: coefficient of x, c: constant term}}

Introduction to Polynomials

Welcome to the world of polynomials! You've actually been using them for years without even knowing it. Think about the simple act of throwing a ball to a friend. The path it travels through the air — rising to a peak and then falling back down — is a perfect parabola. This curve can be precisely described by a polynomial, specifically a quadratic one.

An equation like h(t) = -5t² + 20t + 1.5 could represent the height (h) of the ball at any given time (t). Polynomials are the mathematical language we use to model relationships like these, from the trajectories of objects to the profits of a business. In this chapter, we'll decode this language, starting with understanding their structure, their "zeros," and the powerful connections between their parts.

What is a Polynomial?

A polynomial is an algebraic expression consisting of variables (like x or y), coefficients (the numbers multiplying the variables), and constants, combined using only addition, subtraction, and multiplication. The key rule is that the exponents of the variables must be non-negative integers (0, 1, 2, 3, ...).

The degree of a polynomial is the highest exponent of its variable.

• 3x + 1 is a linear polynomial (degree 1).
• 5y² - 3y + 2 is a quadratic polynomial (degree 2).
• x³ - 8 is a cubic polynomial (degree 3).

This page focuses on quadratic polynomials, the workhorses of algebra.

Definitions & Formulas

Let's define the key terms and formulas for a standard quadratic polynomial p(x) = ax² + bx + c, where a ≠ 0.

The Relationship Between Zeros and Coefficients: A Derivation

Have you ever wondered why the sum of zeros is -b/a? It's not magic; it's a logical consequence of how polynomials are built. Let's derive this relationship step-by-step.

1. Consider a general quadratic polynomial, p(x).

   p(x) = ax² + bx + c
   

2. Let's assume its two zeros are α and β. By the definition of a zero, if x = α or x = β, then p(x) = 0. This means that (x - α) and (x - β) must be the factors of the polynomial.

3. So, we can write the polynomial in its factored form. We include a constant k because the factors could be a multiple of the simplest form (e.g., 2x² + 4x + 2 has the same zeros as x² + 2x + 1, but is scaled by k=2).

   p(x) = k(x - α)(x - β)
   

4. Now, let's expand this factored form by multiplying the terms.

   p(x) = k[x(x - β) - α(x - β)]
   p(x) = k[x² - βx - αx + αβ]
   p(x) = k[x² - (α + β)x + αβ]
   

5. Distribute the constant k into the expression.

   p(x) = kx² - k(α + β)x + k(αβ)
   

6. Now we have two forms of the same polynomial. Let's compare the coefficients of our original form (ax² + bx + c) with this expanded form.

   • The x² term: a = k
   • The x term: b = -k(α + β)
   • The constant term: c = k(αβ)

7. From these comparisons, we can find the relationships.

   • For the sum of zeros: Since b = -k(α + β) and a = k, we can substitute a for k. b = -a(α + β) Rearranging gives us the formula for the sum of zeros:

     α + β = -b/a
     

   • For the product of zeros: Since c = k(αβ) and a = k, we substitute a for k. c = a(αβ) Rearranging gives us the formula for the product of zeros:

     αβ = c/a
     

And there you have it! The relationship is a direct result of comparing the standard and factored forms of a polynomial.

{{KEY: type=concept | title=The Core Relationship | text=For any quadratic polynomial ax² + bx + c, the sum of its zeros is always -b/a, and the product of its zeros is always c/a. This fundamental link allows you to find information about the zeros without actually calculating them.}}

Solved Examples

Let's put this theory into practice with examples of increasing difficulty.

Example 1: Finding Zeros and Verifying the Relationship (Easy)

Given: The quadratic polynomial p(x) = x² + 7x + 10.

To Find: The zeros of the polynomial and verify the relationship between the zeros and the coefficients.

Solution:

1. First, find the zeros by factoring the polynomial (middle term splitting). We need two numbers that multiply to 10 and add up to 7. These are 5 and 2.

   x² + 5x + 2x + 10 = 0
   x(x + 5) + 2(x + 5) = 0
   (x + 2)(x + 5) = 0
   

2. Set each factor to zero to find the zeros.

   x + 2 = 0  =>  x = -2
   x + 5 = 0  =>  x = -5
   

   So, the zeros are α = -2 and β = -5.

3. Now, let's verify the relationship. First, identify the coefficients from x² + 7x + 10. Here, a = 1, b = 7, c = 10.

4. Verification of Sum of Zeros: Sum from our found zeros: α + β = (-2) + (-5) = -7. Sum from the formula: -b/a = -(7)/1 = -7. The sums match.

5. Verification of Product of Zeros: Product from our found zeros: αβ = (-2) × (-5) = 10. Product from the formula: c/a = 10/1 = 10. The products match.

Final Answer: The zeros are -2 and -5. The relationship is verified as Sum = -7 and Product = 10.

Example 2: Working with a Different Form (Medium)

Given: The quadratic polynomial p(s) = 4s² - 4s + 1.

To Find: The zeros of the polynomial and verify the relationship.

Solution:

1. Identify the coefficients. The polynomial is already in standard form. a = 4, b = -4, c = 1.

2. Find the zeros. This polynomial is a perfect square trinomial: (2s - 1)².

   (2s - 1)² = 0
   (2s - 1)(2s - 1) = 0
   

3. This gives us two equal roots.

   2s - 1 = 0  =>  s = 1/2
   

   So, the zeros are α = 1/2 and β = 1/2.

4. Verification of Sum of Zeros: Sum from zeros: α + β = 1/2 + 1/2 = 1. Sum from formula: -b/a = -(-4)/4 = 4/4 = 1. The sums match.

5. Verification of Product of Zeros: Product from zeros: αβ = (1/2) × (1/2) = 1/4. Product from formula: c/a = 1/4. The products match.

Final Answer: The zeros are 1/2 and 1/2. The relationship is verified as Sum = 1 and Product = 1/4.

Example 3: Constructing a Polynomial from its Zeros (Hard)

Given: The sum of the zeros is 1/4 and the product of the zeros is -1.

To Find: The quadratic polynomial.

Solution:

1. Recall the general form of a quadratic polynomial constructed from its sum (S) and product (P) of zeros.

   p(x) = k[x² - (S)x + P]
   

   Here, S = α + β and P = αβ. k is any non-zero real number.

2. Substitute the given values of S and P into the formula.

   S = 1/4
   P = -1
   

   p(x) = k[x² - (1/4)x + (-1)]
   p(x) = k[x² - x/4 - 1]
   

3. To write the polynomial with integer coefficients, we can eliminate the fraction. Let's find a common denominator, which is 4.

   p(x) = k[(4x² - x - 4) / 4]
   

4. We can choose a convenient value for k to simplify the expression. If we let k = 4, the denominator cancels out.

   p(x) = 4 × [(4x² - x - 4) / 4]
   p(x) = 4x² - x - 4
   

   Note: Any multiple of this polynomial, like 8x² - 2x - 8 (where k=8), is also a valid answer as it would have the same zeros.

Final Answer: The quadratic polynomial is 4x² - x - 4.

Example 4: Finding a Missing Coefficient (Tricky)

Given: In the polynomial p(x) = (k - 1)x² + kx + 1, one of the zeros is -3.

To Find: The value of k.

Solution:

1. The definition of a zero is a value of x that makes the polynomial equal to 0. So, if x = -3 is a zero, then p(-3) = 0.

2. Substitute x = -3 into the polynomial equation.

   p(-3) = (k - 1)(-3)² + k(-3) + 1 = 0
   

3. Now, solve this linear equation for k.

   (k - 1)(9) - 3k + 1 = 0
   9k - 9 - 3k + 1 = 0
   6k - 8 = 0
   

4. Isolate k.

   6k = 8
   k = 8/6
   

5. Simplify the fraction.

   k = 4/3
   

Final Answer: The value of k is 4/3.

Tips & Tricks

Here are a few shortcuts to speed up your work with quadratic polynomials.

Common Mistakes

Avoid these common pitfalls when dealing with zeros and coefficients.

Brain-Teaser Questions

Test your understanding with these higher-order thinking problems.

1. If α and β are the zeros of the polynomial p(x) = 2x² - 7x + 3, what is the value of 1/α + 1/β?

   💡 Answer: To find 1/α + 1/β, first combine the fractions: (β + α) / αβ. This is just the (Sum of zeros) / (Product of zeros). For 2x² - 7x + 3, a=2, b=-7, c=3. Sum: α + β = -b/a = -(-7)/2 = 7/2. Product: αβ = c/a = 3/2. So, (α + β) / αβ = (7/2) / (3/2) = 7/3.

2. If the sum of the squares of the zeros of the polynomial p(x) = x² - 8x + k is 40, find the value of k.

   💡 Answer: We are given α² + β² = 40. We know the identity α² + β² = (α + β)² - 2αβ. From the polynomial x² - 8x + k, we have a=1, b=-8, c=k. Sum: α + β = -b/a = -(-8)/1 = 8. Product: αβ = c/a = k/1 = k. Substitute these into the identity: 40 = (8)² - 2(k). 40 = 64 - 2k 2k = 64 - 40 2k = 24, so k = 12.

3. One zero of the quadratic polynomial p(x) = 4x² - 8kx - 9 is the negative of the other. Find the value of k.

   💡 Answer: Let the zeros be α and -α. The sum of the zeros is α + (-α) = 0. From the polynomial, the sum of zeros is also given by the formula -b/a. Here a = 4, b = -8k, c = -9. Sum = -(-8k) / 4 = 8k / 4 = 2k. Since the sum of zeros is both 0 and 2k, we can equate them: 2k = 0. Therefore, k = 0.

Mini Cheatsheet

Here's a summary of the key formulas from this page. Screenshot this for quick revision!

Geometrical Meaning of the Zeroes of a Polynomial — Part 1

Chapter 2: Polynomials

Page 2/5: Geometrical Meaning of the Zeroes of a Polynomial — Part 1

Concept Introduction

Have you ever thrown a ball to a friend? The path it takes through the air is a beautiful arc. This arc can be described mathematically by a polynomial, specifically a quadratic polynomial. The point where you threw the ball from and the point where your friend catches it could be thought of as the "zeroes" if we set the ground as our reference line.

In algebra, we often solve equations like x - 2 = 0 or x² - 4 = 0. The solutions we find are called zeroes. But what do these numbers look like? This is where geometry gives us a powerful new perspective. By plotting a polynomial as a graph on a coordinate plane, we can visually identify its zeroes. The zeroes are simply the points where the graph crosses the horizontal line, the x-axis. This visual connection makes abstract algebra tangible and intuitive.

{{FORMULA: expr=y = p(x) | symbols=p(x):a polynomial in variable x, y:the value of the polynomial for a given x}}

Definitions & Formulas

Let's clarify the key terms we'll be using. A polynomial p(x) has a corresponding graphical representation y = p(x).

{{KEY: type=concept | title=The Fundamental Connection | text=The zeroes of a polynomial p(x) are the x-coordinates of the points where the graph of y = p(x) intersects the x-axis.}}

The Logic: Connecting Algebra to Geometry

How can we be sure that the x-intercepts of a graph are the zeroes of the polynomial? The logic is straightforward and builds on the definition of a graph.

1. The Graph: The graph of y = p(x) is a collection of all points (x, y) that satisfy this equation. For every x value we input, we get a corresponding y value.

2. The x-axis: The x-axis is a special line in the coordinate plane. What do all points on the x-axis have in common? Their y-coordinate is always 0. A point on the x-axis is of the form (k, 0).

3. The Intersection: When the graph of y = p(x) intersects the x-axis, it must be at a point that lies on both the graph and the x-axis.

4. Finding the Point: Let's say this intersection point is (k, 0). Since this point is on the graph of y = p(x), it must satisfy the equation. We substitute x = k and y = 0 into y = p(x).

   0 = p(k)
   

5. The Conclusion: This equation, p(k) = 0, is the very definition of a zero. It means that k is a zero of the polynomial p(x). Thus, the x-coordinate of the intersection point is the zero of the polynomial.

Solved Examples

Let's apply this concept to some graphs.

Example 1: Linear Polynomial (Easy)

Given: The graph of a linear polynomial y = p(x).

{{VISUAL: diagram: A Cartesian plane showing a straight line passing through (-2, 0) and (0, 4). The line slopes upwards from left to right. The point (-2, 0) is clearly marked as the x-intercept.}}

To Find: The zero of the polynomial p(x).

Solution:

1. The zero of a polynomial is the x-coordinate of the point where its graph intersects the x-axis.

2. Observe the given graph. The straight line, which represents the polynomial, cuts the x-axis at a single point.

3. The coordinates of this intersection point are (-2, 0).

4. The x-coordinate of this point is -2. Therefore, the zero of the polynomial is -2.

Final Answer:

The zero of the polynomial is -2.

Example 2: Quadratic Polynomial - Parabola (Medium)

Given: The graph of a quadratic polynomial y = f(x).

{{VISUAL: diagram: A Cartesian plane showing a U-shaped parabola opening upwards. The parabola intersects the x-axis at x = -1 and x = 4. The vertex is shown below the x-axis at approximately (1.5, -6.25).}}

To Find: The zeroes of the polynomial f(x).

Solution:

1. The zeroes of the polynomial are the x-coordinates of the points where its graph intersects the x-axis.

2. Looking at the graph, the parabola intersects the x-axis at two distinct points.

3. Let's identify the coordinates of these points. The first point is (-1, 0) and the second point is (4, 0).

4. The x-coordinates of these points are -1 and 4.

Final Answer:

The zeroes of the polynomial are -1 and 4.

Example 3: Finding Zeroes from the Equation (Hard)

Given: The quadratic polynomial p(x) = x² - x - 6.

To Find: The zeroes of the polynomial and determine how many times its graph will intersect the x-axis.

Solution:

1. To find the zeroes, we set the polynomial equal to zero. This is the algebraic method.

   p(x) = 0
   

   x² - x - 6 = 0
   

2. Now, we factorize the quadratic expression by splitting the middle term. We need two numbers that multiply to -6 and add to -1. These numbers are -3 and +2.

   x² - 3x + 2x - 6 = 0
   

3. Factor by grouping.

   x(x - 3) + 2(x - 3) = 0
   

   (x + 2)(x - 3) = 0
   

4. This gives us two possible solutions for x.

   x + 2 = 0   OR   x - 3 = 0
   

   x = -2   OR   x = 3
   

5. The zeroes of the polynomial are -2 and 3.

6. Since there are two distinct real zeroes, the graph of the polynomial (a parabola) will intersect the x-axis at exactly two points: (-2, 0) and (3, 0).

Final Answer:

The zeroes are -2 and 3. The graph will intersect the x-axis 2 times.

{{VISUAL: diagram: A Cartesian plane with a U-shaped parabola for y = x² - x - 6. It opens upwards and clearly intersects the x-axis at x = -2 and x = 3.}}

Example 4: Interpreting Complex Graphs (Tricky)

Given: The graph of a polynomial y = g(x).

{{VISUAL: diagram: A Cartesian plane with a W-shaped curve, characteristic of a quartic polynomial. It crosses the x-axis at four points: approximately x = -2.5, x = -0.5, x = 1, and x = 3.}}

To Find: The number of zeroes of the polynomial g(x).

Solution:

1. The question asks for the number of zeroes, not their specific values.

2. The number of zeroes is equal to the number of times the graph intersects the x-axis.

3. Carefully trace the curve and count every point where it crosses or touches the x-axis.

4. The graph intersects the x-axis at four distinct points.

5. Therefore, the polynomial g(x) has four zeroes.

Final Answer:

The polynomial g(x) has 4 zeroes.

Tips & Tricks

Use these shortcuts to quickly analyze polynomial graphs.

Common Mistakes

Many students make these simple errors. Be careful to avoid them!

Brain-Teaser Questions

Test your understanding with these slightly more challenging problems.

1. The graph of a quadratic polynomial p(x) = x² - 8x + 16 is drawn. How many times will it intersect the x-axis?

   💡 Answer: The polynomial is p(x) = (x-4)². Setting p(x) = 0 gives (x-4)² = 0, so x=4 is the only zero. Since there is only one real zero, the graph will not cross the x-axis but will touch it at exactly one point, (4, 0).

2. If the graph of y = x² - 4 is shifted upwards by 5 units, what will be the new polynomial, and how many zeroes will it have?

   💡 Answer: Shifting upwards by 5 units means adding 5 to the polynomial. The new polynomial is y = (x² - 4) + 5 = x² + 1. To find its zeroes, we set x² + 1 = 0, which gives x² = -1. There is no real number whose square is negative. Therefore, the new polynomial has zero real zeroes, and its graph will not intersect the x-axis at all.

3. A student claims the graph of a circle x² + y² = 9 represents a polynomial. Are they correct? Why or why not?

   💡 Answer: No, they are incorrect. A polynomial graph must pass the "Vertical Line Test" – any vertical line drawn on the graph can intersect it at most once. A circle fails this test (a vertical line can cut it twice). This means for a single x value, there are two y values, which is not allowed for a function y = p(x).

Mini Cheatsheet

Here’s a quick summary of this page for your revision. Screenshot this!

Geometrical Meaning of the Zeroes of a Polynomial — Part 2

Chapter 2: Polynomials

Page 3 of 5: Geometrical Meaning of the Zeroes of a Polynomial — Part 2

Concept Introduction

In our last session, we saw that the zero of a linear polynomial corresponds to a single point where its straight-line graph crosses the x-axis. But what about more complex polynomials? What story do their graphs tell?

Imagine you're standing in a field and you throw a cricket ball up to a friend. The path the ball takes is not a straight line; it's a beautiful, symmetric curve called a parabola. This parabolic arc is the graph of a quadratic polynomial. The points on the ground where the ball would have started and landed (if thrown from ground level) are the "zeroes" of this polynomial. They represent the moments when the ball's height is zero. By studying the shape of this graph, we can visually understand solutions that are not just abstract numbers but have real-world meaning, like distance, time, or position. This section explores this powerful connection between algebra and geometry for quadratic and cubic polynomials.

{{VISUAL: diagram: A simple landscape showing a person throwing a ball. The path of the ball is traced as a downward-opening parabola. The x-axis represents the ground level, and the points where the ball starts (hand-level) and lands are marked, with the x-intercepts highlighted as theoretical start/end points if thrown from ground level.}}

Definitions & Formulas

Understanding the visual language of polynomials starts with a few key terms.

The Logic: Interpreting the Graphs

Let's build upon our understanding from linear polynomials to see what the graphs of quadratic and cubic polynomials reveal about their zeroes.

1. Recap: Linear Polynomials The graph of a linear polynomial p(x) = ax + b is a straight line. It intersects the x-axis at exactly one point, (-b/a, 0). Thus, a linear polynomial has exactly one zero.

2. Introducing Quadratic Polynomials and Parabolas The graph of a quadratic polynomial, p(x) = ax² + bx + c, is a parabola. This curve is shaped like a 'U'. The orientation of this 'U' depends on the sign of the leading coefficient, a.

3. The Role of the Leading Coefficient 'a'

   • If a > 0, the parabola opens upwards (like a cup holding water). It has a minimum point.
   • If a < 0, the parabola opens downwards (like an umbrella). It has a maximum point.

4. Connecting Zeroes to x-intercepts Just like with linear polynomials, the zeroes of a quadratic polynomial are the x-coordinates of the points where its graph (the parabola) intersects the x-axis. At these points, the y-coordinate is 0, which means p(x) = 0.

5. Three Possibilities for a Quadratic Polynomial Because a parabola is a curve, it doesn't have to intersect the x-axis. This leads to three distinct cases for the number of zeroes:

   • Case 1: Two Distinct Zeroes. The parabola intersects the x-axis at two different points. This means the quadratic polynomial has two unique real zeroes.
   • Case 2: Exactly One Zero. The parabola touches the x-axis at exactly one point (at its vertex). This means the quadratic polynomial has one real zero (also called a repeated root).
   • Case 3: No Real Zeroes. The parabola is entirely above or entirely below the x-axis. It never intersects the x-axis. This means the quadratic polynomial has no real zeroes.

{{VISUAL: diagram: Three coordinate planes side-by-side. The first shows a parabola y = ax² + bx + c (with a>0) intersecting the x-axis at two distinct points (α, 0) and (β, 0). The second shows a parabola touching the x-axis at one point (α, 0). The third shows a parabola entirely above the x-axis, not intersecting it.}}

6. A Glimpse into Cubic Polynomials The graph of a cubic polynomial p(x) = ax³ + bx² + cx + d is a continuous curve that typically has one "hump" and one "valley". Such a graph can intersect the x-axis at most three times. Therefore, a cubic polynomial can have a maximum of three zeroes. It will always have at least one real zero.

{{KEY: type=concept | title=The Fundamental Connection | text=The number of real zeroes of a polynomial is equal to the number of times its graph intersects the x-axis.}}

Solved Examples

Let's apply these concepts to solve some problems, starting from easy and moving to more challenging ones.

Example 1: Counting Zeroes from a Graph (Easy)

Given: The graph of a polynomial y = p(x).

{{VISUAL: diagram: A coordinate plane showing the graph of the parabola y = x² - 4. The parabola opens upwards, its vertex is at (0, -4), and it intersects the x-axis at (-2, 0) and (2, 0).}}

To Find: The number of zeroes of the polynomial p(x).

Solution:

1. Recall that the zeroes of a polynomial are the x-coordinates of the points where its graph intersects the x-axis.

2. Observe the given graph. We need to count how many times the curve crosses or touches the horizontal x-axis.

3. The graph intersects the x-axis at two distinct points: x = -2 and x = 2.

4. Since there are two points of intersection, the polynomial has two zeroes.

Final Answer:

The polynomial p(x) has 2 zeroes.

Example 2: Finding Zeroes and Verifying Geometrically (Medium)

Given: The quadratic polynomial p(x) = x² - 2x - 8.

To Find: The zeroes of the polynomial and describe the geometrical representation.

Solution:

1. First, we find the zeroes algebraically by setting p(x) = 0.

   x² - 2x - 8 = 0
   

2. We can solve this quadratic equation by splitting the middle term. We need two numbers that multiply to -8 and add to -2. These numbers are -4 and +2.

   x² - 4x + 2x - 8 = 0
   

3. Factor by grouping.

   x(x - 4) + 2(x - 4) = 0
   

4. Take (x - 4) as the common factor.

   (x - 4)(x + 2) = 0
   

5. This gives us two possible solutions for x.

   x - 4 = 0  OR  x + 2 = 0
   

   x = 4   OR   x = -2
   

6. The zeroes are 4 and -2. Geometrically, this means the graph of y = x² - 2x - 8 is a parabola that intersects the x-axis at the points (4, 0) and (-2, 0). Since the coefficient of x² is 1 (which is > 0), the parabola opens upwards.

Final Answer:

The zeroes are 4 and -2. The graph is an upward-opening parabola that cuts the x-axis at x = 4 and x = -2.

Example 3: Interpreting a Tangent Graph (Hard)

Given: The graph of a quadratic polynomial p(x) touches the x-axis at the point (3, 0) and passes through the point (1, 8).

To Find: The number of zeroes and the direction in which the parabola opens.

Solution:

1. The problem states that the graph touches the x-axis at exactly one point, (3, 0).

2. According to our understanding (Case 2), when the graph of a quadratic polynomial touches the x-axis at a single point, the polynomial has exactly one real zero (a repeated root).

3. Therefore, the number of zeroes is 1. The zero is x = 3.

4. To determine the direction, we can use the two given points: the vertex (3, 0) and another point on the parabola (1, 8).

5. The point (1, 8) has a y-coordinate of 8, which is positive (above the x-axis). The vertex (3, 0) is on the x-axis.

6. Since a point (1, 8) on the parabola is "above" its vertex (3, 0), the parabola must be opening upwards. If it opened downwards, all other points would have y-coordinates less than or equal to 0.

Final Answer:

The polynomial has exactly one zero. The parabola opens upwards.

Example 4: Deducing Polynomial Type from a Graph (Tricky)

Given: The graph of a polynomial y = p(x) is shown below.

{{VISUAL: diagram: A coordinate plane with the graph of a cubic polynomial y = p(x). The curve starts from the bottom-left, goes up crossing the x-axis at x = -3, reaches a local maximum, comes down crossing the x-axis again at x = 1, reaches a local minimum, and then goes up, crossing the x-axis a third time at x = 4 before moving to the top-right.}}

To Find: The number of zeroes of p(x) and the minimum possible degree of the polynomial.

Solution:

1. To find the number of zeroes, we count the number of distinct points where the graph intersects the x-axis.

2. Observing the graph, the curve crosses the x-axis at three distinct points. Let's assume they are x = -3, x = 1, and x = 4.

3. Since there are three points of intersection, the polynomial p(x) has 3 zeroes.

4. Now, let's determine the minimum possible degree.

   • A linear polynomial has at most 1 zero.
   • A quadratic polynomial has at most 2 zeroes.
   • A cubic polynomial has at most 3 zeroes.
   • A polynomial of degree n has at most n zeroes.

5. Since our polynomial has 3 zeroes, its degree must be at least 3. A quadratic or linear polynomial cannot have 3 zeroes.

6. Therefore, the minimum possible degree for this polynomial is 3. It could be a cubic polynomial.

Final Answer:

The polynomial has 3 zeroes. The minimum possible degree of the polynomial is 3.

Tips & Tricks

Use these shortcuts to analyze graphs quickly and accurately.

Common Mistakes

Many students stumble on similar points. Here’s a guide on what to avoid.

Brain-Teaser Questions

Test your deeper understanding with these tricky problems.

1. The graph of a quadratic polynomial p(x) = ax² + bx + c does not intersect the x-axis at all. If the coefficient a is negative (a < 0), what can you definitively say about the sign of c?

   💡 Answer: If a < 0, the parabola opens downwards. If it does not intersect the x-axis, the entire parabola must be below the x-axis. The y-intercept is the point where x = 0, which is p(0) = c. Since the entire graph is below the x-axis, the y-intercept must also be below the x-axis. Therefore, c must be negative (c < 0).

2. A polynomial of degree 5 is drawn. It is known that it has some non-real (imaginary) zeroes. What are the possible numbers of real zeroes this polynomial can have?

   💡 Answer: For polynomials with real coefficients, non-real zeroes always occur in conjugate pairs. This means there must be an even number of them (2, 4, 6...). A polynomial of degree 5 has a total of 5 zeroes (real + non-real).

   • If there are 2 non-real zeroes, there must be 5 - 2 = 3 real zeroes.
   • If there are 4 non-real zeroes, there must be 5 - 4 = 1 real zero. It cannot have 0 non-real zeroes because the question states it has some. It cannot have 6 non-real zeroes as the degree is only 5. So, the possible number of real zeroes are 1 or 3.

3. The graph of y = p(x) is a parabola with its vertex at the point (-1, -5). How many real zeroes does this polynomial have?

   💡 Answer: The vertex of a parabola is its minimum or maximum point. Here, the vertex is at (-1, -5). This point is below the x-axis. A parabola can either open upwards or downwards.

   • If it opens upwards, its minimum point is (-1, -5). To go up from there, it must cross the x-axis twice.
   • If it opens downwards, its maximum point is (-1, -5). It will continue to go down and will never cross the x-axis. The question doesn't specify if it opens up or down. Wait, any point (x, y) on the parabola is given by y=p(x). A parabola is the graph of a quadratic polynomial. Its general vertex form is y = a(x-h)² + k. Here (h,k) = (-1,-5). So, y = a(x+1)² - 5. This parabola must have a vertex at (-1, -5). If a > 0, it opens up and crosses the x-axis twice. If a < 0, it opens down and never crosses the x-axis. Hmm, the question is ambiguous. Let's assume a standard context. A-ha! In CBSE, "parabola" in this chapter almost always implies a quadratic polynomial. A single point vertex is not enough information. But wait, if it is a polynomial graph, does that help? Let's re-read. Ah, it's a brain-teaser. The key is that it is a parabola. Its vertex is its lowest or highest point. Since the vertex (-1, -5) is below the x-axis, if the parabola opens upwards, it must cross the x-axis to reach positive y-values. If it opens downwards, it will only go further down and never cross. The question is a bit ill-posed without knowing the direction. Let's provide a complete answer. If the parabola opens upwards, it has two real zeroes. If it opens downwards, it has no real zeroes. A better question would specify the sign of a. Let's assume the question implies there's a definite answer. If a question is posed this way, there might be a subtle assumption. But based purely on the given information, both are possible. Let's refine the answer for a student: The question is tricky! A parabola has a vertex, which is its minimum point (if it opens up) or maximum point (if it opens down). The vertex is at y = -5. Case A: If the parabola opens upwards, its lowest point is -5. It must rise from there and will cross the x-axis (where y=0) twice. So, 2 zeroes. Case B: If the parabola opens downwards, its highest point is -5. It will only go lower and will never reach the x-axis. So, 0 zeroes. Without knowing the direction, we cannot give a single answer. However, in many contexts, such a question implies a general case, where an upward-opening parabola is often assumed if not specified. Let's stick with the complete answer: The polynomial has either 2 real zeroes (if it opens upwards) or 0 real zeroes (if it opens downwards).

Mini Cheatsheet

Screenshot this table for a quick revision of the key geometrical ideas from this page.

Relationship between Zeroes and Coefficients of a Polynomial — Part 1

{{FORMULA: expr=α + β = -b/a, αβ = c/a | symbols=α,β: zeroes of the polynomial, a,b,c: coefficients of ax²+bx+c}}

Relationship between Zeroes and Coefficients of a Polynomial — Part 1

Welcome to one of the most fascinating topics in algebra! We've learned what polynomials are and how to find their "zeroes". But is there a hidden connection between those zeroes and the numbers (coefficients) in the polynomial itself? The answer is a resounding YES!

Imagine launching a rocket. Its path through the air can be modeled by a quadratic polynomial. The "zeroes" of this polynomial represent the launch point and the landing point — the two moments when the rocket's height is zero. The coefficients (a, b, and c) in the equation determine the shape of this path: how high the rocket goes and how wide its arc is. By understanding the relationship between these coefficients and the zeroes, we can predict the landing spot just by looking at the launch equation, without having to trace the entire path. This powerful idea forms the core of our lesson today.

Definitions & Formulas

Let's start by defining the key terms for a quadratic polynomial, which has the general form p(x) = ax² + bx + c, where a ≠ 0.

{{VISUAL: diagram: A parabola (U-shaped curve) intersecting the x-axis at two points labeled α and β, representing the zeroes of the quadratic polynomial.}}

The Logic: Deriving the Relationship

Where do these magic formulas α + β = -b/a and αβ = c/a come from? It's not magic, it's pure logic! Let's derive them step-by-step.

1. Start with the factors. If α and β are the zeroes of a quadratic polynomial p(x), then by the Factor Theorem, (x - α) and (x - β) must be its factors.

2. Construct the polynomial from factors. This means we can write the polynomial as a product of its factors. We add a constant k because the factors could be scaled by any non-zero number (e.g., x² - 4 has factors (x-2)(x+2), but 2x² - 8 has the same zeroes and factors 2(x-2)(x+2)).

   p(x) = k(x - α)(x - β)
   

3. Expand the factored form. Let's multiply the factors out.

   p(x) = k[x(x - β) - α(x - β)]
   p(x) = k[x² - βx - αx + αβ]
   p(x) = k[x² - (α + β)x + αβ]
   

4. Distribute the constant k. Now, multiply k through the expression.

   p(x) = kx² - k(α + β)x + kαβ
   

5. Compare with the standard form. We know the standard form of the polynomial is p(x) = ax² + bx + c. Let's put our two forms side-by-side:

   ax² + bx + c kx² - k(α + β)x + kαβ

   For these to be the same polynomial, the coefficients of each power of x must be equal.

6. Equate the coefficients.

   • Comparing the x² terms: a = k
   • Comparing the x terms: b = -k(α + β)
   • Comparing the constant terms: c = kαβ

   Now, substitute a for k in the other two equations:

   • b = -a(α + β) → (α + β) = -b/a
   • c = a(αβ) → (αβ) = c/a

And there we have it! The sum of the zeroes is -b/a and the product of the zeroes is c/a.

{{KEY: type=concept | title=The Core Relationship for Quadratic Polynomials | text=For any quadratic polynomial ax² + bx + c, the sum of its zeroes (α + β) is equal to the negative of the coefficient of x divided by the coefficient of x² (-b/a). The product of its zeroes (αβ) is equal to the constant term divided by the coefficient of x² (c/a).}}

Solved Examples

Let's put this theory into practice with some examples, moving from simple to more complex problems.

Example 1: Finding Zeroes and Verifying the Relationship (Easy)

Given: The quadratic polynomial p(x) = x² + 7x + 10.

To Find: The zeroes of the polynomial and verify the relationship between the zeroes and coefficients.

Solution:

1. First, find the zeroes by splitting the middle term. We need two numbers that add up to 7 and multiply to 10. These numbers are 5 and 2.

   x² + 5x + 2x + 10 = 0
   x(x + 5) + 2(x + 5) = 0
   (x + 2)(x + 5) = 0
   

2. Set each factor to zero to find the values of x.

   x + 2 = 0  =>  x = -2
   x + 5 = 0  =>  x = -5
   

   So, the zeroes are α = -2 and β = -5.

3. Now, let's verify the relationship. From p(x) = x² + 7x + 10, we have a=1, b=7, c=10.

4. Verify the Sum of Zeroes:

   • From our found zeroes: α + β = (-2) + (-5) = -7
   • From the formula: -b/a = -(7)/1 = -7 The values match!

5. Verify the Product of Zeroes:

   • From our found zeroes: αβ = (-2) × (-5) = 10
   • From the formula: c/a = 10/1 = 10 The values match! The relationship is verified.

Final Answer: The zeroes are -2 and -5. The sum of zeroes (α + β) is -7 and the product of zeroes (αβ) is 10, which matches the formulas -b/a and c/a respectively.

Example 2: Forming a Polynomial from its Zeroes (Medium)

Given: The sum of zeroes is -3 and the product of zeroes is 2.

To Find: The quadratic polynomial.

Solution:

1. Recall the formula for constructing a polynomial from its zeroes: p(x) = k[x² - (sum of zeroes)x + (product of zeroes)]. We can assume k=1 for the simplest polynomial.

   p(x) = x² - (α + β)x + αβ
   

2. Substitute the given sum and product into the formula.

   • Sum (α + β) = -3
   • Product (αβ) = 2

   p(x) = x² - (-3)x + (2)
   

3. Simplify the expression.

   p(x) = x² + 3x + 2
   

Final Answer: The required quadratic polynomial is p(x) = x² + 3x + 2.

Example 3: Finding an Unknown Coefficient (Hard)

Given: The polynomial p(x) = 3x² + kx - 2 has one zero equal to -2.

To Find: The value of the coefficient k and the other zero.

Solution:

1. Since -2 is a zero of the polynomial, it must satisfy the equation p(x) = 0. So, p(-2) = 0.

   p(-2) = 3(-2)² + k(-2) - 2 = 0
   

2. Solve this equation for k.

   3(4) - 2k - 2 = 0
   12 - 2k - 2 = 0
   10 - 2k = 0
   10 = 2k
   k = 5
   

3. Now we know the full polynomial is p(x) = 3x² + 5x - 2. Let the zeroes be α and β. We are given α = -2. We need to find β.

4. We can use the product of zeroes relationship: αβ = c/a. Here a=3, b=5, c=-2.

   (-2) × β = -2 / 3
   

5. Solve for β.

   β = (-2 / 3) ÷ (-2)
   β = (-2 / 3) × (-1 / 2)
   β = 2 / 6 = 1/3
   

   (Alternatively, we could use the sum of zeroes: α + β = -b/a → -2 + β = -5/3 → β = -5/3 + 2 = 1/3)

Final Answer: The value of k is 5 and the other zero is 1/3.

Example 4: Evaluating Expressions of Zeroes (Tricky)

Given: α and β are the zeroes of the quadratic polynomial p(x) = 2x² - 7x + 3.

To Find: The value of the expression α² + β².

Solution:

1. First, identify the coefficients: a=2, b=-7, c=3.

2. Calculate the sum and product of the zeroes using the formulas. We don't need to find the actual zeroes.

   Sum: α + β = -b/a = -(-7)/2 = 7/2
   Product: αβ = c/a = 3/2
   

3. The expression we need to find is α² + β². We need to rewrite this using α + β and αβ. Recall the algebraic identity: (x + y)² = x² + y² + 2xy.

4. Rearrange the identity to isolate x² + y².

   x² + y² = (x + y)² - 2xy
   

   In our case, this becomes:

   α² + β² = (α + β)² - 2αβ
   

5. Substitute the values of the sum and product we found in step 2.

   α² + β² = (7/2)² - 2(3/2)
   

6. Calculate the final value.

   α² + β² = (49/4) - 3
   α² + β² = 49/4 - 12/4
   α² + β² = 37/4
   

Final Answer: The value of α² + β² is 37/4.

Tips & Tricks

Here are a few shortcuts to speed up your problem-solving.

Common Mistakes

Be careful! Many students lose marks due to these simple errors.

Brain-Teaser Questions

Test your understanding with these slightly trickier problems.

1. If one zero of the polynomial p(x) = (k² + 4)x² + 13x + 4k is the reciprocal of the other, what is the value of k?

   💡 Answer: Let the zeroes be α and 1/α. The product of the zeroes is α × (1/α) = 1. From the formula, Product = c/a = 4k / (k² + 4). So, 4k / (k² + 4) = 1 → 4k = k² + 4 → k² - 4k + 4 = 0. This is (k - 2)² = 0, which gives k = 2.

2. If the sum of the squares of the zeroes of the polynomial p(x) = x² - 8x + k is 40, find the value of k.

   💡 Answer: We are given α² + β² = 40. From the polynomial, a=1, b=-8, c=k. Sum: α + β = -(-8)/1 = 8. Product: αβ = k/1 = k. We know α² + β² = (α + β)² - 2αβ. Substitute the given values: 40 = (8)² - 2(k). 40 = 64 - 2k → 2k = 64 - 40 → 2k = 24 → k = 12.

3. If α and β are zeroes of x² - x - 2, find a polynomial whose zeroes are (2α + 1) and (2β + 1).

   💡 Answer: For the original polynomial, α + β = -(-1)/1 = 1 and αβ = -2/1 = -2. For the new polynomial, let the new zeroes be α' and β'. New Sum = α' + β' = (2α + 1) + (2β + 1) = 2(α + β) + 2 = 2(1) + 2 = 4. New Product = α'β' = (2α + 1)(2β + 1) = 4αβ + 2α + 2β + 1 = 4αβ + 2(α + β) + 1. Substitute the old values: 4(-2) + 2(1) + 1 = -8 + 2 + 1 = -5. The new polynomial is x² - (New Sum)x + (New Product) = x² - 4x - 5.

Mini Cheatsheet

Here’s a summary of today's key formulas. Screenshot this for last-minute revisions!

Relationship between Zeroes and Coefficients of a Polynomial — Part 2 & Summary

{{FORMULA: expr=p(x) = ax³ + bx² + cx + d | symbols=a:cubic coefficient, b:quadratic coefficient, c:linear coefficient, d:constant term}}

Relationship between Zeroes and Coefficients of a Polynomial — Part 2 & Summary

Welcome back! In our last session, we uncovered the elegant connection between the zeroes and coefficients of a quadratic polynomial. Now, we'll extend this understanding to cubic polynomials and master a fundamental tool in algebra: the Division Algorithm for Polynomials.

Imagine you're an engineer designing a storage tank. The volume of the tank is described by a cubic polynomial V(x) = x³ - 6x² + 11x - 6. You need to find the dimensions (length, breadth, height) for a specific volume. Finding the zeroes of this polynomial gives you the possible values for x where the volume would be zero—critical points for your design. Understanding the relationship between these zeroes and the coefficients (-6, 11, -6) can provide shortcuts to finding these dimensions, making your design process more efficient. This is where today's concepts come to life!

Definitions & Formulas

Let's formally define the key terms for a cubic polynomial and the division algorithm. For a cubic polynomial p(x) = ax³ + bx² + cx + d, let its zeroes be α, β, and γ.

Derivation: The Cubic Relationship

How do we know the relationship between the zeroes (α, β, γ) and the coefficients (a, b, c, d) of a cubic polynomial? Let's derive it.

1. If α, β, and γ are the zeroes of a cubic polynomial p(x), then (x-α), (x-β), and (x-γ) must be its factors.

2. Therefore, we can write the polynomial as a product of these factors, with a constant k.

   p(x) = k(x-α)(x-β)(x-γ)
   

3. Let's expand the right-hand side. First, multiply (x-β) and (x-γ).

   (x-β)(x-γ) = x² - (β+γ)x + βγ
   

4. Now, multiply this result by (x-α).

   (x-α)[x² - (β+γ)x + βγ] = x³ - (α+β+γ)x² + (αβ+βγ+γα)x - αβγ
   

5. So, the polynomial becomes:

   p(x) = k[x³ - (α+β+γ)x² + (αβ+βγ+γα)x - αβγ]
   

   This can be rewritten as:

   p(x) = kx³ - k(α+β+γ)x² + k(αβ+βγ+γα)x - k(αβγ)
   

6. Now, we compare the coefficients of this expanded form with the standard form p(x) = ax³ + bx² + cx + d.

   • a = k
   • b = -k(α+β+γ) → α+β+γ = -b/k = -b/a
   • c = k(αβ+βγ+γα) → αβ+βγ+γα = c/k = c/a
   • d = -k(αβγ) → αβγ = -d/k = -d/a

This gives us the three fundamental relationships for a cubic polynomial!

{{KEY: type=concept | title=Division Algorithm for Polynomials | text=If p(x) and g(x) are any two polynomials with g(x) ≠ 0, then we can find polynomials q(x) and r(x) such that p(x) = g(x) × q(x) + r(x), where r(x) = 0 or degree(r(x)) < degree(g(x)). This is the polynomial equivalent of the familiar Dividend = Divisor × Quotient + Remainder.}}

Solved Examples

Let's apply these concepts to some problems, starting from easy and moving to tricky.

Example 1: Verifying Zeroes and Relationships (Easy)

Given: Polynomial p(x) = 2x³ + x² - 5x + 2 and its zeroes are ½, 1, and -2.

To Find: Verify the relationship between the given zeroes and the coefficients.

Solution:

1. First, let's identify the coefficients and the zeroes. a = 2, b = 1, c = -5, d = 2. α = ½, β = 1, γ = -2.

2. Verify Sum of Zeroes (α + β + γ = -b/a). LHS: α + β + γ = ½ + 1 + (-2) = 1.5 - 2 = -0.5 = -½ RHS: -b/a = -(1)/2 = -½ Since LHS = RHS, the relationship is verified.

3. Verify Sum of Products of Zeroes (αβ + βγ + γα = c/a). LHS: (½)(1) + (1)(-2) + (-2)(½) = ½ - 2 - 1 = 0.5 - 3 = -2.5 = -5/2 RHS: c/a = (-5)/2 = -5/2 Since LHS = RHS, the relationship is verified.

4. Verify Product of Zeroes (αβγ = -d/a). LHS: (½)(1)(-2) = -1 RHS: -d/a = -(2)/2 = -1 Since LHS = RHS, the relationship is verified.

Final Answer:

All three relationships between the zeroes and coefficients are successfully verified.

Example 2: Finding a Cubic Polynomial from its Zeroes' Properties (Medium)

Given: The sum, sum of the product of zeroes taken two at a time, and product of the zeroes of a cubic polynomial are 2, -7, and -14 respectively.

To Find: The cubic polynomial.

Solution:

1. Let the zeroes be α, β, and γ. We are given the following values:

   • Sum of zeroes: α + β + γ = 2
   • Sum of products of zeroes: αβ + βγ + γα = -7
   • Product of zeroes: αβγ = -14

2. The general form of a cubic polynomial based on its zeroes is:

   p(x) = k[x³ - (Sum of zeroes)x² + (Sum of products of zeroes)x - (Product of zeroes)]
   

3. Substitute the given values into this formula.

   p(x) = k[x³ - (2)x² + (-7)x - (-14)]
   

4. Simplify the expression.

   p(x) = k[x³ - 2x² - 7x + 14]
   

5. Assuming the simplest case where k = 1 (for the standard polynomial), we get our final answer.

Final Answer:

The required cubic polynomial is p(x) = x³ - 2x² - 7x + 14.

Example 3: Applying the Division Algorithm (Hard)

Given: Dividend p(x) = x⁴ - 3x² + 4x + 5 and Divisor g(x) = x² - x + 1.

To Find: The quotient q(x) and remainder r(x).

Solution:

1. First, check if both polynomials are in standard form (descending powers of x). p(x) is missing the x³ term, so we can write it as x⁴ + 0x³ - 3x² + 4x + 5. g(x) is already in standard form.

2. Perform the long division.

   • Step 1: Divide the first term of the dividend (x⁴) by the first term of the divisor (x²). x⁴ / x² = x². This is the first term of the quotient.
   • Multiply the divisor by x²: x²(x² - x + 1) = x⁴ - x³ + x². Subtract this from the dividend.

     (x⁴ + 0x³ - 3x²) - (x⁴ - x³ + x²) = x³ - 4x²
   

   • Bring down the next term (+4x). The new dividend is x³ - 4x² + 4x.
3. • Step 2: Divide the first term of the new dividend (x³) by the first term of the divisor (x²). x³ / x² = x. This is the second term of the quotient.
   • Multiply the divisor by x: x(x² - x + 1) = x³ - x² + x. Subtract this.

     (x³ - 4x² + 4x) - (x³ - x² + x) = -3x² + 3x
   

   • Bring down the next term (+5). The new dividend is -3x² + 3x + 5.
4. • Step 3: Divide the first term of the new dividend (-3x²) by the first term of the divisor (x²). -3x² / x² = -3. This is the third term of the quotient.
   • Multiply the divisor by -3: -3(x² - x + 1) = -3x² + 3x - 3. Subtract this.

     (-3x² + 3x + 5) - (-3x² + 3x - 3) = 8
   

5. The degree of the remainder (8) is 0, which is less than the degree of the divisor (2). So, we stop.

Final Answer:

Quotient q(x) = x² + x - 3
Remainder r(x) = 8

Example 4: Finding All Zeroes (Tricky)

Given: Two zeroes of the polynomial p(x) = 2x⁴ - 3x³ - 3x² + 6x - 2 are √2 and -√2.

To Find: All the zeroes of the polynomial.

Solution:

1. Since √2 and -√2 are zeroes, (x - √2) and (x + √2) are factors of p(x).

2. Let's find the product of these factors. This product will also be a factor of p(x).

   (x - √2)(x + √2) = x² - (√2)² = x² - 2
   

3. Now, we can use the division algorithm to divide the polynomial p(x) by this factor g(x) = x² - 2. The quotient will give us the remaining factors. The dividend is p(x) = 2x⁴ - 3x³ - 3x² + 6x - 2.

4. Performing long division of p(x) by g(x):

   • 2x⁴ / x² = 2x². 2x²(x² - 2) = 2x⁴ - 4x².
   • Subtracting gives: -3x³ + x² + 6x - 2.
   • -3x³ / x² = -3x. -3x(x² - 2) = -3x³ + 6x.
   • Subtracting gives: x² - 2.
   • x² / x² = 1. 1(x² - 2) = x² - 2.
   • Subtracting gives a remainder of 0.

5. The quotient is q(x) = 2x² - 3x + 1. The remaining zeroes of p(x) are the zeroes of q(x). We can find them by factoring q(x) using the middle term splitting method.

   2x² - 3x + 1 = 2x² - 2x - x + 1
   

   = 2x(x - 1) - 1(x - 1)
   

   = (2x - 1)(x - 1)
   

6. To find the zeroes, set the factors to zero.

   • 2x - 1 = 0 → x = ½
   • x - 1 = 0 → x = 1

Final Answer:

The four zeroes of the polynomial are √2, -√2, ½, and 1.

Tips & Tricks

Here are a few shortcuts to speed up your work with polynomials.

Common Mistakes

Watch out for these common pitfalls! Many students lose marks here.

Brain-Teaser Questions

Time to test your higher-order thinking skills!

1. If the zeroes of the polynomial p(x) = x³ - 3x² + x + 1 are a-d, a, and a+d, what are the values of a and d?

   💡 Answer: The zeroes are in an Arithmetic Progression. Use the sum of zeroes relationship: (a-d) + a + (a+d) = -(-3)/1 3a = 3 → a = 1. Now, since a=1 is a zero, p(1) must be 0. p(1) = (1)³ - 3(1)² + 1 + 1 = 1 - 3 + 1 + 1 = 0. This is correct. Now use the product of zeroes: (a-d)(a)(a+d) = -(1)/1 (1-d)(1)(1+d) = -1 1 - d² = -1 → d² = 2 → d = ±√2.

2. If one zero of the polynomial p(x) = 3x³ - 5x² - 11x - 3 is the reciprocal of another, find all three zeroes.

   💡 Answer: Let the zeroes be α, 1/α, and β. Use the product of zeroes relationship: (α)(1/α)(β) = -(-3)/3 β = 1. So, one zero is 1. Now we can use factor theorem. (x-1) is a factor. Divide p(x) by (x-1) to get the quadratic factor 3x² - 2x - 3. The other two zeroes are the roots of 3x² - 2x - 3 = 0. Using the quadratic formula, we get x = (2 ± √(4 - 4(3)(-3))) / 6 = (2 ± √40) / 6 = (1 ± √10) / 3. The zeroes are 1, (1 + √10)/3, and (1 - √10)/3.

3. If the polynomial x⁴ - 6x³ + 16x² - 25x + 10 is divided by another polynomial x² - 2x + k, the remainder comes out to be x + a. Find the values of k and a.

   💡 Answer: Perform long division. After the first two steps, you'll be left with (8-k)x² - 25x + 10. The next term in the quotient is (8-k). Multiplying (8-k) by (x² - 2x + k) gives (8-k)x² - 2(8-k)x + k(8-k). The remainder is [-25 - (-2(8-k))]x + [10 - k(8-k)]. Simplifying the x-coefficient: -25 + 16 - 2k = -9 - 2k. Simplifying the constant term: 10 - 8k + k². We are given the remainder is x + a. Comparing coefficients: -9 - 2k = 1 → -2k = 10 → k = -5. 10 - 8k + k² = a. Substitute k = -5. 10 - 8(-5) + (-5)² = 10 + 40 + 25 = 75. So, a = 75. Thus, k = -5 and a = 75.

Mini Cheatsheet

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