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High School Math : Polynomials

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Example Questions

Example Question #13 : Pre Calculus

Factor the polynomial if the expression is equal to zero when $x=-6, -4,\ or\ 2$ .

x^3+8x^2+4x-48

Possible Answers:

(x+4)(x-2)(x-6)

(x-4)(x+2)(x-6)

(x+6)(x+4)(x-2)

(x-4)(x-2)(x+6)

Correct answer:

(x+6)(x+4)(x-2)

Explanation:

Knowing the zeroes makes it relatively easy to factor the polynomial.

The expression (x+6)(x+4)(x-2) fits the description of the zeroes.

Now we need to check the answer.

(x+4)(x-2)=x^2-2x+4x-8=x^2+2x-8

$\small (x^2+2x-8)(x+6)=x^3+6x^2+2x^2+12x-8x-48=x^3+8x^2+4x-48$

We are able to get back to the original expression, meaning that the answer is (x+6)(x+4)(x-2) .

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Example Question #14 : Pre Calculus

A polyomial with leading term $x^{3}$ has 6 as a triple root. What is this polynomial?

Possible Answers:

$x ^{3} +216$

$x ^{3} + 18 x^{2}- 108x -216$

$x ^{3} + 18 x^{2}+ 108x +216$

$x ^{3} - 18 x^{2}+ 108x -216$

$x ^{3} -216$

Correct answer:

$x ^{3} - 18 x^{2}+ 108x -216$

Explanation:

Since 6 is a triple root, and the degree of the polynomial is 3, the polynomial is $(x-6)^{3}$ , which we can expland using the cube of a binomial pattern.

$(x-6)^{3} =x ^{3} - 3\cdot x^{2} \cdot 6 + 3\cdot x\cdot 6 ^{2} -6^{3}$

$x ^{3} - 18 x^{2}+ 108x -216$

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Example Question #1 : Express A Polynomial As A Product Of Linear Factors

A polyomial with leading term $x^ {3}$ has 5 and 7 as roots; 7 is a double root. What is this polynomial?

Possible Answers:

$x^{3} - 19x^{2} + 119x -245$

$x^{3}+ 17x^{2} +95x +175$

$x^{3}- 17x^{2} +95x -175$

$x^{3} -12x^{2} + 35x$

$x^{3}+ 19x^{2} +119x +245$

Correct answer:

$x^{3} - 19x^{2} + 119x -245$

Explanation:

Since 5 is a single root and 7 is a double root, and the degree of the polynomial is 3, the polynomial is $(x-5)(x-7)^{2}$ . To put this in expanded form:

$(x-7)^{2} = x^{2} - 2 \cdotx\cdot 7 + 7^{2} = x^{2} - 14x + 49$

$(x-5)(x-7)^{2} =(x-5) \left ( x^{2} - 14x + 49 \right )$

$=x \left ( x^{2} - 14x + 49 \right ) -5\left( x^{2} - 14x + 4 9 \right )$

$= x^{3} - 14x^{2} + 49x - \left( 5x^{2} - 70x + 245 \right )$

$= x^{3} - 14x^{2} + 49x - 5x^{2} + 70x -245$

$= x^{3} - 19x^{2} + 119x -245$

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Example Question #1 : Understanding Zeros Of A Polynomial

What are the solutions to $y = x^{2} + x -6$ ?

Possible Answers:

x = -6, x = 1

x = 6, x = -1

x = -3, x = 2

x = 3, x = -2

x = 5, x = 1

Correct answer:

x = -3, x = 2

Explanation:

When we are looking for the solutions of a quadratic, or the zeroes, we are looking for the values of such that the output will be zero. Thus, we first factor the equation.

$0 = x^{2} + x - 6 = (x+3)(x-2)$

Then, we are looking for the values where each of these factors are equal to zero.

x + 3 = 0 implies x = -3

and x - 2 = 0 implies x = 2

Thus, these are our solutions.

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Example Question #15 : Pre Calculus

Find the zeros of the following polynomial:

$f(x) = x^{4} - 4x^{3} - 7x^{2} + 22x + 24$

Possible Answers:

x = -4, -3, 1, 2

x = -2, -1, 3, 4

x = -4, -1, 2, 3

x = -3, -2, 1, 4

x = -4, -2, 1, 3

Correct answer:

x = -2, -1, 3, 4

Explanation:

First, we need to find all the possible rational roots of the polynomial using the Rational Roots Theorem:

$\frac{\textup{factors of the constant}}{\textup{factors of the leading coefficient}}$

Since the leading coefficient is just 1, we have the following possible (rational) roots to try:

±1, ±2, ±3, ±4, ±6, ±12, ±24

When we substitute one of these numbers for , we're hoping that the equation ends up equaling zero. Let's see if is a zero:

$f(-1)=(-1)^{4}-4(-1)^{3}-7(-1)^{2}+22(-1)+24$

f(-1)=1+4-7-22+24

f(-1)=0

Since the function equals zero when is , one of the factors of the polynomial is (x+1) . This doesn't help us find the other factors, however. We can use synthetic substitution as a shorter way than long division to factor the equation.

$\textup{-1 } | \textup{ 1 -4 -7 22 24}$

$\small \textup{ -1 }\textup{ 5 }\textup{ 2 }\textup{ -24}$

$\frac{ }{\textup{ 1 -5 -2 24 }\textup{ 0}}$

Now we can factor the function this way:

$f(x)=(x+1)(x^{3}-5x^{2}-2x+24)$

We repeat this process, using the Rational Roots Theorem with the second term to find a possible zero. Let's try :

$f(-2)=[(-2)+1][(-2)^{3}-5(-2)^{2}-2(-2)+24]$

f(-2)=(-1)(-8-20+4+24)=0

When we factor using synthetic substitution for x=-2 , we get the following result:

$f(x)=(x+1)(x+2)(x^{2}-7x+12)$

Using our quadratic factoring rules, we can factor completely:

f(x)=(x+1)(x+2)(x-4)(x-3)

Thus, the zeroes of f(x) are x=-2, -1, 3, 4.

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Example Question #16 : Pre Calculus

Simplify the following polynomial:

$(\frac{-a^{-2}b^{3}c^{-1}}{3a^{-4}b^{-1}c^{3}})^{-2}$

Possible Answers:

$\frac{c^{8}}{9a^{4}b^{8}}$

$\frac{9b^{8}}{c^{8}a^{4}}$

$\frac{9a^{4}}{c^{8}b^{4}}$

$\frac{9a^{4}}{c^{8}b^{8}}$

$\frac{9c^{8}}{a^{4}b^{8}}$

Correct answer:

$\frac{9c^{8}}{a^{4}b^{8}}$

Explanation:

To simplify the polynomial, begin by combining like terms:

$(\frac{-a^{-2}b^{3}c^{-1}}{3a^{-4}b^{-1}c^{3}})^{-2}$

$(\frac{-a^{2}b^{4}}{3c^{4}})^{-2}$

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Example Question #1 : Simplifying Polynomial Functions

Simplify the following polynomial function:

$ab^{-2}(a^{2}b + a^{-1}b^{3}-a^{3}b^{-1})$

Possible Answers:

$\frac{a^{3}}{b}+\frac{a^{4}}{b^{3}}$

$\frac{a^{3}}{b}-\frac{a^{4}}{b^{3}}$

$\frac{a^{3}}{b}+b^{2}-\frac{a^{4}}{b^{3}}$

$\frac{a^{3}}{b}-b+\frac{a^{4}}{b^{3}}$

$\frac{a^{3}}{b}+b-\frac{a^{4}}{b^{3}}$

Correct answer:

$\frac{a^{3}}{b}+b-\frac{a^{4}}{b^{3}}$

Explanation:

First, multiply the outside term with each term within the parentheses:

$ab^{-2}(a^{2}b + a^{-1}b^{3}-a^{3}b^{-1})$

$a^{3}b^{-1} + b-a^{4}b^{-3}$

Rearranging the polynomial into fractional form, we get:

$\frac{a^{3}}{b}+b-\frac{a^{4}}{b^{3}}$

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High School Math : Polynomials

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #13 : Pre Calculus

Example Question #14 : Pre Calculus

Example Question #1 : Express A Polynomial As A Product Of Linear Factors

Example Question #1 : Understanding Zeros Of A Polynomial

Example Question #15 : Pre Calculus

Example Question #16 : Pre Calculus

Example Question #1 : Simplifying Polynomial Functions

All High School Math Resources

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