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High School Math : Understanding Zeros of a Polynomial

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Example Questions

Example Question #11 : Polynomial Functions

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

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

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 #2 : Understanding Zeros Of A Polynomial

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

Possible Answers:

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

$x ^{3} -216$

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

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

$x ^{3} +216$

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 #4 : 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:

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

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

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

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

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

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 #11 : Pre Calculus

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

Possible Answers:

x = 5, x = 1

x = -3, x = 2

x = -6, x = 1

x = 3, x = -2

x = 6, 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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High School Math : Understanding Zeros of a Polynomial

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #11 : Polynomial Functions

Example Question #2 : Understanding Zeros Of A Polynomial

Example Question #4 : Express A Polynomial As A Product Of Linear Factors

Example Question #11 : Pre Calculus

All High School Math Resources

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