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Algebra 1 : Factoring Polynomials

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Example Questions

Example Question #4 : Quadratic Equations

Simplify:

$\frac{2x-5}{x-3}\div \frac{5-2x}{x-3}$

Possible Answers:

$\frac{\left ( 2x-5 \right )^{2}}{\left ( x-3 \right )^{2}}$

$\frac{x-3}{2x-5}$

$\frac{5-2x}{3-x}$

Correct answer:

Explanation:

Change division into multiplication by the reciprocal which gives us the following

$\frac{\left ( 2x-5 \right )\left ( x-3 \right )}{\left ( x-3 \right )\left ( 5-2x \right )}$

Now $\left ( 5-2x \right )=-(2x-5)$

this results in the following:

$\frac{\left ( 2x-5 \right )\left ( x-3 \right )}{\left ( x-3 \right )\left ( -1 \right )\left ( 2x-5 \right )}$

Simplification gives us

$\frac{1}{-1}$ which equals

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Example Question #11 : How To Factor A Polynomial

Simplify $\frac{3x+4}{6x^{2}+5x-4}$ .

Possible Answers:

$\frac{1}{6x^{2}+2x}$

$\frac{1}{3x+7}$

$\frac{1}{2x+3}$

$\frac{1}{2x-1}$

2x-1

Correct answer:

$\frac{1}{2x-1}$

Explanation:

Here, we simply need to identify that the numerator, 3x+4 , is a factor of the denominator. Let's start by factoring $6x^{2}+5x-4$ . The reverse FOIL method shows us that (3x+4)(2x-1) multiplies to give us $6x^{2}+5x-4$ , so we can rewrite the fraction as $\frac{(3x+4)}{(3x+4)(2x-1)}$ . Canceling the common term gives us our answer of $\frac{1}{2x-1}$ .

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Example Question #12 : How To Factor A Polynomial

Factor the polynomial completely: $16x^{2} + 64$

Possible Answers:

16 (x+2) (x-2)

The polynomial cannot be factored further.

$16 (x+2)^{2}$

$16 (x^{2}+4)$

$16(x-2)^{2}$

Correct answer:

$16 (x^{2}+4)$

Explanation:

The coefficients 16 and 64 have greatest common factor 16; there is no variable that is shared by both terms. Therefore, we can distribute out 16:

$16x^{2} + 64 = 16 \cdot x^{2 } + 16 \cdot 4 = 16 (x^{2}} + 4)$

$x^{2} + 4$ cannot be factored further, so $16 (x^{2}+4)$ is as far as we can go.

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Example Question #11 : Factoring Polynomials

$Solve\;for\;x,4x^2-28x=0$

Possible Answers:

0,-7

0,7

Correct answer:

0,7

Explanation:

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Example Question #11 : How To Factor A Polynomial

$Factor \;y^2 + 13y + 42$

Possible Answers:

(y+3)(y+14)

(y-6)(y-7)

(y+5)(y+8)

(y+6)(y+7)

Correct answer:

(y+6)(y+7)

Explanation:

$Find \;two \; numbers \;that \;multiply \;to \; get \;42 \; and \;add \;to \;get \; 13.$

$6\times7=42$

6+7=13

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Example Question #12 : How To Factor A Polynomial

$Factor \;p^2 - 5p - 14$

Possible Answers:

(p-7)(p-2)

(p+7)(p-2)

(p-7)(p+2)

(p+7)(p+2)

Correct answer:

(p-7)(p+2)

Explanation:

$Find\; two\; numbers\; that \; multiply \; to \; get\; -14 \; and\; add \; to\; get \; -5.$

$-7\times2=-14$

-7+2=-5

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Example Question #14 : Factoring Polynomials

Factor completely:

$5x^{2 } - 50x$

Possible Answers:

5(x+1)(x-10)

5x(x-10)

5(x+10)(x-10)

5(x-1)(x+10)

x(x-50)

Correct answer:

5x(x-10)

Explanation:

First, take out the greatest common factor of the terms. The GCF of 5 and 50 is 5 and the GCF of $x^{2}$ and is , so the GCF of the terms is .

When is distributed out, this leaves 5x (x -10) .

x-10 is linear and thus prime, so no further factoring can be done.

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Example Question #333 : Variables

Factor the following polynomial.

$16x^{2}y^{4} - 9w^{4}z^{6}$

Possible Answers:

$(4xy - 3wz^{3})(4xy^{3} + 3w^{3}z^{3})$

$(4xy^{2} - 3w^{2}z^{3})(4xy^{2} + 3w^{2}z^{3})$

$(4xy^{2} + 3w^{2}z^{3})(4xy^{2} + 3w^{2}z^{3})$

$(4xy^{2} - 3w^{2}z^{3})(4xy^{2} - 3w^{2}z^{3})$

$(4xy^{2} - 4w^{2}z^{3})(3xy^{2} + 3w^{2}z^{3})$

Correct answer:

$(4xy^{2} - 3w^{2}z^{3})(4xy^{2} + 3w^{2}z^{3})$

Explanation:

This polynomial is a difference of two squares. The below formula can be used for factoring the difference of any two squares.

a^2-b^2=(a-b)(a+b)

Using our given equation as $\small a^2-b^2$ , we can find the values to use in our factoring.

a^2-b^2=16x^2y^4-9w^4z^6

$(4xy^2)^2=16x^2y^4\rightarrow 4xy^2=a$

$(3w^2z^3)^2=9w^4z^6\rightarrow 3w^2z^3=b$

(a-b)(a+b)=(4xy^2-3w^2z^3)(4xy^2+3w^2z^3)

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Example Question #333 : Polynomials

Factor $x^{2}+2x-35$

Possible Answers:

$\left ( x-5 \right )\left ( x+7 \right )$

(x+5)(x-8)

(x-7)(x-5)

(x+5)(x+8)

(x+5)(x+7)

Correct answer:

$\left ( x-5 \right )\left ( x+7 \right )$

Explanation:

When factoring a polynomial that has no coefficient in front of the $x^{2}$ term, you begin by looking at the last term of the polynomial, which is . You then think of all the factors of that when added together equal , the coefficient in front of the term. The only combination of factors of that can satisfy this condition is and . Thus, the factors of the polynomial are (x-5)(x+7) .

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Example Question #335 : Variables

$Factor \;x^2 -4x -32$

Possible Answers:

(x+4)(x+8)

(x-16)(x+2)

(x-4)(x+8)

(x+4)(x-8)

Correct answer:

(x+4)(x-8)

Explanation:

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Algebra 1 : Factoring Polynomials

Study concepts, example questions & explanations for Algebra 1

All Algebra 1 Resources

Example Questions

Example Question #4 : Quadratic Equations

Example Question #11 : How To Factor A Polynomial

Example Question #12 : How To Factor A Polynomial

Example Question #11 : Factoring Polynomials

Example Question #11 : How To Factor A Polynomial

Example Question #12 : How To Factor A Polynomial

Example Question #14 : Factoring Polynomials

Example Question #333 : Variables

Example Question #333 : Polynomials

Example Question #335 : Variables

All Algebra 1 Resources

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