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Algebra 1 : Factoring Polynomials

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Example Questions

Example Question #21 : How To Factor A Polynomial

Factor the following expression.

x^2-3x+2

Possible Answers:

(x-3)(x+1)

(x+2)(x-2)

(x-2)(x-1)

(x+2)(x-1)

(x-2)(x+1)

Correct answer:

(x-2)(x-1)

Explanation:

x^2-3x+2

The factored form of this equation should be in the format (A+B)(C+D) .

To yield the first term in our original equation ( $\small x^2$ ), $\small A=x$ and $\small C=x$ .

(x+B)(x+D)

To yield the last term in our original equation ( $\small +2$ ), we can set $\small B=-2$ and $\small D=-1$ .

$-2*-1=2\ \text{and}\ -2+(-1)=-3$

(x-2)(x-1)

We can check our answer by using FOIL to expand back to the original expression.

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Example Question #334 : Polynomials

Factor the following expression.

3x^2-2x-5

Possible Answers:

(3x-5)(x+1)

(x-2)(5x+2)

(3x-1)(x+5)

(x-5)(3x+1)

(3x-5)(3x-1)

Correct answer:

(3x-5)(x+1)

Explanation:

3x^2-2x-5

The factored form of this equation should have the format (A+B)(C+D) .

To yield the first term in our original equation ( $\small 3x^2$ ), $\small A=3x$ and $\small C=x$ .

(3x+B)(x+D)

To yield the final term in our original equation ( $\small -5$ ), we can set $\small B=-5$ and $\small D=1$ .

$-5*1=-5\ \text{and}\ -5+1(3)=-2$

(3x-5)(x+1)

If you are unsure of your answer, you can check using FOIL to end up with the original equation.

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Example Question #335 : Polynomials

Factor the following expression.

x^3+7x^2+3x+21

Possible Answers:

(x+7)(x+3)

(x+7)(x^2+3)

(x+21)(x^2-1)

x(x+3)(x+7)

(7x-4)(3x^2+5)

Correct answer:

(x+7)(x^2+3)

Explanation:

x^3+7x^2+3x+21

To factor an expression like this, we can use factoring by grouping. The greatest common factor in the first two terms is $\small x^2$ , so $\small x^2$ should be factored out.

x^2(x+7)+3x+21

Now, look at the last two terms. The greatest common factor in these terms is $\small 3$ . Factor out the $\small 3$ .

x^2(x+7)+3(x+7)

Now, both terms in parentheses are $\small x+7$ . We can group $\small x^2$ and $\small 3$ together, and multiply it by $\small x+7$ .

(x+7)(x^2+3)

We can check the answer using FOIL to end up with the original expression.

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Example Question #5 : Solving By Factoring

Factor the following expression.

125x^3-216

Possible Answers:

(5x-6)(25x^2+30x+36)

(5x+6)(25x^2-30x+36)

x(5x+6)(5x-6)

(x-13)(2x^2-7x+3)

(5x-3)(5x^2-6x+3)

Correct answer:

(5x-6)(25x^2+30x+36)

Explanation:

125x^3-216

This expression involves the difference of two cubic terms. To factor an expression in this format, we can use a special formula.

A^3-B^3=(A-B)(A^2+AB+B^2)

Before we can use this formula, we need to manipulate our original expression to identify $\small A$ and $\small B$ .

125x^3-216=(5x)^3-6^3

Comparing this with the formula, $\small A=5x$ and $\small B=6$ . Now we can use the formula to factor.

A^3-B^3=(A-B)(A^2+AB+B^2)

(5x)^3-(6)^3=(5x-6)((5x)^2+(5x)(6)+6^2)

(5x)^3-(6)^3=(5x-6)(25x^2+30x+36)

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Example Question #341 : Variables

Factor completely:

$5t^{3}+125t$

Possible Answers:

$\left (t + 5 \right )^{3}$

$5t (t^{2}+25)$

5t (t+5)(t-5)

$5t (t-5)^{2}$

$5t (t+5)^{2}$

Correct answer:

$5t (t^{2}+25)$

Explanation:

can be factored out of each term, so do this first:

$5t^{3}+125t$

$= 5t \cdot t^{2}+5t \cdot 25$

$= 5t ( t^{2}+ 25)$

$t^{2}+ 25$ is the sum of squares and cannot be factored, so $5t ( t^{2}+ 25)$ is the correct factorization.

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Example Question #342 : Variables

Factor completely:

$6t^{3} - 150t$

Possible Answers:

$6t (t-5)^{2}$

$6t (t+5)^{2}$

6t (t+5)(t-5)

$(6t-5)(t^{2}+ 30)$

$(6t+5)(t^{2}- 30)$

Correct answer:

6t (t+5)(t-5)

Explanation:

can be factored out of each term, so do this first:

$6t^{3} - 150t$

$=6t \cdot t^{2} - 6t \cdot 25$

$=6t \left ( t^{2} - 25 \right )$

The second factor can be factored as the difference of two squares:

$=6t \left ( t^{2} - 5^{2} \right )$

$=6t \left ( t+5 \right )\left ( t-5 \right )$

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Example Question #343 : Variables

Factor completely:

$9t^{2} - 100$

Possible Answers:

$\left (3t + 10 \right )\left (3t - 10 \right )$

$\left (3t + 10 \right )^{2}$

$\left (3t - 10 \right )^{2}$

The polynomial is prime.

$3t \left ( t-10\right )$

Correct answer:

$\left (3t + 10 \right )\left (3t - 10 \right )$

Explanation:

This is a difference of squares and can therefore be factored using the appropriate pattern:

$9t^{2} - 100$

$=\left ( 3t \right )^{2} - 10^{2}$

$= \left (3t + 10 \right )\left (3t - 10 \right )$

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Example Question #344 : Variables

Factor the following polynomial:

20x^4 - 5x^3 + 10x^2

Possible Answers:

5(4x^4 - x^3 + 2x^2)

5x^2(4x^2 - x + 2)

5(4x^4 - x^3 + 2x^2)

5x^2(4x^2 + x + 2)

5x(4x^3 - x^2 + 2x)

Correct answer:

5x^2(4x^2 - x + 2)

Explanation:

The coefficients are 20, 5, and 10. The greatest common factor of those three numbers is 5. Factor out the 5, making sure to keep the signs the same inside the parentheses:

20x^4 - 5x^3 + 10x^2

=5(4x^4 - x^3 + 2x^2)

Now factor out the x^2 term:

=5(4x^4 - x^3 + 2x^2)

=5x^2(4x^2 - x + 2)

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Example Question #345 : Variables

Factor by grouping.

27ab+12b-9a-4

Possible Answers:

(a+4)(b-1)

(6a+4)(3b-1)

(9a+4)(3b-1)

9a (3b-1)

Not factorable

Correct answer:

(9a+4)(3b-1)

Explanation:

Looking at the first two terms, the greatest common factor is 3b, hus we can factor out 3b.

27ab+12b-9a-4

3b(9a+4)-9a-4

Looking at the last two terms, the greatest common factor is –1. Factor out the –1.

3b(9a+4)-1(9a+4)

Notice that you can factor out the .

(9a+4)(3b-1)

This is your final answer.

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Example Question #5 : Solving By Factoring

Factor the following expression.

343x^3-64y^3

Possible Answers:

(7x+4y)(49x^2-28xy+16y^2)

(7x-4y)(49x^2+28xy+16y^2)

-(7x-4y)(49x^2+28xy+16y^2)

(49x-28y)(7x^2+4xy+4y^2)

Not factorable

Correct answer:

(7x-4y)(49x^2+28xy+16y^2)

Explanation:

This problem involves the difference of two cubic terms. We need to use a special factoring formula that will allow us to factor this equation.

a^3-b^3=(a-b)(a^2+ab+b^2)

But before we can use this formula, we need to manipulate to make it more similar to the left hand side of the special formula. We do this by making the coefficients (343 and 64) part of the cubic power.

343x^3-64y^3=(7x)^3-(4y)^3

Comparing this with , and .

Plug these into the formula.

(7x)^3-(4y)^3=(7x-4y)((7x)^2+(7x)(4y)+(4y)^2)

(7x)^3-(4y)^3=(7x-4y)(49x^2+28xy+16y^2)

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Algebra 1 : Factoring Polynomials

Study concepts, example questions & explanations for Algebra 1

All Algebra 1 Resources

Example Questions

Example Question #21 : How To Factor A Polynomial

Example Question #334 : Polynomials

Example Question #335 : Polynomials

Example Question #5 : Solving By Factoring

Example Question #341 : Variables

Example Question #342 : Variables

Example Question #343 : Variables

Example Question #344 : Variables

Example Question #345 : Variables

Example Question #5 : Solving By Factoring

All Algebra 1 Resources

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