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High School Math : Factoring Polynomials

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Example Questions

Example Question #1 : Factoring Polynomials

Factor $x^{3}-64$

Possible Answers:

(x-4)(x^2+4x+16)

(x-4)(x^2-4x-16)

(x+4)(x^2-4x+16)

Cannot be Factored

Correct answer:

(x-4)(x^2+4x+16)

Explanation:

Use the difference of perfect cubes equation:

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

In $x^{3}-64$ ,

a=x and $b=\sqrt[3]{64}=4$

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

(x-4)(x^2+4x+16)

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Example Question #2 : Factoring Polynomials

Factor the polynomial completely and solve for .

Possible Answers:

8x(4x^2 -1)

$x = \left \{ -\frac{1}{2}, 0, \frac{1}{2} \right \}$

4x (8x^2 - 2) = 0

$x = \left \{0, \frac{1}{2} \right \}$

8x(2x+1)(2x-1)

$x = \left \{-\frac{1}{2}, 0, \frac{1}{2} \right \}$

32x(x+1)(x-1)

$x = \left \{ -1, 0,1\right \}$

8(4x^3- x)

x = 0

Correct answer:

8x(2x+1)(2x-1)

$x = \left \{-\frac{1}{2}, 0, \frac{1}{2} \right \}$

Explanation:

To factor and solve for in the equation

Factor out of the equation

$\rightarrow 8x(4x^2-1)$

Use the "difference of squares" technique to factor the parenthetical term, which provides the completely factored equation:

$\rightarrow 8x(2x+1)(2x-1)$

Any value that causes any one of the three terms , 2x+1 , and 2x -1 to be will be a solution to the equation, therefore

$x = \left \{-\frac{1}{2}, 0, \frac{1}{2} \right \}$

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Example Question #1 : Factoring Polynomials

Factor the following expression:

x^2+xy+x=

Possible Answers:

x(x+y+1)

x(x+y)

x+y

x^2+x(y+1)

x+y+1

Correct answer:

x(x+y+1)

Explanation:

You can see that each term in the equation has an "x", therefore by factoring "x" from each term you can get that the equation equals x(x+y+1) .

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Example Question #3 : Factoring Polynomials

Factor this expression:

x^2+7x+12

Possible Answers:

(x+4)(x+3)

(x+2) (x+6)

(x+4)(x-3)

(x-2) (x+6)

(x-4)(x+3)

Correct answer:

(x+4)(x+3)

Explanation:

First consider all the factors of 12:

1 and 12

2 and 6

3 and 4

Then consider which of these pairs adds up to 7. This pair is 3 and 4.

Therefore the answer is (x+4)(x+3) .

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Example Question #41 : Intermediate Single Variable Algebra

Find the zeros.

$f(x)=x^{3}-1$

Possible Answers:

x=-1

x=10

x=1

x=1,-1

x=0

Correct answer:

x=1

Explanation:

This is a difference of perfect cubes so it factors to (x-1)(x^2+x+1) . Only the first expression will yield an answer when set equal to 0, which is 1. The second expression will never cross the -axis. Therefore, your answer is only 1.

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Example Question #42 : Intermediate Single Variable Algebra

Find the zeros.

f(x)=2x^5-8x^3+8x

Possible Answers:

$x=0, \sqrt{2}, -\sqrt{2 }$

x=2,0

$x=\sqrt{2}$

x=0

$x= \sqrt{-2}}$

Correct answer:

$x=0, \sqrt{2}, -\sqrt{2 }$

Explanation:

Factor the equation to 2x(x^4-4x^2+4) . Set 2x=0 and get one of your 's to be . Then factor the second expression to (x^2-2) (x^2-2) . Set them equal to zero and you get $\pm \sqrt{2}$ .

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Example Question #4 : Factoring Polynomials

Factor the following polynomial:

54a^3b-2b^4

Possible Answers:

2b(3a-b)(9a^2+2ab+b^2)

3b(3a-b)(9a^2+3ab+b^2)

2b(3a-b)(9a^2+3ab+b^2)

2b(2a-b)(9a^2+3ab+b^2)

3b(2a-b)(9a^2+3ab+b^2)

Correct answer:

2b(3a-b)(9a^2+3ab+b^2)

Explanation:

Begin by extracting from the polynomial:

54a^3b-2b^4

=2b(27a^3-b^3)

Now, factor the remainder of the polynomial as a difference of cubes:

=2b(3a-b)(9a^2+3ab+b^2)

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Example Question #2 : Factoring Polynomials

Factor the following polynomial:

5ab+b^2+5ac-c^2

Possible Answers:

(b+c)(5a+b+c)

(b-c)(5a+b-c)

(b-c)(5a-b+c)

(b+c)(5a-b-c)

(b+c)(5a+b-c)

Correct answer:

(b+c)(5a+b-c)

Explanation:

Begin by rearranging like terms:

5ab+b^2+5ac-c^2

=5ab+5ac+b^2-c^2

Now, factor out like terms:

=5a(b+c)+(b+c)(b-c)

Rearrange the polynomial:

=(b+c)(5a+b-c)

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Example Question #3 : Factoring Polynomials

Factor the following polynomial:

m^3-3m^2a+3ma^2-a^3

Possible Answers:

m-a

(m-a)^3

(m-a)^4

(m-a)^2

Correct answer:

(m-a)^3

Explanation:

Begin by rearranging like terms:

m^3-3m^2a+3ma^2-a^3

=m^3-a^3-3m^2a+3ma^2

Now, factor out like terms:

=(m-a)(m^2+ma+a^2)-3ma(m-a)

Rearrange the polynomial:

=(m-a)(m^2-2ma+a^2)

Factor:

=(m-a)(m-a)^2

=(m-a)^3

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Example Question #2 : Factoring Polynomials

Factor the following polynomial:

12n^2-7n-10

Possible Answers:

(3n-5)(3n+3)

(4n-5)(3n+2)

(4n+5)(3n-2)

(4n-5)(4n+2)

(3n-5)(3n+2)

Correct answer:

(4n-5)(3n+2)

Explanation:

Begin by separating -7n into like terms. You do this by multiplying and , then finding factors which sum to

12n^2-7n-10

12n^2-15n+8n-10

Now, extract like terms:

3n(4n-5)+2(4n-5)

Simplify:

(4n-5)(3n+2)

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High School Math : Factoring Polynomials

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #1 : Factoring Polynomials

Example Question #2 : Factoring Polynomials

Example Question #1 : Factoring Polynomials

Example Question #3 : Factoring Polynomials

Example Question #41 : Intermediate Single Variable Algebra

Example Question #42 : Intermediate Single Variable Algebra

Example Question #4 : Factoring Polynomials

Example Question #2 : Factoring Polynomials

Example Question #3 : Factoring Polynomials

Example Question #2 : Factoring Polynomials

All High School Math Resources

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