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Algebra 1 : Factoring Polynomials

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Example Questions

Example Question #41 : Factoring Polynomials

Factor the following polynomial: x^2-2x-35 .

Possible Answers:

(x+5)(x-7)

(x+5)(x+7)

(x-1)(x+35)

(x-5)(x-7)

(x-35)(x-1)

Correct answer:

(x+5)(x-7)

Explanation:

Because the x^2 term doesn’t have a coefficient, you want to begin by looking at the term ( ax^2+bx+c ) of the polynomial: . Find the factors of that when added together equal the second coefficient (the term) of the polynomial.

There are only four factors of : 1, 5, 7, 35 , and only two of those factors, 5, 7 , can be manipulated to equal when added together and manipulated to equal when multiplied together: 5, -7 (i.e., 5-7=-2 ).

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Example Question #11 : How To Factor A Variable

Factor the following polynomial: x^2+13x+30 .

Possible Answers:

(x-10)(x-3)

(x-5)(x-6)

(x+5)(x+6)

(x-10)(x+3)

(x+10)(x+3)

Correct answer:

(x+10)(x+3)

Explanation:

Because the x^2 term doesn’t have a coefficient, you want to begin by looking at the term ( ax^2+bx+c ) of the polynomial: .

Find the factors of that when added together equal the second coefficient (the term) of the polynomial: .

There are seven factors of : 1, 2, 3, 5, 6, 10, 15 , and only two of those factors, 3, 10 , can be manipulated to equal when added together and manipulated to equal when multiplied together: 3, 10

$3+10=13, 3\times10=30$

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Example Question #13 : How To Factor A Variable

Solve for when $x\neq-5$ : $\frac{x^2+2x-15}{x+5}=10$

Possible Answers:

x=15

x=5

x=13

x=7

x=-5

Correct answer:

x=13

Explanation:

First, factor the numerator: (x+5)(x-3) .

Now your expression looks like

$\frac{(x+5)(x-3)}{x+5}$

Second, cancel the "like" terms - (x+5) - which leaves us with x-3=10 .

Third, solve for , which leaves you with x=13 .

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Example Question #4652 : Algebra 1

Factor the following polynomial: 2x^2-5x-3 .

Possible Answers:

(2x+1)(x+3)

(2x+1)(x-3)

(2x-1)(x+3)

(2x+3)(x-1)

(2x+3)(x+1)

Correct answer:

(2x+1)(x-3)

Explanation:

Because the x^2 term has a coefficient, you begin by multiplying the and the terms ( ax^2+bx+c ) together: $2\times-3=-6$ .

Find the factors of that when added together equal the second coefficient (the term) of the polynomial: .

There are four factors of : 1, 2, 3, 6 , and only two of those factors, 1, 6 , can be manipulated to equal when added together and manipulated to equal when multiplied together: 1, -6

$1+-6=-5, 1\times-6=-6$

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Example Question #4653 : Algebra 1

Factor: x^6+2x^3+2x

Possible Answers:

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

$\frac{1}{x^6}(x^{-6}+2x^{-2}+2x^{-5})$

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

x^3(x^3+4)

$6x(x^5+2x+\frac{1}{3})$

Correct answer:

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

Explanation:

For each term in this expression, we will notice that each shares a variable of . This can be pulled out as a common factor.

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

There are no more common factors, and this is the reduced form.

The answer is: x(x^5+2x^2+2)

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Example Question #4654 : Algebra 1

Factor: x^6+2x^3-3x

Possible Answers:

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

$6x(x^5+2x-\frac{1}{18})$

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

$\frac{1}{x^6}(x^{-6}+2x^{-2}-3x^{-5})$

x^3(x^3-8)

Correct answer:

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

Explanation:

For each term in this expression, we will notice that each shares a variable of . This can be pulled out as a common factor.

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

There are no more common factors, and this is the reduced form.

The answer is: x(x^5+2x^2-3)

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Algebra 1 : Factoring Polynomials

Study concepts, example questions & explanations for Algebra 1

All Algebra 1 Resources

Example Questions

Example Question #41 : Factoring Polynomials

Example Question #11 : How To Factor A Variable

Example Question #13 : How To Factor A Variable

Example Question #4652 : Algebra 1

Example Question #4653 : Algebra 1

Example Question #4654 : Algebra 1

All Algebra 1 Resources

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