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Algebra 1 : Polynomials

Study concepts, example questions & explanations for Algebra 1

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10 Diagnostic Tests 557 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #331 : Variables

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

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+5)(y+8)

(y-6)(y-7)

(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 #334 : Polynomials

Factor completely:

$5x^{2 } - 50x$

Possible Answers:

5x(x-10)

5(x+1)(x-10)

x(x-50)

5(x-1)(x+10)

5(x+10)(x-10)

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 #13 : How To Factor A Polynomial

Factor the following polynomial.

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

Possible Answers:

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

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

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

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

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

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 #336 : Polynomials

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

Possible Answers:

(x+5)(x-8)

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

(x-7)(x-5)

(x+5)(x+7)

(x+5)(x+8)

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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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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Algebra 1 : Polynomials

Study concepts, example questions & explanations for Algebra 1

All Algebra 1 Resources

Example Questions

Example Question #331 : Variables

Example Question #11 : How To Factor A Polynomial

Example Question #12 : How To Factor A Polynomial

Example Question #334 : Polynomials

Example Question #13 : How To Factor A Polynomial

Example Question #336 : Polynomials

Example Question #335 : Variables

Example Question #21 : How To Factor A Polynomial

Example Question #334 : Polynomials

Example Question #335 : Polynomials

All Algebra 1 Resources

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