Algebra 1 : Variables

Study concepts, example questions & explanations for Algebra 1

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Example Questions

Example Question #331 : Variables

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

Possible Answers:

\displaystyle 0,-7

\displaystyle -7

\displaystyle 0,7

\displaystyle 7

\displaystyle 0

Correct answer:

\displaystyle 0,7

Explanation:
\(\displaystyle 4x^2-28 x = 0\)\(\displaystyle 4x(x-7) = 0\)\(\displaystyle x (x-7) = 0\)\(\displaystyle \fn_cm x-7 = 0 \;or\; x = 0\)
\(\displaystyle First\;equation:x-7 = 0\)
\(\displaystyle x=7\)

\(\displaystyle x=0,7\)

Example Question #332 : Variables

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

Possible Answers:

\displaystyle (y+6)(y+7)

\displaystyle (y-6)(y-7)

\displaystyle (y+5)(y+8)

\displaystyle (y+3)(y+14)

Correct answer:

\displaystyle (y+6)(y+7)

Explanation:

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

\displaystyle 6\times7=42

\displaystyle 6+7=13

Example Question #333 : Variables

 \displaystyle Factor \;p^2 - 5p - 14

Possible Answers:

\displaystyle (p+7)(p+2)

\displaystyle (p-7)(p+2)

\displaystyle (p-7)(p-2)

\displaystyle (p+7)(p-2)

Correct answer:

\displaystyle (p-7)(p+2)

Explanation:

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

\displaystyle -7\times2=-14

\displaystyle -7+2=-5

Example Question #334 : Variables

Factor completely:

\displaystyle 5x^{2 } - 50x

Possible Answers:

\displaystyle 5x(x-10)

\displaystyle 5(x+1)(x-10)

\displaystyle x(x-50)

\displaystyle 5(x+10)(x-10)

\displaystyle 5(x-1)(x+10)

Correct answer:

\displaystyle 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 \displaystyle x^{2} and \displaystyle x is \displaystyle x, so the GCF of the terms is \displaystyle 5x.

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

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

Example Question #13 : How To Factor A Polynomial

Factor the following polynomial.

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

Possible Answers:

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

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

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

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

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

Correct answer:

\displaystyle (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.

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

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

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

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

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

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

Example Question #335 : Variables

Factor \displaystyle x^{2}+2x-35

Possible Answers:

\displaystyle (x-7)(x-5)

\displaystyle (x+5)(x-8)

\displaystyle (x+5)(x+7)

\displaystyle (x+5)(x+8)

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

Correct answer:

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

Explanation:

When factoring a polynomial that has no coefficient in front of the \displaystyle x^{2} term, you begin by looking at the last term of the polynomial, which is \displaystyle -35.  You then think of all the factors of \displaystyle -35 that when added together equal \displaystyle 2, the coefficient in front of the \displaystyle x term.  The only combination of factors of \displaystyle -35 that can satisfy this condition is \displaystyle -5 and \displaystyle 7.  Thus, the factors of the polynomial are \displaystyle (x-5)(x+7).

Example Question #337 : Variables

 \displaystyle Factor \;x^2 -4x -32

Possible Answers:

\displaystyle (x+4)(x+8)

\displaystyle (x-4)(x+8)

\displaystyle (x+4)(x-8)

\displaystyle (x-16)(x+2)

Correct answer:

\displaystyle (x+4)(x-8)

Explanation:
\(\displaystyle Factor \;x^2-4 x-32\)\(\displaystyle Find \; two \; numbers \; whose \; product \; is \; -32 \; and \; sum \; is -4.\)\(\displaystyle \fn_cm Those \; numbers \; are \; 4 \; and \; -8.\)
\(\displaystyle Therefore, \; x^2-4 x-32 = (x+4)(x-8)\)
\(\displaystyle Answer: \;(x+4)(x-8)\)

 

Example Question #338 : Variables

Factor the following expression.

\displaystyle x^2-3x+2

Possible Answers:

\displaystyle (x-3)(x+1)

\displaystyle (x+2)(x-2)

\displaystyle (x-2)(x-1)

\displaystyle (x-2)(x+1)

\displaystyle (x+2)(x-1)

Correct answer:

\displaystyle (x-2)(x-1)

Explanation:

\displaystyle x^2-3x+2

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

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

\displaystyle (x+B)(x+D)

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

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

\displaystyle (x-2)(x-1)

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

Example Question #21 : Factoring Polynomials

Factor the following expression.

\displaystyle 3x^2-2x-5

Possible Answers:

\displaystyle (3x-1)(x+5)

\displaystyle (3x-5)(3x-1)

\displaystyle (3x-5)(x+1)

\displaystyle (x-2)(5x+2)

\displaystyle (x-5)(3x+1)

Correct answer:

\displaystyle (3x-5)(x+1)

Explanation:

\displaystyle 3x^2-2x-5

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

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

\displaystyle (3x+B)(x+D)

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

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

\displaystyle (3x-5)(x+1)

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

Example Question #22 : Factoring Polynomials

Factor the following expression.

\displaystyle x^3+7x^2+3x+21

Possible Answers:

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

\displaystyle (x+7)(x+3)

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

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

\displaystyle x(x+3)(x+7)

Correct answer:

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

Explanation:

\displaystyle 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 \displaystyle \small x^2, so \displaystyle \small x^2 should be factored out.

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

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

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

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

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

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

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