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GED Math : Solving by Factoring

Study concepts, example questions & explanations for GED Math

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Example Questions

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Example Question #332 : Algebra

What is the value of if is a positive integer?

t^2-6t-40=0

Possible Answers:

Correct answer:

Explanation:

Start by factoring the equation.

t^2-6t-40=0

We will need two numbers that multiply up to and add up to . These two numbers are $-10 \text{ and }4$ .

Thus, we can factor the equation.

(t-10)(t+4)=0

Solving the equation will give the following solutions:

t-10=0

t=10

and

t+4=0

t=-4

Since the question states that must be a positive integer, can only equal to .

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Example Question #22 : Quadratic Equations

x^2 + 4x + 3 = 0 . Solve for .

Possible Answers:

3,1

-3,1

-3, -1

Correct answer:

-3, -1

Explanation:

In order to solve this equation, you need to factor the expression,

x^2 + 4x + 3 .

To do so, you need to find two factors of that have a sum of .

The two factors are and and the correct factoring is

(x + 1)(x +3) ,

so you know that

(x+ 1)(x +3) = 0 .

Therefore, will be equal to any values that give the inside of the parentheses a value of .

So, is equal to both and .

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Example Question #891 : Ged Math

Solve for :

x^2-14x+33=0

Possible Answers:

x=11, x=3

x=-11, x=3

x=11, x=-3

x=-11, x=-3

Correct answer:

x=11, x=3

Explanation:

Start by factoring the equation:

x^2-14x+33=0

For this equation, we will want two numbers that add up to and multiply up to . In this case, the two numbers are and .

We can then write the following:

(x-3)(x-11)=0

Now, set each factor equal to zero and solve for .

x-3=0

x=3

and

x-11=0

x=11

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Example Question #331 : Algebra

For what value of will the following equation be true?

(x+b)(x+4)=x^2+x-12

Possible Answers:

Correct answer:

Explanation:

(x+b)(x+4)=x^2+x-12

Start by factoring x^2+x-12 . Since one factor is already given to you, you just need to figure out what number when multiplied by will give and when added to will give . The only number that fits both criteria is . must be equal to .

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Example Question #25 : Quadratic Equations

Solve by factoring

x^2+8x+15=0

Possible Answers:

x=-3 or x=-5

x=-3 or x=5

Not enough information

x=3 or x=5

x=0

Correct answer:

x=-3 or x=-5

Explanation:

We must start by factoring x^2+8x+15

We must think of two numbers that multiply to be 15 and add to be 8. We come up with 3 and 5

Then

x^2+8x+15=(x+3)(x+5)

So we have

(x+3)(x+5)=0

Which means either (x+3)=0 or (x+5)=0

So, x=-3 or x=-5

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Example Question #21 : Quadratic Equations

Solve for x by factoring:

$9x^{2}+24x+16=0$

Possible Answers:

$x=\frac{3}{4}$

$x=\frac{4}{3}$ or $x=-\frac{4}{3}$

$x=-\frac{4}{3}$

$x=-\frac{3}{4}$

$x=\frac{4}{3}$

Correct answer:

$x=-\frac{4}{3}$

Explanation:

Looking at $9x^{2}+24x+16$ , we notice that it is a perfect square trinomial.

*A perfect square trinomial is given by the form $a^{2}+2ab+b^{2}=(a+b)^{2}$ (where "a" represents a variable term and "b" represents a constant term).

*Comparing this to our trinomial, we find...

a=3x

b=4

2ab=2(3x)(4)=24x

*So, we confirm it is, indeed, a perfect square trinomial.

$9x^{2}+24x+16=(3x+4)^{2}$

$9x^{2}+24x+16=0$

$(3x+4)^{2}=0$

(3x+4)(3x+4)=0

We have one solution:

3x+4=0

3x=-4

$x=\frac{-4}{3}$

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Example Question #27 : Quadratic Equations

Factor completely:

$x^{5}+2x^{4}-48x^{3}$

Possible Answers:

$x^{3}(x^{2}+2x-48)$

$x^{3}(x+8)(x-6)$

$x^{3}(x-6)(x+8)$

$x(x^{4}+2x^{3}-48x^{2})$

$(x^{2}+2x-48)(x^{3}+2x-48x)$

Correct answer:

$x^{3}(x-6)(x+8)$

Explanation:

1) First to note in this problem is a common factor of $x^{3}$ in $x^{5}+2x^{4}-48x^{3}$

2) Factoring an $x^{3}$ , we have $x^{3}(x^{2}+2x-48)$

3) We now have $x^{3}$ and a factorable trinomial.

4) $x^{3}(x^{2}+2x-48)$

5) Two numbers that add to 2 and have a product of -48 are: 8 and -6

6) $x^{3}(x^{2}+2x-48)=x^{3}(x-6)(x+8)$

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Example Question #28 : Quadratic Equations

Find the values of x in the following:

$x^{2}-3x-28=0$

Possible Answers:

x=-4

x=7

x=4 or x=-7

No solutions

x=7 or x=-4

Correct answer:

x=7 or x=-4

Explanation:

This question can be answered by factoring

$x^{2}-3x-28=0$

Our factors will be in the form (x+a)(x+b)=0

We need to find and such that a+b=-3 and (a)(b)=-28

We notice that a=-7 and b=4 fit those criteria

Then:

(x-7)(x+4)=0

We need to consider both binomials can equal 0 and satisfy the equation

x-7=0 x+4=0

x=7 x=-4

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Example Question #21 : Solving By Factoring

Factor completely:

$16x^{6}-64$

Possible Answers:

$(4x^{3}-8)(4x^{3}+8)$

$4(x^{2}-2)(x^{2}+2)$

$(4x^{3}-8)^{2}$

$4(x^{6}-16)$

$16(x^{2}-2)(x^{2}+2)$

Correct answer:

$16(x^{2}-2)(x^{2}+2)$

Explanation:

$16x^{6}-64$ is in the form of $a^{2}-b^{2}$ which is a perfect square binomial

$a^{2}-b^{2}=(a+b)(a-b)$

*** $16x^{6}-64=(4x^{3})^{2}-(8)^{2}$

$(4x^{3})^{2}-(8)^{2}=(4x^{3}+8)(4x^{3}-8)$

Factor a 4 from each binomial

$(4x^{3}+8)(4x^{3}-8)=4(x^{3}+2)\cdot4(x^{3}-2)$

multiplying 4 x 4 gives the result $16(x^{3}+2)(x^{3}-2)$

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GED Math : Solving by Factoring

Study concepts, example questions & explanations for GED Math

All GED Math Resources

Example Questions

Example Question #332 : Algebra

Example Question #22 : Quadratic Equations

Example Question #891 : Ged Math

Example Question #331 : Algebra

Example Question #25 : Quadratic Equations

Example Question #21 : Quadratic Equations

Example Question #27 : Quadratic Equations

Example Question #28 : Quadratic Equations

Example Question #21 : Solving By Factoring

All GED Math Resources

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