Solving by Factoring

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GED Math › Solving by Factoring

Questions 1 - 10

Solve the equation by factoring:
$\small 2x^2-2x-4=0$

$\small x=-1\ or\ x=2$

$\small x=1\ or\ x=2$

$\small x=-1\ or\ x=-2$

$\small x=1\ or\ x=-2$

Explanation

$\small 2x^2-2x-4=2(x^2-x-2)=2(x+1)(x-2)=0$

Therefore:

$\small x+1=0\ or\ x-2=0$

$\small x+1-1=0-1\ or\ x-2+2=0+2$

$\small x=-1\ or\ x=2$

Factor: $2x^{2}+4x-70$

Explanation

Begin by factoring out a 2:

$2x^{2}+4x-70=2(x^{2}+2x-35)$

Then, we recognize that the trinomial can be factored into two terms, each beginning with :

$2(x\ \ \ \ )(x\ \ \ \ )$

Since the last term is negative, the signs of the two terms are going to be opposite (i.e. one positive and one negative):

$2(x+\ \ )(x- \ \ )$

Finally, we need two numbers whose product is negative thirty-five and whose sum is positive two. The numbers and fit this description. So, the factored trinomial is:

Solve for .

$y-7=\frac{4x+6y}{3}$

$y=-\frac{4}{3}x-7$

$y=\frac{2}{3}x-7$

$y=-\frac{2}{3}x-7$

$y=\frac{4}{3}x-7$

Explanation

$y-7=\frac{4x+6y}{3}$

Multiply both sides by 3:

$\frac{3}{1}(y-7)=(\frac{4x+6y}{3})\frac{3}{1}$

Distribute:

Subtract from both sides:

Add the terms together, and subtract from both sides:

Divide both sides by :

$\frac{-4x-21=3y}{3}$

Simplify:

$y=-\frac{4}{3}x-7$

Solve for :

Explanation

You can factor this trinomial by breaking it up into two binomials that lead with :

$(x \ \ \ \ \ \) (x \ \ \ \ \ \) = 0$

You will fill in the binomials by finding two factors of 36 that add up to 5. This is achieved with positive 9 and negative 4:

You can then set each of the two binomials equal to 0 and solve for :

$x+9=0 \ \ \ \ \ \ x-4=0$

$x=-9 \ \ \ \ \ \ x=4$

Solve for x by factoring:

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

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

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

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

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

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

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...

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

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

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

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

We have one solution:

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

What is the value of if is a positive integer?

Explanation

Start by factoring the equation.

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.

Solving the equation will give the following solutions:

and

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

Solve for :

and

Explanation

For quadratic equations, you need to factor in order to solve for your variable. You do this after the equation is set equal to zero. Luckily, this is already done for you! Thus, start by factoring:

into

Then, you set each factor equal to . Solve each "small" equation:

BOTH of these are answers to the equation.

Factor completely:

$16x^{6}-64$

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

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

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

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

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

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 $16(x^{3}+2)(x^{3}-2)$

. Solve for .

Explanation

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

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

so you know that

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

So, is equal to both and .

Solve for : $x^{2}+16x=-28$

Explanation

This is a factoring problem so we need to get all of the variables on one side and set the equation equal to zero. To do this we subtract from both sides to get $x^{2}+16x+28=0$

Think of the equation in this format to help with the following explanation. $ax^{2}+bx+c=0$

We must then factor to find the solutions for . To do this we must make a factor tree of which is 28 in this case to find the possible solutions. The possible numbers are $1\cdot 28$ , $2\cdot 14$ , $4\cdot 7$ .