You need to find two numbers that multiply to give 72 and add up to give 18

easiest way: write the multiples of 72:

1, 72

2, 36

3, 24

4, 18

6, 12: these add up to 18

 (x + 6)(x + 12)

Nothing common cancels at the beginning. To factor this, we need to find two numbers that multiply to 9 * 4 = 36 and sum to 12. 6 and 6 work.

So 9x² + 12x + 4 = 9x² + 6x + 6x + 4

Let's look at the first two terms and last two terms separately to begin with. 9x² + 6x can be simplified to 3x(3x + 2) and 6x + 4 can be simplified into 2(3x + 2). Putting these together gets us 

9x² + 12x + 4

= 9x² + 6x + 6x + 4

= 3x(3x + 2) + 2(3x + 2) 

= (3x + 2)(3x + 2)

This is as far as we can factor. 

The numerator on the left can be factored so the expression becomes \(\displaystyle \dpi{100} \small \frac{\left ( x+3 \right )\times \left ( x-3 \right )}{\left ( x+3 \right )}=5\), which can be simplified to \(\displaystyle \dpi{100} \small \left ( x-3 \right )=5\)

Then you can solve for \(\displaystyle \dpi{100} \small x\) by adding 3 to both sides of the equation, so \(\displaystyle \dpi{100} \small x=8\)

First, factor.

\(\displaystyle \small x^2+3x+2=(x+2)(x+1)=0\)

Set each factor equal to 0

\(\displaystyle \small x+2=0; x=-2\)

\(\displaystyle \small x+1=0; x=-1\)

Therefore,

\(\displaystyle \dpi{100} \small x=-2\ or-1\)

Let's try to factor x² – y² – z² + 2yz.

Notice that the last three terms are very close to y² + z² – 2yz, which, if we rearranged them, would become y² – 2yz+ z². We could factor y² – 2yz+ z² as (y – z)², using the general rule that p² – 2pq + q² = (p – q)² .

So we want to rearrange the last three terms. Let's group them together first.

x² + (–y² – z² + 2yz)

If we were to factor out a –1 from the last three terms, we would have the following:

x² – (y² + z² – 2yz)

Now we can replace y² + z² – 2yz with (y – z)².

x² – (y – z)²

This expression is actually a differences of squares. In general, we can factor p² – q² as (p – q)(p + q). In this case, we can substitute x for p and (y – z) for q.

x² – (y – z)² = (x – (y – z))(x  + (y – z))

Now, let's distribute the negative one in the trinomial x – (y – z)

(x – (y – z))(x  + (y – z)) 

(x – y + z)(x + y – z)

The problem said that factoring x² – y² – z² + 2yz would result in two polynomials in the form (ax + by + cz)(dx + ey + fz), where a, b, c, d, e, and f were all integers, and a > 0.

(x – y + z)(x + y – z) fits this form. This means that a = 1, b = –1, c = 1, d = 1, e = 1, and f = –1. The sum of all of these is 2.

The answer is 2. 

\(\displaystyle 64y^{2} - 16\) is a difference of squares.

The difference of squares formula is \(\displaystyle a^{2} - b^{2} = (a - b)(a + b)\).

Therefore, \(\displaystyle \frac{64y^{2} - 16}{8y + 4}\) = \(\displaystyle \frac{(8y + 4)(8y - 4)}{8y + 4} = 8y - 4\).

We can first factor out \(\displaystyle -3\):

\(\displaystyle -3(4x^{2}-9)\)

This factors further because there is a difference of squares:

\(\displaystyle -3(2x+3)(2x-3)\)

You need to factor to find the possible values for x. You need to fill in the blanks with two numbers with a sum of -12 and a product of 36. In both sets of parenthesis, you know you will be subtracting since a negative times a negative is a positive and a negative plus a negative is a negative

(x –__)(x –__).

You should realize that 6 fits into both blanks.

You must now set each set of parenthesis equal to 0.

x – 6 = 0; x – 6 = 0

Solve both equations: x = 6 

We first expand the right hand side as x²+2x+tx+2t and factor out the x terms to get x²+(2+t)x+2t. Next we set this equal to the original left hand side to get x²+rx +6=x²+(2+t)x+2t, and then we subtract x²  from each side to get rx +6=(2+t)x+2t. Since the coefficients of the x terms on each side must be equal, and the constant terms on each side must be equal, we find that r=2+t and 6=2t, so t is equal to 3 and r is equal to 5.

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

First, add 4 to both sides:

\(\displaystyle 2x^2=12\)

Divide both sides by 2:

\(\displaystyle x^2=6\)

\(\displaystyle x=\pm \sqrt{6}\)