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 can solve this problem by substituting in \(\displaystyle 50^{\circ}\) for \(\displaystyle T_{Fahrenheit}\) and solving for \(\displaystyle T_{Celsius}\):

\(\displaystyle T_{Celsius}=(50-32)\cdot \frac{5}{9}\)

\(\displaystyle T_{Celsius}=18\cdot \frac{5}{9}\)

\(\displaystyle T_{Celsius}=\frac{18}{1}\cdot\frac{5}{9}=\frac{18\cdot5}{1\cdot9}=\frac{90}{9}=10\)

That means that \(\displaystyle 50^{\circ}\) Fahrenheit is the same as \(\displaystyle 10^{\circ}\) Celsius.

Group all the terms with the \(\displaystyle x\) variable.

\(\displaystyle x^3+2x^2+a+bx^2+2= (x^3+2x^2+bx^2)+a+2\)

Pull out an \(\displaystyle x^2\) term from parentheses.

\(\displaystyle (x^3+2x^2+bx^2)+a+2 = x^2(x+2+b)+a+2\)

There are no more common factors.

The correct answer is:  \(\displaystyle x^2(x+2+b)+a+2\)

This word problem can be broken down into a basic algebraic equation:

A half-loaded semi weighs 37,000 lbs. Subtracting the weight of the truck, we can determine that a half load weighs 17,000 lbs:

\(\displaystyle \frac{1}{2}x=37,000\:lbs-20,000\:lbs=17,000\:lbs\)

From that, we can determine that a full load = 34,000 lbs: 

\(\displaystyle 2\cdot \frac{1}{2}x=17,000\:lbs\cdot2=34,000\:lbs\)

Knowing the weight of a full load, we can calculate the weight of a 3/4 load:

\(\displaystyle \frac{3}{4}\cdot x=\frac{3}{4}\cdot34,000\:lbs=25,500\:lbs\)

Add the weight of the truck to get the total weight:

\(\displaystyle Weight=20,000\:lbs\cdot25,500\:lbs=45,500\:lbs\)