PSAT Math : How to factor a variable
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Example Questions
Example Question #1 : Algebra
Factor the following variable
(x2 + 18x + 72)
(x + 6) (x + 12)
(x – 6) (x – 12)
(x + 6) (x – 12)
(x – 6) (x + 12)
(x + 18) (x + 72)
(x + 6) (x + 12)
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)
Example Question #2 : Variables
Factor 9x2 + 12x + 4.
(9x + 4)(9x + 4)
(9x + 4)(9x – 4)
(3x – 2)(3x – 2)
(3x + 2)(3x – 2)
(3x + 2)(3x + 2)
(3x + 2)(3x + 2)
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.
Example Question #1 : Factoring Polynomials
If 
6
–6
8
–8
0
8
The numerator on the left can be factored so the expression becomes 
Then you can solve for 
Example Question #1 : Algebra
Solve for x:
First, factor.
Set each factor equal to 0
Therefore,
Example Question #1 : Factoring
When 
What is the value of 
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.
Example Question #1 : How To Factor A Variable
Factor and simplify:
The difference of squares formula is 
Therefore, 
Example Question #2 : Factoring Polynomials
Factor:
We can first factor out 
This factors further because there is a difference of squares:
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