All PSAT Math Resources
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 : How To Factor A Variable
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 #2 : Algebra
If , and , what is the value of ?
8
–8
0
–6
6
8
The numerator on the left can be factored so the expression becomes , which can be simplified to
Then you can solve for by adding 3 to both sides of the equation, so
Example Question #1 : Algebra
Solve for x:
First, factor.
Set each factor equal to 0
Therefore,
Example Question #3 : Factoring
When is factored, it can be written in the form , where , , , , , and are all integer constants, and .
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 #3 : Algebra
Factor and simplify:
is a difference of squares.
The difference of squares formula is .
Therefore, = .
Example Question #51 : Variables
Factor:
We can first factor out :
This factors further because there is a difference of squares:
Example Question #1 : How To Factor A Polynomial
What is a possible value for x in x2 – 12x + 36 = 0 ?
–6
There is not enough information
6
2
6
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
Example Question #2 : How To Factor A Polynomial
If r and t are constants and x2 +rx +6=(x+2)(x+t), what is the value of r?
5
7
6
It cannot be determined from the given information.
5
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.
Example Question #5 : Algebra
Solve for :
First, add 4 to both sides:
Divide both sides by 2:
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