How to factor a variable - PSAT Math
Card 0 of 7
Factor the following variable
(x2 + 18x + 72)
Factor the following variable
(x2 + 18x + 72)
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)
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)
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When 
is factored, it can be written in the form 
, where 
, 
, 
, 
, 
, and 
are all integer constants, and 
.
What is the value of 
?
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.
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.
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Factor 9_x_2 + 12_x_ + 4.
Factor 9_x_2 + 12_x_ + 4.
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 9_x_2 + 12_x_ + 4 = 9_x_2 + 6_x_ + 6_x_ + 4
Let's look at the first two terms and last two terms separately to begin with. 9_x_2 + 6_x_ can be simplified to 3_x_(3_x_ + 2) and 6_x_ + 4 can be simplified into 2(3_x_ + 2). Putting these together gets us
9_x_2 + 12_x_ + 4
= 9_x_2 + 6_x_ + 6_x_ + 4
= 3_x_(3_x_ + 2) + 2(3_x_ + 2)
= (3_x_ + 2)(3_x_ + 2)
This is as far as we can factor.
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 9_x_2 + 12_x_ + 4 = 9_x_2 + 6_x_ + 6_x_ + 4
Let's look at the first two terms and last two terms separately to begin with. 9_x_2 + 6_x_ can be simplified to 3_x_(3_x_ + 2) and 6_x_ + 4 can be simplified into 2(3_x_ + 2). Putting these together gets us
9_x_2 + 12_x_ + 4
= 9_x_2 + 6_x_ + 6_x_ + 4
= 3_x_(3_x_ + 2) + 2(3_x_ + 2)
= (3_x_ + 2)(3_x_ + 2)
This is as far as we can factor.
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Factor and simplify:
Factor and simplify:
is a difference of squares.
The difference of squares formula is 
.
Therefore, 
= 
.
The difference of squares formula is
Therefore,
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If 
, and 
, what is the value of 
?
If
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 
The numerator on the left can be factored so the expression becomes
Then you can solve for
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Factor:
Factor:
We can first factor out 
:
This factors further because there is a difference of squares:
We can first factor out
This factors further because there is a difference of squares:
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Solve for x:
Solve for x:
First, factor.
Set each factor equal to 0
Therefore,
First, factor.
Set each factor equal to 0
Therefore,
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