Who is asking: Student
Level: Secondary
Question:
I'm having trouble factoring expressions that aren't monic. I can do
things like
x2-9x+8, but problems like 12x2+5x-3 have me stumped.
I also have a
question about factoring out common factors. In a problem like
x2-18x+81,
wouldn't you divide by 9? But what happens to the
x2, is it x2/9?
These are problems off my homework, but I want to know how to do them as well as just the answers.
I'm a little behind my class in regards to quadratics and such, so I'd like to have a good understanding of it. Thanks so much!
Hi,
Factoring 12x2+5x-3 is like solving a mystery. There are lots of clues, you just have to see what they are. The major clues are the coefficient of x2 and the constant term, that is the 12 and the -3.
The goal is to write
12x2+5x-3 = (ax + b)(cx + d)
Lets look at the constant term first, the -3. From the right side above you can see that
b
d = -3
Thus b and d are 1 and 3 but they have opposite signs since the product is -3. Hence this clue gives us
12x2+5x-3 = (ax + 3)(cx - 1)
or
12x2+5x-3 = (ax - 3)(cx + 1).
Now the second clue. Where does the 12 come from? In either of the two cases above
12 = a
But 12 factors as
12 = 12
12 = 6
12 = 4
Now you need to do the "leg work". In each of the two possible factorizations of 12x2+5x-3 above you need to try
a = 12, c = 1
a = 1, c = 12
a = 6, b = 2
a = 2, b = 6
.
.
.
To see which one gives the coefficient of x as 5.
For your second problem, x2 - 18x + 81, you are correct that if you attempt to factor out a 9 you get the coefficient of x2 as 1/9. But this isn't very nice as you want to work with integers, not fractions. Ther is no common factor here.
Conside the clues as in the first problem. Since the coefficient of x2 is 1, I know that
x2 - 18x + 81 = (x + a)(x + b)
Since the constant term is 81 I know that
a
Since 81 is positive, a and b have the same sign. But the coefficient of x is negative so
x2 - 18x + 81 = (x - a)(x - b)
What are a and b so that a b
= 81 and the coefficient of x is -18?
Penny