Factoring
- Thread starter cad
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There are several ways to go about this. For example, there is a theorem which basically says that if you have a polynomial p(x)
p(x) = an xn + an-1 xn-1 + ... + a1 x + a0
an is not zero and the roots of p(x) (x's where p(x) is zero) are
{x0, x1, x2, ..., xn}
then p(x) can be written as
p(x) = an (x - x0) (x - x1) (x - x2) ... (x - xn-1)(x - xn)
So you could find the root to
p(x) = 2 x2 - 9 x + 7
which are 1 and 3.5 from the answer given and write
p(x) = 2 (x - 1) (x - 3.5) = (x - 1) (2x - 7)
which is equivalent to what you have given.
The other way is an easier way most of the time after you become familiar with problems like this (after a lot of practice). Since
(a x + b) (cx + d) = ac x2 + (ad + bc) x + bd
you need to find factors a and c of the coefficient of x2, 2 in our problem, and factors of the constant coefficient b and d, 7 in our problem) so that
(ad + bc) is equal to the coefficient of x, -9 in our problem. That sounds like a mouthful but really is easier a lot of the time and especially as you get more and more practice.
Now for your specific problem: The factors of 2 are 2 and 1 or 1 and 2 )or you can use the negative of these, i.e. -2 and -1 or -1 and -2). The factors of 7 are 1 and 7 or 7 and 1. We note that 7 plus 2 is 9 and so we need to use the 1 and 7 and the 1 and 2 but for one set we need to use the negatives because we want -9 not +9. This leads to (following your answer)
a = -1
b = 1
c = -2
d = 7
so that
ac = 2
ad + bc = -9
cd = 7
or
p(x) = (-x + 1) (-2x + 7) = (1 - x) (7 - 2x)
Note that we could also have chosen
a = 1
b = -1
c = 2
d = -7
which I think is the usual way the factors would be written
p(x) = (x - 1) (2x - 7)
The plus sign in front of the "7" means the signs are the same. Additionally, the minus sign in front
of the "9x" means the signs are both minus. Either start with the \(\displaystyle \ 2x^2 \ \) term or the constant 7 term.
For instance:
(2x -\(\displaystyle \ \ \ \) )(x -\(\displaystyle \ \ \ \) )
You have two choices for the positive factors of 7 (1 & 7) to be placed. One of the choices gives the
correct factorization in this case. (It happens to factor.)
I don't know why that form of the answer was given, but if you multiply each of the factors
above in this post by -1, you can manipulate them to the answer you presented.
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