# algebra

posted by
**Fidelia** on
.

Okay, how do you factor a polynomial when there is a number in front of the highest x term.

For example:

how would you factor 2x^2-11x+15

i know the answer is (2x-5)(x-3), right? So you put the 2x on one side and then you put an x on the other. And then you know that 15*2=30, so you find numbers that multiply to equal 30 and add to equal -11 and you get -6 and -5 and then you take have of the -6 to get a -3, and you put it on the x side because x is half of 2x. Is this why (2x-5)(x-3) is the answer? Is that how you do it for every problem when there is a number in front of the x^2?

(2x-5)(x-3) is the answer because

2*1 = 2, -6 + -5 = -11, and -5*-x = 15, the three coefficients of the binomial answer. I don't agree with or understand your reasoning. There is a bit of trial and error in factoring. If nothing seems to work, use the quadratic equation

[-b +/- sqrt (b^2 - 4ac)]/2a for the roots.

can you plz explain how to do it then?

Fidelia, it looks like you are learning a method called decomposition of the middle term.

I will try to explain it in the simplest way I know how

First you multiply the first number by the last, 2*15=30, you did that.

now make a column of pairs of numbers which when multiplied give you +30

They must be either both positive or both negative to get a positive 30, but the middle number is -11, so obviously the pair of numbers are both negative

-2 -15

-3 -10

-5 -6

-15 -1

Which pair adds up to -11?

it is -5 and -6

so now we know that -11x must be split up into -5x and -6x, so....

2x^2 - 11x + 15

=2x^2 - 5x - 6x + 15

Now use common factoring in "pairs"

=x(2x - 5) - 3(2x - 5) , now we see another common factor of (2x-5)

=(2x-5)(x-3)

Now isn't that easy?