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How do I factor polynomials?

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2 Answers

It's basically the opposite of solving a problem that looks like this (4x+5)(3x-2) and getting 12x+15x-8x-10 which you reduce to a polynomial which would be 12x^2+7x-10.  You start with the polynomial, and have to find the factors which will be in the form I started with.  So, if your problem is 12x^2+7x-10, I see it as having a middle number, a front number and an end number.  You need to find numbers that can be multiplied to get 12, the front number, and numbers that multiply to get -10, the end number.  It is important to keep track of you negatives--the best thing is to think of something like 7x-10 as 7x + (-10) so you don't forget it is negative! Now, 3x4 and 2x6 and 1x12 are all 12, and 2x5 and 1x10 are both 10, but you need to pick the ones that when multiplied and added will equal your middle number.  Try putting them in the answer format.  (2x+1)(6x-10) will give you 12x^2-20x+6x-10 or 12x^2-14x-10, which is not what you started with

(x-5)(12x+2) will give you 12x^2+12x-60x-10, or 12x^2-58x-10, which is also wrong

so you will end up going with(4x+5)(3x-2) eventually.

If you have a trinomial (ax^2 + bx + c)

you have to break it down into two binomials 

For example x^2 + 5x + 6

If a = 1 then the product of the two numbers equaling c must add up b

6 = 2*3 ............... so 2+3 = 5

thus (x+3)(x+2)

If a is not equal to one, you then have to take into consideration the factors of a as well as c

For example: 2x^2 + 7x + 3

a = 2 and c = 3 so the product of two numbers equaling a and products of two numbers equaling c must equal b in some way of multiplying

2 * 1 = 2 ....

Because those are the only two factors we can start writing the binomials as (2x + ?)(x + ?)

now c = 3 and the only two factors would be 3*1 = 3 

so you have 2 and 1  as well as 3 and 1. 

If the middle number is 7, I can take 2*3 = 6  and 1*! = 1 so 6+1=7

If I put them in factored form, I get (2x+1)(x+3). 

Does that make sense?