SOLUTION: factoring polynomials of the form ax^2+bx+c 4x^2-27x+18 I can figure the factors of 4 and the factors of 18 but I'm lost with the trial factors and middle term not really s

Algebra ->  Distributive-associative-commutative-properties -> SOLUTION: factoring polynomials of the form ax^2+bx+c 4x^2-27x+18 I can figure the factors of 4 and the factors of 18 but I'm lost with the trial factors and middle term not really s      Log On


   

Question 46540: factoring polynomials of the form ax^2+bx+c

4x^2-27x+18
I can figure the factors of 4 and the factors of 18
but I'm lost with the trial factors and middle term
not really sure as to its purpose
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Well, there'e a bit of trial-and-error processing that occurs when you try to facto a trinomial whose x%5E2 coefficient is greater than 1.
Here's the way I go about it:
I know that the factors of 4x%5E2 are:
%284x++%29%28x++%29 or %282x++%29%282x++%29
The factors of 18 are: (1)(18) or (2)(9) or (3)(6)
I also notice that the middle term of the trinomial is negative and the last term is positive. This tells me that the contants part of the factors are both negative numbers.
Why? Because a negative times a negative is a positive (last term is positive) and a negative plus a negative is a negative (middle term is negative)
Try the first pair of factors with the x's
%284x+-+1%29%28x+-+18%29+=+4x%5E2+-72x+-+x+%2B+18 = 4x%5E2+-+73x+%2B+18No go! The middle term is wrong.
Try:
%284x+-+2%29%28x+-+9%29+=+4x%5E2+-36x+-2x+%2B+18 = 4x%5E2+-+38x+%2B+18 No go! The middle term is wrong.
Try:
%284x+-+3%29%28x+-+6%29+=+4x%5E2+-+24x+-+3x+%2B+18 = 4x%5E2+-+27x+%2B+18 Bingo!
I hope this helps.