Method for factorizing when "a" not equal to 1?

Question

Can u help with a method to factor a quadratic equation when the x square coefficent is NOT1. example like 2xsquare plus 3x minus 2

I want a method to apply to other problems. not the solution to this one.

Leaving it in general terms of a,b,c

Answer 1

multiply a*c (ignoring signs)
now list all the factors and do the problem the same way you do problems when a=1.
{EXAMPLE 6x^2 + 13x + 6 ===> 6*6. look for factors of 36 that add to 13 ===> they are 9 and 4}
now do the usual
( x - BLAH) ( x - BLUH)
then divide each BLAH-BLAH by a
reduce the fraction
eliminate fraction
(x+9) (x+4)
now divide by 6 ===> 9/6 = 3/2 and 4/6 = 2/3
change now
(x+3/2)(x+2/3) ==> eliminate fractions
(2x+3) (3x+2)

=]

good luck to you!

Answer 2

I will use this one as an example:

2x^2 + 3x - 2

Multiply the coefficient of x^2 with the constant term
2*(-2) = -4

Find two number whose product is -4 and the sum is +3 (coefficient of x)

-4 = 4*(-1) ------ 4+(-1)=3

Split up the 3x term into 4x-x

2x^2 + 3x - 2
=2x^2 + 4x - x -2
Take common factors for the first two terms and the last two terms
2x(x+2) -1(x+2) -----> note that the term in brackets will be common

This factorises to
(x+2)(2x-1)

Answer 3

well really you still do the exact same thing, except make the coeffiecients of each x term equal a. Like in your example
2x^2 + 3x - 2
becomes
(2x ) (x )
then from there you figure out factors of the last term, the -2, and determine +'s and -'s
(2x - 1 ) (x + 2)
as you can see, if you use foil on those two terms, it makes your example correct. really its just a lot of guess and check. there is no exact formula

Answer 4

ax^2 + bx + c =
a(x^2 + (b/a)x + (b/(2a))^2 + c/a - (b/(2a))^2) =
a(x + b/(2a))^2 - (b^2 - 4ac)/(4a^2) =
a(x + b/2 - √((b^2 - 4ac)/(4a^2))(x + b/2 - √((b^2 - 4ac)/(4a^2))