# Factorsing help!

• Sep 20th 2008, 09:49 AM
Mathematix
Factorsing help!
how do i factorise this ??? Can someone show me how..

5x^2 -10x = 4x + 3
So, 5x^2 - 14x - 3 = 0

Is that right so far, & need to know how to factorise it ! (Worried)

• Sep 20th 2008, 09:53 AM
Peritus
• Sep 20th 2008, 09:55 AM
skeeter
\$\displaystyle 5x^2 - 14x - 3 = 0\$

\$\displaystyle (5x + 1)(x - 3) = 0\$
• Sep 20th 2008, 09:57 AM
Mathematix
^ thanks, can you please explain HOW you did that..
• Sep 20th 2008, 11:08 AM
skeeter
Quote:

can you please explain HOW you did that..
lots of practice!

the factors of the first term 5x^2 are 5x and x, and the factors of the last term -3 are 1 and 3, where each must have opposite signs.

with that info, start with (5x ... )(x ... ), place 3 and 1 and remember one has to (+) and the other (-) ... just play with the placement so that your middle term comes out to be -14x when you FOIL.
• Sep 20th 2008, 12:25 PM
sus-annah
5x^2 - 10x = 4x + 3
5x^2 - 10x -4x - 3 = 0
5x^2 - 14x -3 = 0

In the equation 5x^2-14x-3 the coefficient of x^2 is 5 so the factors must be of the form
(5x + c)(x + d) = 5x^2 + 5xd + cx + cd = 5x^2 + (5d+c)x +cd
so 5d + c = -14, cd = -3. The pair of factors of 3 are 1 and 3,
but for cd = -3 one of c and d must be negative and one positive.
To satisfy 5d + c = -14, we require c = 1, d = -3

Therefore 5x^2 - 14 -3 = (5x + 1)(x - 3)
Therefore the required factorised equation is (5x + 1)(x - 3) = 0