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)

Thanks in advance..

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- Sep 20th 2008, 09:49 AMMathematixFactorsing 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)

Thanks in advance..

- Sep 20th 2008, 09:53 AMPeritus
- Sep 20th 2008, 09:55 AMskeeter
$\displaystyle 5x^2 - 14x - 3 = 0$

$\displaystyle (5x + 1)(x - 3) = 0$ - Sep 20th 2008, 09:57 AMMathematix
^ thanks, can you please explain HOW you did that..

- Sep 20th 2008, 11:08 AMskeeterQuote:

can you please explain HOW you did that..

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 PMsus-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