Question

factor: 6x^2-7x-3

Answer 1

\[(2x-3)(3x+1)\]
Do you need help on how this answer was obtained?

Answer 2

(3x+1)(2x-3)

Answer 3

what's the point in answering a question with the same reply??

Answer 4

yes please

Answer 5

We are looking for factors of 6 and factors of -3 which cross multiply and then add to give -7.

Factors of 6: 3 and 2
Factors of -3: -3 and 1

3x-3 = -9
3x1 = +2

-9+2 = -7

Then you put the -3 and +1 into the brackets with 3x and 2x on both sides.
Hard to exlplain, I did my best - try watching a youtube video.

Answer 6

how 3 * 1 = 2 ??

Answer 7

and do you a good video that explains this?

Answer 8

http://www.youtube.com/watch?v=ISPxJ6JXT8o&feature=relmfu
This guy is good :)

Answer 9

A second degree polynomial of the form\[ax^2-bx+c\]can be factored using the formula \[ax^2-bx+c=a(x-x_1)(x-x_2)\] where \[x_1,x_2\]the roots of the equation \[ax^2-bx+c=0\]
So, let's start by solving the equation
\[6x^2-7x-3=0\]\[Δ=b^2-4ac=(-7)^2-4\times6(-3)=121\]\[x_{1,2}=\frac{-b \pm \sqrt{Δ}}{2a}\]\[x_{1,2}=\frac{-(-7)\pm \sqrt{121}}{2\times6}=\frac{7\pm11}{12}\]\[x_1=\frac{7+11}{12}=\frac{18}{12}=\frac{3}{2}\]\[x_2=\frac{7-11}{12}=\frac{-4}{12}=-\frac{1}{3}\]
Now, let's put into practise the formula given above\[6x^2-7x-3=6(x-\frac{3}{2})(x-(-\frac{1}{3}))=\]\[(6x-9)(x+\frac{1}{3})=(2x-3)3(x+\frac{1}{3})=\]\[(2x-3)(3x+1)\]
And there you have, \[6x^2-7x-3\]when factored is\[(2x-3)(3x+1)\]Hope this helped. It may look tricky at first, but once you get the hang of it, it becomes really easy.