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February 8, 2016
Posted by **lisa** on Wednesday, December 19, 2012 at 7:15pm.

- algebra 2 -
**Matt**, Thursday, December 20, 2012 at 11:39amAre you wanting it factorised?

If so:

multiply the constant, in this case '-3' by the co-efficient of x^2, so:

(2x-3=6)

Take the 2 off the co-efficient and replace the constant, -3, with -6, so you have:

x^2 -5x -6

Find products of -6 that also add to make -5.

The products, in this case, are -6 and 1:

-6 multiplied by 1 = -6

-6 + 1 = -5

You now have both co-efficients of x, so remove the -5 in front of the x in the ORIGINAL equation: '2x^2-5x-3' and replace with '-6x + x' so it becomes:

2x^2-6x+x-3

Split the equation between the two co-efficients of x:

'2x^2-6x' and 'x-3'

Factorise these seperately e.g.

'2x^2-6x' becomes 2x(x-3).

and,

'x-3' becomes 1(x-3).

As you can see the equation inside both brackets are the same so now take the numbers in front of both brackets:'2x' and '+1' and form a seperate bracket: (2x+1)

Discard one of the (x-3) brackets and keep the other.

so in brackets: (2x+1) (x-3).

x=-1/2 or x=3.

Matt

- algebra 2 -
**Matt**, Thursday, December 20, 2012 at 11:42amIn short:

2x^2-5x-3 becomes 2x^2-6x+x-3.

Split and factorise for:

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

Which becomes:

(2x+1)(x-3)

x=-1/2 or x=3