# Math Help - Factorising a quadratic trinomial

1. ## Factorising a quadratic trinomial

Factorise the following:

6x^2-11x-10

2. Originally Posted by Tessarina
Factorise the following:

6x^2-11x-10
$6x^2-11x-10$

$=6x^2-15x+4x-10$

$=3x(2x-5)+2(2x-5)$

$=(2x-5)(3x+2)$

3. Hello Tessarina
Originally Posted by Tessarina
Factorise the following:

6x^2-11x-10
$(3x+2)(2x-5)$. Was there a particular problem with this one?

4. Originally Posted by Tessarina
Factorise the following:
6x^2-11x-10
Hi Tessarina,
this might assist:

The factors of 6 (call them a & c)
1 x 6 = 6
2 x 3 = 6

The factors of 10 (call them b & d)
1 x 10 = 10
2 x 5 = 10
Since -10 is negative you know that one of the signs will be.

You are looking for this form:
(ax - b)(cx + d) or (ax - d)(cx + b)

We already know the factors (coefficients) that will give: 6x^2 & -10

We are seeking the middle part
such that
ax*d - cx*b = -11x
or
ax*b - cx*d = -11x

We are trying to find -11x in the following:
--------------------------------------
plugging in the factors (1,6 & 1,10)
(1x - 1)(6x + 10)
1x*10 - 6x*1 = 4x (does not equal -11x)
or
(1x - 10)(6x + 1)
1x*1 - 6x*10 = -59 (does not equal -11x)

plugging in the factors (1,6 & 2,5)
(1x - 2)(6x + 5)
1x*5 - 6x*2 = -1x (does not equal -11x)
or
(1x - 5)(6x + 2)
1x*2 - 6x*5 = -28x (does not equal -11x)

plugging in the factors (2,3 & 1,10)
(2x - 1)(3x + 10)
2x*10 - 3x*1 = 17x (does not equal -11x)
or
(2x - 10)(3x + 1)
2x*1 - 3x*10 = -28 (does not equal -11x)

plugging in the factors (2,3 & 2,5)
(2x - 2)(3x + 5)
2x*5 - 3x*2 = 4x (does not equal -11x)
or
(2x - 5)(3x + 2)
2x*2 - 3x*5 = -11x (what is needed ! )

The result wanted:
(2x - 5)(3x + 2) = 6x^2 -11x - 10
::
After you do this a few times alexmahone's method will be easy to use and generate the result;
after a few more times you will be able to do it immediately the way Grandad can.

The asterisk (*) used above denotes multiplication.
.

5. Originally Posted by Tessarina
Factorise the following:

6x^2-11x-10
Of related interest: http://www.mathhelpforum.com/math-he...tml#post385857