# Factorising

• Aug 14th 2008, 09:13 AM
Joker37
Factorising
Hi! I am having trouble factorising these:

6xsquared - 11x - 35

6xsquared - 37x + 45

Can someone please show me worked examples to these. And sorry, I don't know how to use latex but I will learn it soon to make mathematical interpretations easier.

Any help will be appreciated heaps!!!
• Aug 14th 2008, 09:21 AM
o_O
Here's a thread that demonstrates how to factor general quadratic equations: Tips on factoring, a not equal to 1

Go through that and see what you can come up with. If you have any troubles, feel free to post again.
• Aug 14th 2008, 09:24 AM
skeeter
just have to play with them a bit ...

$\displaystyle 6x^2 - 11x - 35 = (3x + 5)(2x - 7)$

$\displaystyle 6x^2 - 37x + 45 = (3x - 5)(2x - 9)$
• Aug 14th 2008, 09:29 AM
masters
Quote:

Originally Posted by Joker37
Hi! I am having trouble factorising these:

6xsquared - 11x - 35

6xsquared - 37x + 45

Can someone please show me worked examples to these. And sorry, I don't know how to use latex but I will learn it soon to make mathematical interpretations easier.

Any help will be appreciated heaps!!!

$\displaystyle 6x^2-11x-35$

Here's what I do:

$\displaystyle 6\times35=210$

Find 2 factors of 210 that add up to -11.

$\displaystyle 210=2\cdot3\cdot5\cdot7$

How about -21 and 10?

Replace -11x with -21x+10x

$\displaystyle 6x^2-21x+10x-35$

Group the first two terms and the last two terms

$\displaystyle (6x^2-21x)+(10x-35)$

Factor out common factors in each group

$\displaystyle 3x(2x-7)+5(2x-7)$

$\displaystyle \boxed{(3x+5)(2x-7)}$
• Aug 14th 2008, 09:43 AM
Joker37
Quote:

Originally Posted by masters

$\displaystyle \boxed{(3x+5)(2x-7)}$

Will it still be considered correct if you put (2x-7)(3x+5) instead of (3x+5)(2x-7)?
• Aug 14th 2008, 09:48 AM
masters
Quote:

Originally Posted by Joker37
Will it still be considered correct if you put (2x-7)(3x+5) instead of (3x+5)(2x-7)?

Absolutely. Multiplication is commutative.

$\displaystyle a \cdot b=b \cdot a$