I need help with factorising the following:

a) (c+d)^2 - d^2

I got c^2 + 2cd + d^2 - d^2 which is equal to c^2 + 2cd, is this correct please?

b) 2w^2 + w - 3

How do you do this again please (need simple and clear advice)? Thanks

Printable View

- Feb 5th 2011, 09:03 AMNatasha1Factorising a quadratic
I need help with factorising the following:

a) (c+d)^2 - d^2

I got c^2 + 2cd + d^2 - d^2 which is equal to c^2 + 2cd, is this correct please?

b) 2w^2 + w - 3

How do you do this again please (need simple and clear advice)? Thanks - Feb 5th 2011, 09:14 AMTheEmptySet
For the first one I think they wanted you to use the difference of squares.

$\displaystyle a^2-b^2=(a-b)(a+b)$ so in your case

$\displaystyle \underbrace{(c+d)^2}_{a^2}-\underbrace{d^2}_{b^2}=\underbrace{(c+d-d)}_{(a-b)}\underbrace{(c+d+d)}_{(a+b)}=c(c+2d)$

For b notice that

$\displaystyle 2w^2+w-3=2w^2-2w+3w-3$ and now factor by grouping.

Notice to break up the middle term we found factors of $\displaystyle 2(-3)=-6$ that add up to the middle term $\displaystyle -2w+3w=w$ - Feb 5th 2011, 09:25 AMArchie Meade
$\displaystyle 2w^2=2w(w)$

$\displaystyle (2w+a)(w+b)=2w^2+w-3$

Examine "a" and "b" to find out how we get a single "w" after multiplying out the factors....

$\displaystyle ab=-3\Rightarrow\ a=3,\;b=-1;\;\;\;a=-3,\;b=1,\;\;\;a=1,\;b=-3,\;\;a=-1,\;\;b=3$

Since "b" will be multiplied by 2, we see that it needs to be $\displaystyle -1$

$\displaystyle 2w^2+w-3=(2w+3)(w-1)$ - Feb 5th 2011, 10:25 AMskoker
here is how i did the first equation. $\displaystyle (c+d)^{2}$ is a positive perfect square.

$\displaystyle (c+d)^{2}-d^{2}$

$\displaystyle (c+d)(c+d)-d^{2} $

$\displaystyle ({c}^{2}+2cd+{d}^{2})-{d}^2$ expand perfect square.

$\displaystyle {c}^{2}+2cd$ collect d squared terms they cancel.

$\displaystyle c(c+2d)$ factor out c.