• Feb 5th 2011, 10:03 AM
Natasha1
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, 10:14 AM
TheEmptySet
Quote:

Originally Posted by Natasha1
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

For the first one I think they wanted you to use the difference of squares.

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

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

For b notice that

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

Notice to break up the middle term we found factors of $2(-3)=-6$ that add up to the middle term $-2w+3w=w$
• Feb 5th 2011, 10:25 AM
Quote:

Originally Posted by Natasha1
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?

It's almost complete, notice that "c" is common...

c(c+2d) is fully factored.

Also, the method shown by TheEmptySet is very useful.

b) 2w^2 + w - 3

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

$2w^2=2w(w)$

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

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

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

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

$2w^2+w-3=(2w+3)(w-1)$
• Feb 5th 2011, 11:25 AM
skoker
here is how i did the first equation. $(c+d)^{2}$ is a positive perfect square.

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

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

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

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

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