# Factorising a quadratic

• Feb 5th 2011, 09:03 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, 09:14 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.

\$\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 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

\$\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 AM
skoker
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.