how do i factor 4r^2+4rs+s^2-16

Question

help me please

Answer 1

To factor the expression 

4r^2 + 4rs + s^2 - 16

We first take the first 3 terms which reduces to ( 2r +s)²

and 16 = 4²

 

So given expression is ( 2r +s)² -4² 

= (2r+s -4 ) (2r+s +4 ) using (a^2 - b^2 ) = (a+b)(a-b)

 

 

Answer 2

Hang in there Valarie; this is a stout one.

 

Since there are four terms and 2 variables, none of the factoring patterns for the form ax^2 + bx + c will work.

Four terms suggest a factor by grouping approach.

 

4r^2 + 4rs + s^2-16

 

= 4r^2 + 4rs     + s^2 - 16      Break the 4 terms into two binomials and factor each pair.

 

= 4r(r + s) + (s + 4)(s - 4)      Hmm ... the 2nd pair of terms don't factor.

                                              If an (r + s) term had factored out, we'd have been one step away

                                                  from being done. But we don't have it. Sooooo ...

 

If your're sure that you've written the problem down correctly, then the answer is

that it's prime, a one word way to say it can't be factored.

 

Ah-ha! Mr. P saw something I didn't. Clever man; seeing the patterns for a perfect square and the difference of two squares in the same problem.  Hats off to you Mr. P!

 

Mr. G