Factor polynomial completely

Question

a)    216x³– y³

 

b)-   x² +16

 

c)    16x³y – 8x²y²+ 20xy³
 

Answer 1

Hi Silvana.

a) is the difference of perfect cubes (6x)³-(y)³

 

So there is a formula for this it goes

 

a³-b³=(a-b)(a²+ab+b²)

 

So for us, a=6x and b=y

 

Using the formula this factors into

 

(6x)³-(y)³ = (6x-y)[(6x)²+6xy+y²)  or

 

(6x-y)(36x²+6xy+y²)

 

 

For part b, is that -x²  or just  x²  ?

 

If it is just x²+16 it is prime and does  not factor further.

 

If it is  -x²+16, then this is the difference of perfect squares and there is a formula for that.  It goes

 

a²-b²=(a-b)(a+b)

 

Rewriting -x²+16 as 16-x² = (4)²-(x)²

 

then our a=4 and our b=x

 

and we get

 

16-x²=(4-x)(4+x)

 

 

Part c, it would be best to factor out any common terms and see where we stand.  So there is a 4, an "x", and a "y" in every term.  So factor those out and get

 

4xy(4x²-2xy+5y²)

 

And I don't think that the (4x²-2xy+5y²) will factor, so part (c) should be finished with

 

4xy(4x²-2xy+5y²)