algebra help??

Question

working with factors:
3p^3 - 12p + 15
to do this here is an ex.: x^2-16= (x+4)(x-4)
answer and please explain how you did so

Answer 1

3 is common to all 3p^3 and 12p and 15

so.... 3 (p^3-4p+5)

basically factoring is reverse multiplication....what can i multiply together to get the original piece. in ur example u will see that (x+4) times (x-4) is x^2 -16....so in ur question if u distribute the 3 that I pulled out u will get the original piece....algebra is hard at first but believe me it is important. If u r having trouble w/ this plz get help or u will put urself in a bad position later on bcuz it all builds on previous material

Answer 2

3p^3 - 12p + 15
>>3( p^3 - 4p + 5)
>>3( p(p^2 - 4) + 5 )
>>3( p(p-2)(p+2) + 5)
>>3p(p-2)(p+2) + 15

I dont know precisely how you want to answer the question, but i hope this helps.

Answer 3

I think you mean
3p^2- 12p + 15


Answer; (3p+3)(p-5)
Since (3p+3)(p-5)= 3p^2+3p-15p-15=3p^2-12p+15

or if you mean

3p^3 - 12p + 15 then
3p^3 - 12p + 15
=3( p^3 - 4p + 5)
=3( p(p^2 - 4) + 5 )
=3( p(p-2)(p+2) + 5)
=3p(p-2)(p+2) + 15

Answer 4

(x-4)(x+5)

Answer 5

3(p^3-4p+5)