SOLUTION: I believe I have done this one right but if someone could please confirm? Multiply as required and collect terms: (x - 4) (2x^2 + 4) + (x + 4)(x^3 - 1) - (x^2 - x -n 1) OK

Algebra.Com
Question 172919: I believe I have done this one right but if someone could please confirm?
Multiply as required and collect terms:
(x - 4) (2x^2 + 4) + (x + 4)(x^3 - 1) - (x^2 - x -n 1)
OK! This is what I have:
(x - 4)+(x +4)(x^3 - 1) =
x^5 - 1)(2x^2 + 4)-(x^2 - x - 1)=
(4x^7 + 3)-(x^2 - x - 1) =
(4x^7 + 3)-(x - 1)=
4x^6 +2
Is this correct? Thanks for any help available?

Answer by monika_p(71)   (Show Source): You can put this solution on YOUR website!
(x - 4) (2x^2 + 4) + (x + 4)(x^3 - 1) - (x^2 - x - 1)
First you have do multiplication using distributive property (a+b)(c+d)=ac+ad+bc+bd
(x*2x^2+x*4-4*2x^2-4*4) + (x*x^3+x*(-1)+4*x^3+4*(-1))-(x^2-x-1)= remove brackets and remember to change sign if there is - before (...)
2x^3+4x-8x^2-16+x^4-x+4x^3-4-x^2+x+1= add like items
x^4+6x^3-9x^2+4x-19

RELATED QUESTIONS

I was instructed to "multiply as required and collect terms" I am really not sure what it (answered by vleith,solver91311)
Multiply as required and collect terms ( x – 4)(2x^2 + 4) + ( x + 4)(x^3 – 1) – ( x^2 –... (answered by jim_thompson5910)
This problem stumped me last night and tried to work it out but have no clue how I... (answered by solver91311)
Can someone confirm if I am doing this problem right? I have to provide the answer in... (answered by stanbon,Alan3354)
On this one, I have done it 4 times and each time came up with a different answer. Could... (answered by rossiv53)
Multiply as required and collect terms ( x – 4)(2x2 + 4) + ( x + 4)(x3 – 1) – ( x2 – x – (answered by colliefan)
If I have x/2x would this then equal 1/x?? Also, if I have x^2/x^3 would this equal... (answered by josmiceli)
multiply (3/4 X^3- !/3)(1/5 X^2 + 3/7) I have never done one like this.Would... (answered by Alan3354)
i have done this problem several times...each i do it i get a different answer!i'm not... (answered by THANApHD)