binomials plzz help.?

Question

multiply the bionomials and write in standard form
(2x-1)(x-4)(x+4)


divide the polynomials

(x^5 - 5x^3 + 9x^2 - 3x + 18) / (x-1)

Answer 1

(2x-1)(x-4)(x+4)
= (2x-1)(x^2-16)
= 2x^3-x^2 -32 x + 16

we add and subtract add to make (x-1) as a afctor and subtarct to compesnate

x^5-5x^3+9x^2-3x+18
= x^5-x^3 - 4x^3 + 4x^2 +5x^2 - 5x +2x -2 + 20
= x^3(x^2-1) - 4x^2(x-1) + 5x(x-1) + 2(x-1) + 20
= (x-1) (x3(x+1)- 4x^2+5x+2) + 20
= (x-1)(x^4+x^3-4x^2+5x+2) + 20

quotient = (x^4+x^3-4x^2+5x+2)
remainder = 20

Answer 2

Answer 1:
(2x-1)(x-4)(x+4)
Ignore (2x-1) for now.
(x-4)(x+4) = x^2 - 4^2
= x^2 - 16
Now bring (2x-1) into the frame to complete the process
(2x-1)(x^2 - 16) = 2x(x^2 - 16) -1(x^2 - 16)
= 2x^3 - 32x - x^2 +16
Standard form = 2x^3 -x^2 -32x + 16

Answer 2:
If you need the quotient and remainder, there is no other way but long division. If you need just the remainder, here's what to do:
Take x - 1 =0
You get x = 1
Substitute for x in: x^5 - 5x^3 + 9x^2 - 3x + 18
As x = 1, the polynomial becomes:
1^5 - 5(1) + 9(1) - 3 + 18
= 1-5+9-3+18
=1+9+18-5-3
=28-8
=20 = Remainder

Answer 3

With the division problem, do it like long division
i.e. (x-1) into x^5 etc goes x^4 then substract (x^4)(x-1) from numerator.