# Dividing Polynomials

• Sep 18th 2013, 03:47 PM
sakonpure6
Dividing Polynomials
Hi, I am struggling with the following question >.<
Quote:

a) 8x^3 + 10x^2 -px - 5 is divisible by 2x-1. There is no remainder. Find the value of p.

b) When x^6+x^4-2x^2+k is divided by 1+x^2, the remainder is 5. Find the value of k.
I tried, but could not come up with anything. Thank you in advance.
• Sep 18th 2013, 04:09 PM
Plato
Re: Dividing Polynomials
Quote:

Originally Posted by sakonpure6
Hi, I am struggling with the following question >.<
I tried, but could not come up with anything. Thank you in advance.

Do you actually know how to do polynomial "long division"?
If not then that is where you start.
• Sep 18th 2013, 04:50 PM
SworD
Re: Dividing Polynomials
You don't need polynomial division. If $8x^3 + 10x^2 -px - 5$ is divisible by $2x-1$, then it must have $x=\frac{1}{2}$ as one of its roots. Plug in $x=\frac{1}{2}$ into $8x^3 + 10x^2 -px - 5 = 0$ and solve for p. Do you see how this works?

Similarly, if $x^6+x^4-2x^2+k$ has a remainder of 5 when divided by $1+x^2$, then $x^6+x^4-2x^2+k - 5$ must have all of the roots which $1+x^2$ has. Plug in $x = i$ OR $x = -i$ into $x^6+x^4-2x^2+k - 5 = 0$ and solve for k.
• Sep 19th 2013, 12:36 PM
Hartlw
Re: Dividing Polynomials
Thought you might like to know why:

if x-r divides p(x), p(x)=(x-r)q(x), and p(r)=0

if x-r divides p(x) with remander R, p(x)=(x-r)q(x) + R, and p(r)=R