Finding value of "p" of a polynomail

**Devi09** · March 25th 2011, 09:46 AM

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

What do I have to do to find p?

**masters** · March 25th 2011, 10:02 AM

Originally Posted by Devi09

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

What do I have to do to find p?

Hi Devi09,

There's a couple of things you can do here. Do you know synthetic division?

You could use $\frac{-1}{2}$ as your divisor and find P that way.

Perhaps an easier method would be to act like $2x + 1$ is a factor of $f(x)=8x^3+10x^2-px-5$ in which case $x = -\frac{1}{2}$ is a root. (Factor Theorem)

Now, set $f(x)=0$ and find $f(-\frac{1}{2})$ . (Remainder Theorem.) $f(x)=0$ means when $-\frac{1}{2}$ is substituted for x, the remainder is 0.

Once you've substituted $-\frac{1}{2}$ for x, you'll can solve for p.

$8(-\frac{1}{2})^3+10(-\frac{1}{2})^2-p(-\frac{1}{2})-5=0$

**Devi09** · March 25th 2011, 10:36 AM

I got another question. Lets say $x^6 + x^4 - 2x^2 + k$ is divided by $1 + x^2$ and the remainder is 5 and you need to find k

So $f(sqrt(-1))$

$(sqrt-1)^6 + (sqrt-1)^4 - 2(sqrt-1)^2 + k = 5$

but -1 can't be square rooted so what would you do?

**masters** · March 25th 2011, 10:46 AM

Originally Posted by Devi09

I got another question. Lets say $x^6 + x^4 - 2x^2 + k$ is divided by $1 + x^2$ and the remainder is 5 and you need to find k

So $f(sqrt(-1))$

$(sqrt-1)^6 + (sqrt-1)^4 - 2(sqrt-1)^2 + k = 5$

but -1 can't be square rooted so what would you do?