# Thread: Finding value of "p" of a polynomail

1. ## Finding value of "p" of a polynomail

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

What do I have to do to find p?

2. 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.

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

3. 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?

4. 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?
Sure you can. $\sqrt{-1}=i$

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

5. The answer to the question according to the answer section is 3. But how do you get that with these imaginary numbers?

6. Originally Posted by masters
Sure you can. $\sqrt{-1}=i$

$(i)^6+(i)^4-2(i)^2+k=5$
Originally Posted by Devi09
The answer to the question according to the answer section is 3. But how do you get that with these imaginary numbers?

$i^6=-1$
$i^4=+1$
$i^2=-1$

$-1+1-2(-1)+k=5$