Polynomial division (I think)

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July 4th 2010, 03:15 AM
blackdragon190

Polynomial division (I think)

I'm absolutely stumped by this problem:

If f(x) = (x + 5) (x - k) and the remainder is 28 when f(x) is divided by x - 2, find k.

I figure that I'm supposed to use synthetic division and then have the remainder equal 28, but that isn't getting me the right answer (which is k = -2). Could someone please help me out?
July 4th 2010, 03:20 AM
Unknown008

You can use the remainder theorem, which states that if f(x) is divided by x-a, and the remainder is b, then f(a) = b.

In your case,

$f(x) = (x+5)(x-k)$

$f(2) = (2+5)(2-k) = 28$

Solve for k (Happy)
July 4th 2010, 03:46 AM
blackdragon190

Thanks! (Happy)
July 4th 2010, 03:49 AM
Unknown008

You're welcome.

I tried using long division too, but it's easy to get the numbers messed up. Both methods work though.

Through long division, I get as equation:

14 - 7k = 28

Then, you get k = -2.
July 4th 2010, 04:02 AM
Soroban

Hello, blackdragon190!

Quote:

$f(x) \:=\: (x + 5) (x - k)$ and the remainder is 28
when $f(x)$ is divided by $x - 2$ .

$\text}Find }k.$

If you must use long division . . .

We have: . $\bigg[x^2 + (5-k)x - 5k\bigg] \div \bigg[x-2\bigg]$

. . $\begin{array}{ccccccccc} & & & & x & + & 7-k \\ & & -- & -- & --- & -- & --- \\ x-2 & | ^& x^2 & + & (5-k)x & - & 5k \\ & & x^2 & - & 2x \\ & & -- & -- & --- \\ &&&& (7-k)x & - & 5k \\ &&&& (7-k)x & - & 2(7-k) \\ & & & & --- & -- & --- \\ &&&&&& \text{-}7k +14 & \leftarrow\text{ remainder}\end{array}$

Hence: . $\text{-}7k+14 \;=\:28 \quad\Rightarrow\quad \text{-}7k \:=\:14 \quad\Rightarrow\quad \boxed{k \:=\:-2}$

Edit: Too slow . . . again!
July 4th 2010, 04:05 AM
Unknown008

Quote:

Originally Posted by Soroban

Edit: Too slow . . . again!

I was trying to post the long division method, but the code tabs weren't behaving and there were some sort of limited spaces... I gave up then, and posted only the final equation.
July 4th 2010, 06:51 AM
blackdragon190

Quote:

Edit: Too slow . . . again!

Thanks anyway! :)