Polynomial division (I think)

Mar 2010
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1
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?
 
May 2010
1,260
410
Mauritius
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,

\(\displaystyle f(x) = (x+5)(x-k)\)

\(\displaystyle f(2) = (2+5)(2-k) = 28\)

Solve for k (Happy)
 
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May 2010
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Mauritius
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.
 

Soroban

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May 2006
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Hello, blackdragon190!

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

\(\displaystyle \text}Find }k.\)

If you must use long division . . .


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


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


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


Edit: Too slow . . . again!
 
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May 2010
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Mauritius
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.