# Dividing Polynomials....

• Feb 24th 2009, 06:39 PM
EyesThatSparkle02
Dividing Polynomials....
I'm having problems with two questions from my hw. I have no idea where to even start to solve the problems. Please help as much as you can!! Thanks!

1. When x^3 - 7x + 4 is divided by the polynomial D(x), the quotient is x^2 - 3x + 2 and the remainder is -2. Find D(x).

2. Find k so that when x^3 + kx^2 + k^2x + 14 is divided by x+2, the remainder is 0.
• Feb 24th 2009, 07:26 PM
Quote:

Originally Posted by EyesThatSparkle02

2. Find k so that when x^3 + kx^2 + k^2x + 14 is divided by x+2, the remainder is 0.

$x^3+kx^2+k^2x+14=Q(x)(x+2)+R$

$(-2)^3+k(-2)^2+k^2(-2)+14=Q(x)(-2+2)+0$

Continue doing and solve for k .
• Feb 24th 2009, 07:49 PM
EyesThatSparkle02
Quote:

$x^3+kx^2+k^2x+14=Q(x)(x+2)+R$

$(-2)^3+k(-2)^2+k^2(-2)+14=Q(x)(-2+2)+0$

Continue doing and solve for k .

Thank you so much!!!

I'm still kind of confused with the simplifying though...

I get: -8 +4k - 2k^2 + 14 = Q(x)

Q(x) = -2 (k^2-2k-3)

Is that right? Or did I make a mistake somewhere?
• Feb 24th 2009, 07:57 PM
Quote:

Originally Posted by EyesThatSparkle02

1. When x^3 - 7x + 4 is divided by the polynomial D(x), the quotient is x^2 - 3x + 2 and the remainder is -2. Find D(x).

I can see that D(x) is a polynomial of degree 1 , (x-a)

Expressing the whole thing in terms of quotient , divisor and remainder :

$x^3-7x+4=(x^2-3x+2)(x-a)-2$

$x^3-7x+4=x^3-x^2a-3x^2+3xa+2x-2a-2$

By comparing ,

3xa+2x=-7x

a=-3

-2a-2=4

a=-3

Thus , D(x) = (x+3)
• Feb 24th 2009, 08:00 PM
Quote:

Originally Posted by EyesThatSparkle02
Thank you so much!!!

I'm still kind of confused with the simplifying though...

I get: -8 +4k - 2k^2 + 14 = Q(x)

Q(x) = -2 (k^2-2k-3)

Is that right? Or did I make a mistake somewhere?

Q(x)(-2+2)+0=0

So -8 +4k - 2k^2 + 14 = 0
• Feb 24th 2009, 08:05 PM
EyesThatSparkle02
Quote:

I can see that D(x) is a polynomial of degree 1 , (x-a)

Expressing the whole thing in terms of quotient , divisor and remainder :

$x^3-7x+4=(x^2-3x+2)(x-a)-2$

$x^3-7x+4=x^3-x^2a-3x^2+3xa+2x-2a-2$

By comparing ,

3xa+2x=-7x

a=-3

-2a-2=4

a=-3

Thus , D(x) = (x+3)

Thank you!! One question though, how did you compare?? I mean, how did you go from that to 3xa+2x = -7x ?
• Feb 24th 2009, 08:16 PM
Quote:

Originally Posted by EyesThatSparkle02
Thank you!! One question though, how did you compare?? I mean, how did you go from that to 3xa+2x = -7x ?

Firstly , you must make sure that the degree of 'x' on both sides are the same .

In this case , it is 1 so we can compare .

(3a+2)x=(-7)x , we can cancel the x and then solve for a .
• Feb 24th 2009, 08:21 PM
EyesThatSparkle02
Quote:

Firstly , you must make sure that the degree of 'x' on both sides are the same .

In this case , it is 1 so we can compare .

(3a+2)x=(-7)x , we can cancel the x and then solve for a .

How did you get 3ax+2x = -7x in the first place??
• Feb 24th 2009, 08:36 PM
EyesThatSparkle02
Quote:

Q(x)(-2+2)+0=0

So -8 +4k - 2k^2 + 14 = 0

I'm so sorry for having so many questions, but you're really helping!!

The answer is supposed to be - 9/2 .

I keep getting k= 3 or -1 ??
• Feb 24th 2009, 08:41 PM
Quote:

Originally Posted by EyesThatSparkle02
How did you get 3ax+2x = -7x in the first place??

Alright , lets come back to this equation .

$x^3-7x+4=x^3-x^2a-3x^2+3xa+2x-2a-2$

DO some regrouping .

$x^3+0x^2-7x+4=x^3-(a+3)x^2+(3a+2)x-21-2$

So now we compare ..

x^3=x^3 ( there is nothing to compare)

0x^2=-(a+3)x^2 , cancel the x^2 and solve for a .

-7x=(3a+2)x , cancel the x and solve for a

the last one , the constant , do the same comparing .

All of these comparisons give a so you will only have to compare one of them .

Hope this helps. Any more doubts ?
• Feb 24th 2009, 08:42 PM
Quote:

Originally Posted by EyesThatSparkle02
How did you get 3ax+2x = -7x in the first place??

Alright , lets come back to this equation .

$x^3-7x+4=x^3-x^2a-3x^2+3xa+2x-2a-2$

DO some regrouping .

$x^3+0x^2-7x+4=x^3-(a+3)x^2+(3a+2)x-2a-2$

So now we compare ..

$x^3=x^3$ ( there is nothing to compare)

$0x^2=-(a+3)x^2$ , cancel the $x^2$ and solve for a .

$-7x=(3a+2)x$ , cancel the x and solve for a

the last one , the constant , do the same comparing .

All of these comparisons give a so you will only have to compare one of them .

Hope this helps. Any more doubts ?
• Feb 24th 2009, 08:52 PM
EyesThatSparkle02
Quote:

Alright , lets come back to this equation .

$x^3-7x+4=x^3-x^2a-3x^2+3xa+2x-2a-2$

DO some regrouping .

$x^3+0x^2-7x+4=x^3-(a+3)x^2+(3a+2)x-2a-2$

So now we compare ..

$x^3=x^3$ ( there is nothing to compare)

$0x^2=-(a+3)x^2$ , cancel the $x^2$ and solve for a .

$-7x=(3a+2)x$ , cancel the x and solve for a

the last one , the constant , do the same comparing .

All of these comparisons give a so you will only have to compare one of them .

Hope this helps. Any more doubts ?

Thank you so much for all your help!!! It's really helped!!!
I had a question about the 2nd problem. The answer is supposed to be -9/2 yet I keep getting 3 or -1.
• Feb 24th 2009, 09:19 PM