remainder theorem

**Punch** · October 8th 2010, 07:13 PM

The expressions $x^3-7x+6$ and $x^3-x^2-4x+24$ have the same remainder when divided by $(x+p)$

Find the possible values of $p$ .

what i did

$ -p^3-7(-p)+6=(-p^3)-(-p)^2-4(-p)+24 $
$ -p^3+7p+6=-p^3-p^2+4p+24 $
$ -p^3+7p+6-[-p^3-p^2+4p+24]=0 $
$ p^2+3p-18=0 $
$ p=3,-6 $

however, ans is 3,-3!!

**Soroban** · October 8th 2010, 07:33 PM

Hello, Punch!

Absolutely correct . . . good work!

**harish21** · October 8th 2010, 07:53 PM

Originally Posted by Punch

The expressions $x^3-7x+6$ and $x^3-x^2-4x+24$ have the same remainder when divided by $(x+p)$

Find the possible values of $p$ .

what i did

$ -p^3-7(-p)+6=(-p^3)-(-p)^2-4(-p)+24 $
$ -p^3+7p+6=-p^3-p^2+4p+24 $
$ -p^3+7p+6-[-p^3-p^2+4p+24]=0 $
$ p^2+3p-18=0 $
$ p=3,-6 $

however, ans is 3,-3!!

It comes out that 3 and -6 are indeed the correct answers.

when you divide the given terms by (x+3), the remainder is 0 in both the cases.

when you divide the given terms by (x-6), you remainder in both the cases is 180

But if you divide the given terms by (x-3), the remainders are not the same..

You can check back by dividing the terms by (x+3), (x-6), and (x-3), and you'll see 3 and -6 are the correct values for p...

**Punch** · October 9th 2010, 01:29 AM

Thanks!! I was confused about why my answer was different from the answer sheet!!! i repeated the question 5 times

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