1. ## remainder theorem

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!!

2. Hello, Punch!

Absolutely correct . . . good work!

3. 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...

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