# Math Help - theorem

1. ## theorem

x^3 + 2x^2 + px -3 is divided by x+1 the remainder is the same as when it is divided by x-2. find value of p

thanks

2. Hello,

Let y be the common remainder. y will be a constant, because you divide by a polynom of degree 1.

$
x^3+2x^2+px-3=(x+1)(ax^2+bx+c)+y$

$
x^3+2x^2+px-3=(x-2)(a'x^2+b'x+c')+y
$

If you develop $(x+1)(ax^2+bx+c)$, you get, by identification :

$a=1$
$b=1$
$c=p-1 (1)$
$c+y=-3 \Longleftrightarrow y+3=-c$

(i leave you the right to do the calculus )

In the same way,
$a'=1$
$b'=4$
$c'=p+8 (2)$
$-2c'+y=-3 \Longleftrightarrow y+3=2c'$

Hence 2c'=-c.

With (1) and (2), we can state that 2(p+8)=-(p-1)

And then, just solve for p

3. Let P(x) and Q(x) be polynomials and k be the root of Q(x)=0.

Then the remainder of the division P(x)/Q(x) is equal to P(k).

$P(x) = x^3 + 2x^2 + px - 3$ and $Q(x)=x+1$
The root of Q(x) = 0 is,
$x+1=0$, $x=-1$

So find P(-1),
$P(x) = x^3 + 2x^2 + px - 3$
$P(-1) = -1 + 2 -p - 3 = -p-2$

Now find the remainder for $Q(x) = x-2$
The root of $Q(x) = 0$ is,
$x-2 =0$, $x=2$

So $P(2) = 8 + 8 + 2p -3 = 2p + 13$

Since the remainders are the same,
$-p-2 = 2p + 13$
$p=-5$

4. Originally Posted by gracey
x^3 + 2x^2 + px -3 is divided by x+1 the remainder is the same as when it is divided by x-2. find value of p

thanks
To streamline things a bit ......

Use the remainder theorem:

Let the remainder be r. Then:

-1 + 2 - p - 3 = r => -2 - p = r .... (1)

8 + 8 + 2p - 3 = r => 13 + 2p = r .... (2)

Equate (1) and (2): -2 - p = 13 + 2p => p = -5.