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
Hello,
Let y be the common remainder. y will be a constant, because you divide by a polynom of degree 1.
$\displaystyle
x^3+2x^2+px-3=(x+1)(ax^2+bx+c)+y$
$\displaystyle
x^3+2x^2+px-3=(x-2)(a'x^2+b'x+c')+y
$
If you develop $\displaystyle (x+1)(ax^2+bx+c)$, you get, by identification :
$\displaystyle a=1$
$\displaystyle b=1$
$\displaystyle c=p-1 (1)$
$\displaystyle c+y=-3 \Longleftrightarrow y+3=-c$
(i leave you the right to do the calculus )
In the same way,
$\displaystyle a'=1$
$\displaystyle b'=4$
$\displaystyle c'=p+8 (2)$
$\displaystyle -2c'+y=-3 \Longleftrightarrow y+3=2c'$
Hence 2c'=-c.
With (1) and (2), we can state that 2(p+8)=-(p-1)
And then, just solve for p
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).
In your question,
$\displaystyle P(x) = x^3 + 2x^2 + px - 3$ and $\displaystyle Q(x)=x+1$
The root of Q(x) = 0 is,
$\displaystyle x+1=0$, $\displaystyle x=-1$
So find P(-1),
$\displaystyle P(x) = x^3 + 2x^2 + px - 3$
$\displaystyle P(-1) = -1 + 2 -p - 3 = -p-2$
Now find the remainder for $\displaystyle Q(x) = x-2$
The root of $\displaystyle Q(x) = 0$ is,
$\displaystyle x-2 =0$, $\displaystyle x=2$
So $\displaystyle P(2) = 8 + 8 + 2p -3 = 2p + 13$
Since the remainders are the same,
$\displaystyle -p-2 = 2p + 13$
$\displaystyle p=-5$
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.