1. ## factorising

the question is: a polynomial is given by $\displaystyle P(x) = x^3 + 3x^2 - 10x - 24$. given that $\displaystyle (x + 2)$ is a factor factorise P(x) completely

now i actually know a way of doing this question but it involves a bit of trial and error so i was wondering if there was a quicker way. i would start by using the idea that 2ab = -24 leaving ab = -12. but then i'd have to go through 1, 2, 3, 4, 6, 12 and their minuses to find the right one. could someone tell me if there's a quicker way or this it?

thanks, Mark

2. Originally Posted by mark
the question is: a polynomial is given by $\displaystyle P(x) = x^3 + 3x^2 - 10x - 24$. given that $\displaystyle (x + 2)$ is a factor factorise P(x) completely

now i actually know a way of doing this question but it involves a bit of trial and error so i was wondering if there was a quicker way. i would start by using the idea that 2ab = -24 leaving ab = -12. but then i'd have to go through 1, 2, 3, 4, 6, 12 and their minuses to find the right one. could someone tell me if there's a quicker way or this it?

thanks, Mark
Hi

Do the long division ,then you will get a quadratic equation , further factorise it if possible . I think this is the fastest for such questions .

3. Originally Posted by mark
the question is: a polynomial is given by $\displaystyle P(x) = x^3 + 3x^2 - 10x - 24$. given that $\displaystyle (x + 2)$ is a factor factorise P(x) completely

now i actually know a way of doing this question but it involves a bit of trial and error so i was wondering if there was a quicker way. i would start by using the idea that 2ab = -24 leaving ab = -12. but then i'd have to go through 1, 2, 3, 4, 6, 12 and their minuses to find the right one. could someone tell me if there's a quicker way or this it?

thanks, Mark
synthetic division ...

Code:
-2]........1......3......-10......-24
.................-2......-2........24
-------------------------------------
..........1.......1.....-12........0
note the depressed coefficients 1, 1, -12

$\displaystyle (x+2)(x^2 + x - 12)$

now you can factor the quadratic factor rather easily ...

$\displaystyle (x+2)(x+4)(x-3)$

4. P(x) = x^3 + 3x^2 - 10x - 24
Assume that P(x) = (x+2)(Ax^2 + Bx + C)

Equating the coefficient of x^3, x^2, x, you will get
A = 1
2A+ B = 3 => B = 1
C + 2B = -10 => C = -12

p(x) = (x+2)(x^2 + x -12) = (x+2)(x+4)(x-3)

Therefore P(x) = (x+2)(x+4)(x-3)