# factorising

• Sep 13th 2009, 07:32 AM
mark
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
• Sep 13th 2009, 07:35 AM
Quote:

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 .
• Sep 13th 2009, 07:41 AM
skeeter
Quote:

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)\$
• Sep 13th 2009, 07:46 AM
tutor.mathitutor
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)