1. ## factorising

hi, i've done part of a question that i need help finishing. the question is:

use the factor theorem to show that $(x - 2)$ is a factor of $P(x) = x^3 - 3x^2 - 4x + 12$ (which i did)
then it says:
write P(x) as a product of three linear factors.
now obviously (x - 2) is one of them so i found out what the equation was that had to be multiplied by (x - 2) to get $P(x) = x^3 - 3x^2 - 4x + 12$ and i came to $(x^2 - x -6)$ and tried to factorise it further but couldn't. why can't i find two numbers that multiply to give -6 but add to give -1? surely this is the way to do it?

thanks, mark

2. Originally Posted by mark
i came to $(x^2 - x -6)$ and tried to factorise it further but couldn't. why can't i find two numbers that multiply to give -6 but add to give -1? surely this is the way to do it?

thanks, mark
Hi

You mean : find two numbers that multiply to give -6 but add to give +1 ?
-2 and 3

3. i looked at it again and i can't see why its two numbers that multiply to give -6 but add to give +1? the equation does come to $x^2 - x - 6$ to factorise it from there you need to multiply 1 from the $x^2$ by minus 6 (the constant) and then find two numbers that multiply to give -6 and add to give -1 (from the middle term). can anyone point me in the right direction here please?

4. Originally Posted by mark
i can't see why its two numbers that multiply to give -6 but add to give +1?
You're right: You need factors of -6 that add to -1, not +1. So use the numbers provided, but reverse their signs.

5. I am sorry but $a x^2 + bx + c$ has 2 roots whose sum is $-\frac{b}{a}$ and whose product is $+\frac{c}{a}$

In your case $x^2 - x - 6$, the sum is +1 and the product is -6

6. (x + 3)(x - 2) = x^2 + 1x - 6

This is not the original quadratic, so adding to +1 would seem to lead to an invalid result.