Question

Factor fully

Answer 1

\[\huge x^3 + 2x^2 - 18x - 40\]

Answer 2

You can take an x out of the first three terms

Answer 3

and then ?

Answer 4

I don't know what you can do other than that

Answer 5

because (x^2 + 2x - 18) can't be factored

Answer 6

Well, I guess you could take an x out of the first two terms of that

Answer 7

but that doesn't really help you. It also happens to be divisible by (x+4), which would help you, but I don't know how you'd figure that out.

Answer 8

i know :S

Answer 9

Yeah it's a nasty one.

Answer 10

can you do it with grouping of some sort ? :S

Answer 11

like first two terms and then last two

Answer 12

actually, there's a fairly neat technique that I just came across

Answer 13

so, a cubic always has at least one linear factor

Answer 14

(x)

Answer 15

or maybe something else, but either way, since it has at least one linear factor, you can write it like (x + q)(ax^2 + bx+c).

Now, it's easy to see that q*c has to equal the constant term of the cubic. It's also easy to see that -q is a solution to the cubic. So, all you have to do is list out the factors of the constant term of the cubic, and then check each one to see if it's a root of the cubic. If it is, then you can find your linear factor.

Answer 16

Try using trial and error by basically finding factors of 40 and 1, and multiplying them to see if you get x^3 + 2x^2 - 18x - 40. Hint: Try 4 and 10.

Answer 17

(x+4)(x^2-2x-10) might work. Try expanding that to see if we get x^3 + 2x^2 - 18x - 40.

Answer 18

You want to find the roots.

Answer 19

The is the most mathematical solution I can think of:

1) Use the rational root theorem to find the first root.
2) Use synthetic division to divide \(x-r\) from the expression, where \(r\) is the first root.
3) Use the quadratic formula on the remaining quadratic equation to find the other roots.

Answer 20

x^3 + 2x^2 - 18x - 40
=x^3+4x^2-2x^2-8x -10x -40
=x^2(x+4) -2x(x+4) -10(x+4)
=(x+4)(x^2-2x - 10)

Answer 21

@burhan101
follow the steps shown by wio

Answer 22

thankyou @wio that was exactly what i was looking for :)