Factor this please?

Question

2x^3-3x^2-12x

I know it is easy but giving me trouble

Can x(2x^2-3x-12) be factored anymore

Answer 1

First factor out x:

2x³-3x²-12x = x·(2x²-3x-12)

then factor (2x²-3x-12)

An easy way to factor "naughty" quadratics is to factor out the leading coefficients and then use the quadratic formula:

!!! (but do not forget to first factor the leading coefficient [the coeff of x²] out, or you are going to lose this factor in your final answer) !!!

(2x²-3x-12) = 2(x² -3/2x - 6)
x = [3/2 ± √(9/4 + 36 )] / 2
x = [3/2 ± √(153/4)] / 2
x = [3/2 ± (√153)/2)] / 2
x = [3 ± (√153)] / 4
x = [3 ± (3√17)] / 4
x = 3/4(1± √17)

So (2x²-3x-12) = 2(x² -3/2x - 6) = 2(x - 3/4(1+ √17))(x -3/4(1- √17))

And
2x³-3x²-12x = x·(2x²-3x-12) =
2·x·(x - 3/4(1+ √17))·(x -3/4(1- √17))

Not a very elegant result :-/

Either the question really is "funny" or there is a typo :-)

Answer 2

It is a little tricky.
The first thing to do is to factor out one x term from the expression. This will leave you with x(2x^3-3x-12). The next step is to use the quadratic formula which is:
x= -b+/- sqrt(b^2-4ac))/(2a)
In this case, a=2, b=-3 and c=-12.
When you plug in and solve for x, you get;
(x)(4x+(sqrt(105)-3)(4x-sqrt(105)-3) and all of this divided by 8.

Answer 3

First factor out the x: x(2x^2-3x-12)
It looks like this is as far as you can go since the roots are irrational.

Answer 4

Use the quadratic equation:

-b +/- (b^2 - 4ac)^1/2
-------------------------------
2a

Where aX^2 + bx + c = 0


In you problems, you already have it partially factored, the x removed from the equation leaves you


x(2x^2 - 3x - 12)


Using the quadratic equations,

a = 2
b = -3
c = -12


3 +/- ((-3)^2 - 4*2*(-12))^1/2
------------------------------------------
2*2


3 +/- (9 + 96)^1/2
----------------------------
4


3 +/- (105)^1/2
------------------------
4

3 +/- 10.247
--------------------
4


which gives you 2 answers:

3.312 and -1.812


so, it is completely factored as follows:


x (x + 3.312) (x - 1.812)