Question

Factor each polynomial completely over the set of complex numbers.
a(x)=2x^4-3x^3-24x^2+13x+12
Please help, I am very confused.

Answer 1

\[2x^4-3x^3-24x^2+13x+12\]

Answer 2

I recommend you test various rational roots first.

You may have learnt that rational roots (p/q) if they exist have numerator being factor of the constant term, and the denominator being a factor of the leading coefficient.

So factors of constant 12 - gives 1,2,3,4,6,12
factors of leading coeff. 2 - gives 2 and 1
So possible rational roots are 12/2 = 6, 12/1 =12, 6/2 = 3, 6/1 = 6, 4/2 = 2, 4/1 = 4, 2/2 =1, 2/1 = 2 and 1/1 =1, which narrows to 6,4, 3,2 and 1 (in both positive or negative forms)
Try subbing in the smaller numbers first and see if that produces 0.

Answer 3

a(1) = 2(1)^4 - 3(1)^3 - 24(-1)^2 +13(1) +12 = 0
So (x-1) must be a factor of a(x).

You can then try some of the other possible rational factors by subbing in -1, 2,-2, 3,-3, 4, -4 and 6, -6 in, or can do long division with (x-1) as the factor and see if that simplifies it a bit for you.

Answer 4

From Mathematica:
\[(x-4) (x-1) (x+3) (2 x+1) \]

Answer 5

^That's cheating! lol.

But as you can see the rational factors did agree with the possible set I provided, though I somehow forgot about +/- 1/2.
In any case with most polynomials you are given you must find one factor to get started by using factor theorem and then it tends to start chug along slowly but surely.