how do you go about factoring this kind of problem??

Question

2x^4-17x^3+46x^2-43x+12

the answer given is
(2x-1)(x-3)(x-4)(x-1)

how did they get this answer?? can someone show me please.

Answer 1

Use the Factor theorem that says
(x-a) is a factor of P(x) if P(a) =0

So you test for factors using this theorem.
Let P(x) = 2x^4-17x^3+46x^2-43x+12

Then to test to see if (x-1) is a factor evaluate P(1)
If P(1) = 0 then (x-1) is a factor.
And so on.

Try different values until you have two factors, since you have a fourth degree polynomial, then divide these factors into the polynomial to find the remaining quadratic factor which you will factor as a trinomial.

Answer 2

Note that the first term is 2x^4, so it's factors are 2*1*1*1

The last term is +12, so it's factors could be:
1*1*1*12
1*1*2*6
1*1*3*4
1*2*2*3
Note that one of the factors must be one.

Note that the terms are +, -, +, -, +
That tells you that an even number of the factors have negatives.

Try (x-1) because it's easy to divide:
(2x^4-17x^3+46x^2-43x+12) / (x-1) = 2x^3 -15x^2 +31x -12

Then you have to pick another one to try. (x-2) won't work.
Try (x-3)
(2x^3 -15x^2 +31x -12) / (x-3) = 2x^2 - 9x -4
And that factors into (2x-1)(x-4).

It's guess and check, but you can use what you know from the signs and from the first and last terms to reduce it to a reasonable number of possibilities. If (x-1) hadn't worked, I would have tried (x+1) next, then (2x-1) and (2x+1) because at least one of the factors has to be +/- 1

Answer 3

this is a longer way, its a method kinda oldie, first you take away the variable (just for comfort) and then you choose one factor from the last number, you multiply this factor by the first number and then you algebraicly added it to the next number, it's better to understand with numbers, on the left side is the factor choosed, you need to use the factor that takes the last addition to zero.

_ ¡ 2 -17 46 -43 12
4 ! 8 -36 40 -12
-------------------------------------
¡ 2 -9 10 -3 0

_ ! 2 -9 10 -3
3 ¡ 6 -9 3
--------------------------------
! 2 -3 1 0

_ ¡ 2 -3 1
1 ! 2 -1
-------------------------
_ ¡ 2 -1 0

Then you take the factors that worked and you substract every factor to x like this:
for the first operation you'll have: x-4
for the second: x-3
for the last: x-1
and the rest is taken when you can't do the operation anymore, and you put the x in order, in this case it would be: 2x-1 like it is on the last part of the operation, then you multiply every factor obtained doing this and you'll have:
(2x-1)(x-1)(x-3)(x-4) like it is on the answer given