just checking right now?

Question

P(x)=2x^3-5x^2-4x+3

find that complete factorization of P

i want to make sure if i did it right please answer it so i can check my answer with yours to see if i did it right please

Answer 1

x=-1 satisfies the equation, so (x+1) is a factor. Divide P(x) by (x+1). You'll get something like ax^2+bx+c and factor it.
The final answer will be (x+1)(x+a)(x+b).
You can also check you answer this way yourself.
If your factor has say, (x+3) , then in the equation 2x^3-5x^2-4x+3 substitute x=-3 and see if it reduces to 0.
I said (x+1) is a factor. let x=-1
2(-1)^3-5(-1)^2-4(-1)+3
=-2-5+4+3=0

Answer 2

P(x) = 2x^3 - 5x^2 - 4x + 3

NB Use the Factor Theorem, if P(a) = 0 then (x - a) is a factor of the trinomial.
In this case
P(-1) = 2(-1)^3 - 5(-1)^2 - 4(-1) +3
P(-1) = -2 - 5 + 4 + 3 = 0
therefore x - - 1 is a factor : (x + 1)
NNB You have try and error a few numbers (a) , e.g. -2, -1, 0, 1 , 2 etc until you reach P(a) = 0

Then divide through by x+1
x+ 1 ) 2x^3 - 5x^2 - 4x + 3 ( 2x^2 - 7x + 3
2x^3 +2x^2 & subtract
__________
0 - 7x^2 -4x
- 7x^2-7x & subtract
____________
0 + 3x
+ 3x + 3 & subtract
________
0

So 2x^2 - 7x + 3 is another factor .
(2x - 1)(x - 3)
Combining all together we have
(x +1)(2x - 1)(x - 3) the COMPLETE factorization.