Question

Please help me solve the following polynomial:

X^3+x^2+x+6=3x^2+6x

You should get the solutions X= -2,1, 3
I keep getting X=-6,1. Nowhere near what the solutions are.... please help me...

Answer 1

Make sure to subtract 3x^2+6x from both sides.
x^3 + x^2 + x + 6 − (3x^2 + 6x) = 3x^2 + 6x − (3x^2 + 6x)
x^3 − 2x^2 − 5x + 6 = 0
Then you need to factor left side of equation.
(x − 1)(x + 2)(x − 3) = 0
Finally you need to set the factors equal to zero.
x − 1 = 0 or x + 2 = 0 or x − 3 = 0
x = 1 or x = −2 or x =3 < That is how we get the 3 solutions.

Answer 2

I have a very accurate approach I think it would help you!

Answer 3

If simplified correctly, the equation becomes:
\(x^3-2x^2-5x+6=0\)
To solve a cubic (or any polynomial), it is best to start with factorization.
You decide how far to go before trying something else.
First, the leading coefficient is unity (1x^3), so all rational factors are of the form (x+a) or (x-a), that makes life easier.
Then, use Descarte's rule of signs. The coefficients have changed twice (+--+). If all factors are rational, there are two factors of the form (x-a) and one (x+a), i.e. two positive roots and one negative root.
The Factor theorem tells us that, if all roots are rational, they are 1,2,3,6, with signs going either way (remember leading coefficient is unity).
Try the easy factors, I always try (x-1)=0. If all the coefficients add to zero, then (x-1) is a factor. [ indeed the case here, 1-2-5+6=0].
Then try (x+1)=0. Add the coefficients together, but change the sign of those from odd powers. Again, if the sum is zero, (x+1) is a factor. [ -1-2+5+6=8, so not the case here].
It's up to you to try (x+2), (x-2).... but at least you might have picked up easy ones.
Once you have found the first factor, do a synthetic division to reduce the degree of the polynomial. When reduced to degree 2 [ the case here x^2-x-6], it can be easily completed by factors or quadratic equation].

Answer 4

@DivineSolar thanks, I didn't know you had to factor the entire cubic trinomial.... I tried it to where I can factor out quadratics (ax^2+bx+c)to make things simpler but it never led to the right answer.

However is there a certain method on factoring cubic trinomials or is it a trial and error kind of thing?

Answer 5

Thanks to everyone else who answered btw, this actually was my friend's homework problem and I thought I could help solve it but couldn't

Answer 6

Sure