How do you simplify (2x^3 -3x -1)?

Question

How would you factor that into two polynomials?
like (x^2+2x+1) = (x+1)(x+1) (obviously this is simpler!) thanks for help...

thanks for the help, but I still don't understand HOW. How do you know that (x+1) is a root? how do you determine that... and then how do you determine the next root? ... thanks again.

Answer 1

f(x)=2x^3-3x-1
f(-1)=-2+3-1=0
x+1 is a factor
dividing by x+1
2 0 -3 -1
0 -2 2 1
-------------
2x^2-2x-1
so the factors are
(x+1)(2x^2-2x-1)

Answer 2

Because the expression has -1 as the constant, the factors must be of the form (A - 1)(B + 1), where A or B might represent more than one term.

It's probably wise to try the simple approach first, and consider that x may be an integer.
In that case, we need only try the factors (x - 1) and (x + 1), that is, try x = 1 and x = -1, and hopefully, you'll get lucky.

So, substituting x = 1 into the expression gives: 2*1^3 - 3*1 - 1 = -2, so that doesn't work.

Substituting x = -1 gives : 2*(-1)^3 - 3*(-1) - 1 = 0, which does work, so therefore, one of the factors is (x + 1).

It's a good idea to try all the variations, to get as many integer solutions as possible. If one works, don't stop; there may be more.

So now we know that :
2x^3 - 3x - 1 = (x + 1)(2x^2 + Px - 1)

We know that 2x^2 and -1 are correct in the second factor, but we don't know the value of P.
So multiply out the RHS and gather like terms. This will give you :
2x^3 - 3x - 1 = 2x^3 + (P + 2)x^2 + (P - 1)x - 1

There is no x^2 term on the LHS, so P + 2 = 0, that is, P = -2.
Likewise with the terms in x, we have : -3 = P - 1, or, P = -2, so they correlate.
Thus we find that : 2x^3 - 3x - 1 = (x + 1)(2x^2 - 2x - 1)

You could also do this by dividing (x + 1) directly into 2x^3 - 3x - 1.

We've exhausted all the integer values for x, so now things get a bit tougher.
And when that happens, it's good to find that the last factor is a quadratic expression.

So turn it into a quadratic equation : 2x^2 - 2x - 1 = 0.

Using the quadratic formula, we find that x = [1 ± sqrt(3)] / 2.

Thus, 2x^3 - 3x - 1 = (x + 1){x - [1 - sqrt(3)] / 2}{x - [1 + sqrt(3)] / 2}.

Answer 3

(x+1)(2x^2-2x-1)

There is a trick where you focus on p & q's, but it has been many years since I have done that.

For this particular case you have "-1" and there are only so many combinations that can multiple to get -1.

Answer 4

we can see one of the roots is x+1
so i simplified it max to (2x^2-2x-1)(x+1)
in root terms it can be further simplified to
(x-(1+sqrt3)/2)(x-(1-sqrt3)/2)(x+1)
hope this helps

Answer 5

f(x)=2x^3-3x-1
f(-1)=-2+3-1=0
x+1 is a factor
dividing by x+1
2 0 -3 -1
0 -2 2 1
-------------
2x^2-2x-1
THUS
(x+1)(2x^2-2x-1)

Answer 6

well, that kinda cubic equation cannot always be solved.
the only thing to do is to use trial error method to find a root, and then find the other two roots with the help of quadratic formula.

Answer 7

no homework help here, go back to your math book and figure it out....