# For given P(X) solve P(x)=0

helpmedude

## Homework Statement

For the polynomial function P(x)=x^4-3x^3-3x^2+7x+6 solve P(x)=0

## The Attempt at a Solution

I am not really sure how to break this down and factor it should I break it down into a trinomial and binomial?

rock.freak667
Homework Helper
try looking for factors, when factorised in the form (x-a)(x-b)(x-c)(x-d)=0

the last number , 6 in your equation is the producr abcd. Meaning that to find a linear factor try putting x equal to the factors of 6

You can solve it, with grouping the elements of the polynomial function:

$$x^4+x-3x^3-3x^2+6x+6=0$$

$$x(x^3+1)-3x^2(x+1)+6(x+1)=0$$.

I think I helped you enough. I think you can continue out from here.

helpmedude
yes your approach is very logical and so the answer appears to be x=0 with a multiplicity of 2 and x=-1 with a multiplicity of 3 ? but i checked the function using a graphing calculator and the zeros are -1 2 and 3?

Homework Helper
yes your approach is very logical and so the answer appears to be x=0 with a multiplicity of 2 and x=-1 with a multiplicity of 3 ? but i checked the function using a graphing calculator and the zeros are -1 2 and 3?

$$x = 0$$ isn't a solution. Look at

$$x\left(x^3 + 1 \right) - 3x^2 \left(x+1\right) + 6\left(x+1\right) = 0$$

When you substitute 0 into this the left side equals 6.

Try factoring

$$x^3 + 1$$

(sum of two cubes) and see what this does to the form of the equation. I believe this is the path the previous poster intended you to take.

$$x = 0$$ isn't a solution. Look at

$$x\left(x^3 + 1 \right) - 3x^2 \left(x+1\right) + 6\left(x+1\right) = 0$$

When you substitute 0 into this the left side equals 6.

Try factoring

$$x^3 + 1$$

(sum of two cubes) and see what this does to the form of the equation. I believe this is the path the previous poster intended you to take.

Yes, that will be my next step. Try factoring x3+13, or if you don't know how, use the formula (A+B)(A2-AB+B2)=A3+B3 which will give you ... HallsofIvy
Homework Helper
Дьявол's formula makes it very clear that x=-1, not 0, is a root. Knowing that, you can divide the polynomial by x+1 to get a cubic equation which might be simpler.

One thing you can do is use the "rational root theorem" to argue that if there is a rational root, then it must be an integer divisor of 6: 1, -1, 2, -2, 3, -3, 6, or -6.
14- 3(13)- 3(12)+ 7(1)+ 6= 1- 3- 3+ 7+ 6= 8.

(-1)4- 3((-1)3)- 3((-1)2)+ 7(-1)+ 6= 1+ 3- 3- 7+ 6= 0
so -1 is a root- but we knew that thanks to Дьявол.

24- 3(23)- 3(22)+ 7(2)+ 6= 16- 24- 12+ 14+ 6= 0
so 2 is also a root. At this point it might be easier to go ahead and divide the polynomial by x-(-1) and x- 2 to see that x4- 3x3- 3x2+ 7x+ 6= (x+1)(x-2)(x2- 2x+ 3). x2- 2x+ 3= 0 can be solved by completing the square or using the quadratic formula.