- #1

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?

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- Thread starter helpmedude
- Start date

- #1

helpmedude

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

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

- #2

rock.freak667

Homework Helper

- 6,223

- 31

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

- #3

- 365

- 0

[tex]x^4+x-3x^3-3x^2+6x+6=0[/tex]

[tex]x(x^3+1)-3x^2(x+1)+6(x+1)=0[/tex].

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

- #4

helpmedude

- #5

statdad

Homework Helper

- 1,495

- 36

[tex] x = 0 [/tex] isn't a solution. Look at

[tex]

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

[/tex]

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

Try factoring

[tex]

x^3 + 1

[/tex]

(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.

- #6

- 365

- 0

[tex] x = 0 [/tex] isn't a solution. Look at

[tex]

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

[/tex]

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

Try factoring

[tex]

x^3 + 1

[/tex]

(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 x

- #7

HallsofIvy

Science Advisor

Homework Helper

- 41,847

- 966

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.

1

(-1)

so -1 is a root- but we knew that thanks to Дьявол.

2

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 x

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