# Determine an equation of p(x)

1. Sep 24, 2011

### Nelo

1. The problem statement, all variables and given/known data

Determine the equation in simplied form for the family of quartic functions with zeroes of..

5 (order 2) and -1± 2√ 2

2. Relevant equations

3. The attempt at a solution

so.. (x-5)^2 (x-1+2√2) (x-1-2√2)

(x-5) (x-5) (x-1+2√2) (x-1-2√2)

Would be all of it expanded, but I dont know how to foil on this.. theres 2 terms on the left side brackets and 3 terms on the right side brackets.. (x) (-1) (2√2) , I know that X will become x^4, but i dont get how to continue using foil, can anyone explain?

We never learned this.. so.. i also know that the plusminus root is (x+1)^2 + 8 if simplified into that form..

2. Sep 24, 2011

### eumyang

This is wrong. It should be
(x - 5)^2 [x - (-1+2√2)][x - (-1-2√2)]
= (x - 5)(x - 5)[x + 1 - 2√2][x + 1 + 2√2]

FOIL the first two binomials, and then FOIL the last two binomials. In FOILing the last two binomials, it may be helpful to rewrite like this:
[x + 1 - 2√2][x + 1 + 2√2]
= [(x + 1) - 2√2][(x + 1) + 2√2]
.. and then use the sum+difference pattern (a - b)(a + b) = a2 - b2.

Then, multiply the two resulting trinomials (see http://www.purplemath.com/modules/polymult3.htm" [Broken] if you don't know how).

This is also wrong (probably because you were missing signs earlier. It should be
(x+1)^2 - 8.

Last edited by a moderator: May 5, 2017
3. Sep 24, 2011

### Nelo

wat? explain... why are you distributing random negetives into those brackets

4. Sep 24, 2011

### eumyang

Because if a is a root of a polynomial equation f(x) = 0, then
(x - a)
is a factor of that polynomial. So there is a negative outside the roots -1+ 2√ 2 and -1 - 2√ 2:
(x - 5)(x - 5)[x - (-1 + 2√ 2)][x - (-1 - 2√ 2)]

5. Sep 24, 2011

### Nelo

okay..? so you factor inwards...? like (x+1 -2√ 2) and (x+1 +2√ 2) ?

6. Sep 24, 2011

### Nelo

any1?

7. Sep 24, 2011

### SammyS

Staff Emeritus
What do you mean by "factor inwards" ?

Also, eumyang basically gave you the next step.

8. Sep 24, 2011

### HallsofIvy

The simplest thing to do with complex conjugate terms is to use $(a- b)(a+ b)= a^2- b^2$.

If $-1+ 2\sqrt{2}$ and $-1- 2\sqrt{2}$ then $(x- (-1+2\sqrt{2})$ and $(x- (-1-2\sqrt{2}))$ are factors.

$$((x- 1)- 2\sqrt{2})((x-1)+ 2\sqrt{2})= (x- 1)^2- (2\sqrt{2})^2= x^2- 2x+ 1- 8= x^2- 2x- 7$$.

It shouldn't be too difficult to multiply $(x^2- 10x+ 5)(x^2- 2x- 7)$