How Do You Simplify the Equation of a Quartic Function with Given Zeroes?

[Nelo]

Homework Statement

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

5 (order 2) and -1± 2√ 2

Homework Equations

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 don't know how to foil on this.. there's 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 don't 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..

 

[eumyang]

so.. (x-5)^2 (x-1+2√2) (x-1-2√2)

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]

Would be all of it expanded, but I don't know how to foil on this..

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" if you don't know how).

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

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

 

[Nelo]

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

 

[eumyang]

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

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)]

 

[Nelo]

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

 

[Nelo]

any1?

 

[SammyS]

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

What do you mean by "factor inwards" ?

Also, eumyang basically gave you the next step.

 

[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 $(x^2- 10x+ 5)(x^2- 2x- 7)$