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

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To simplify the equation of a quartic function with given zeroes, the correct approach involves using the factors derived from the zeroes. The zeroes include 5 (with multiplicity 2) and -1±2√2, leading to the factors (x-5)² and (x - (-1+2√2))(x - (-1-2√2)). The latter can be simplified using the difference of squares formula, resulting in (x+1)² - 8. The next steps involve expanding the resulting trinomials and multiplying them to obtain the quartic equation. Understanding how to apply the FOIL method and the difference of squares is crucial for completing the simplification.
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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..
 
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Nelo said:
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]

Nelo said:
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).

Nelo said:
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.
 
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wat? explain... why are you distributing random negetives into those brackets
 
Nelo said:
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)]
 
okay..? so you factor inwards...? like (x+1 -2√ 2) and (x+1 +2√ 2) ?
 
any1?
 
Nelo said:
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.
 
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)
 
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