anonymous
  • anonymous
Find a cubic function with the given zeros.
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
mathstudent55
  • mathstudent55
A polynomial equation with roots a, b, c is f(x) = (x - a)(x - b)(x - c)
anonymous
  • anonymous
f(x) = (x - a)(x - b)(x - c)\[f(x) = (x - \sqrt{2})(x - \sqrt{-2})(x - (-2))\]

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anonymous
  • anonymous
anonymous
  • anonymous
ugh im so confused
mathstudent55
  • mathstudent55
You are close, but a little off. This is what you should have: f(x) = (x - a)(x - b)(x - c) \(f(x) = (x - \sqrt{2})(x - (-\sqrt{2}))(x - (-2))\) \(f(x) = (x - \sqrt{2})(x +\sqrt{2})(x + 2)\) Now multiply out the three binomials. Hint: Make sure to multiply the first two binomials first because since they are the product of a sum and a difference, you can use the short cut: \((a + b)(a - b) = a^2 - b^2\) instead of having to use FOIL.
anonymous
  • anonymous
f(x)=(x−√2)(x+√2)(x+2) f(x)=2-2+2?
anonymous
  • anonymous
its either A f(x) = x^3 + 2x^2 - 2x + 4 B f(x) = x^3 + 2x^2 + 2x - 4 C f(x) = x^3 - 2x^2 - 2x - 4 D f(x) = x^3 + 2x^2 - 2x - 4
anonymous
  • anonymous
@Hero i need help too please
anonymous
  • anonymous
anonymous
  • anonymous
(a + b^)3 = (a + b)(a^2 + 2ab + b^2) = a^3 + 3a^2b + 3ab^2 + b^3??
mathstudent55
  • mathstudent55
\(f(x) = \color{red}{(x - \sqrt{2})(x +\sqrt{2})}(x + 2)\) becomes \(f(x) = \color{red}{(x^2 - 2)}(x + 2)\) Ok so far?
anonymous
  • anonymous
ok
mathstudent55
  • mathstudent55
Now we use FOIL for the two binomials above: \(f(x) = (x^2 - 2)(x + 2)\) \(f(x) = x^3 + 2x^2 - 2x - 4\)
anonymous
  • anonymous
ohhh ok
anonymous
  • anonymous
so i do like
anonymous
  • anonymous
x^2*x and so on

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