## anonymous one year ago Find a cubic function with the given zeros.

1. anonymous

2. mathstudent55

A polynomial equation with roots a, b, c is f(x) = (x - a)(x - b)(x - c)

3. anonymous

f(x) = (x - a)(x - b)(x - c)$f(x) = (x - \sqrt{2})(x - \sqrt{-2})(x - (-2))$

4. anonymous

?? @mathstudent55

5. anonymous

ugh im so confused

6. 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.

7. anonymous

f(x)=(x−√2)(x+√2)(x+2) f(x)=2-2+2?

8. 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

9. anonymous

@Hero i need help too please

10. anonymous

@mathstudent55

11. anonymous

(a + b^)3 = (a + b)(a^2 + 2ab + b^2) = a^3 + 3a^2b + 3ab^2 + b^3??

12. 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?

13. anonymous

ok

14. mathstudent55

Now we use FOIL for the two binomials above: $$f(x) = (x^2 - 2)(x + 2)$$ $$f(x) = x^3 + 2x^2 - 2x - 4$$

15. anonymous

ohhh ok

16. anonymous

so i do like

17. anonymous

x^2*x and so on