Question

Find a cubic function with the given zeros.

Answer 1

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

Answer 2

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

Answer 3

??
@mathstudent55

Answer 4

ugh im so confused

Answer 5

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.

Answer 6

f(x)=(x−√2)(x+√2)(x+2)
f(x)=2-2+2?

Answer 7

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

Answer 8

@Hero i need help too please

Answer 9

@mathstudent55

Answer 10

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

Answer 11

\(f(x) = \color{red}{(x - \sqrt{2})(x +\sqrt{2})}(x + 2)\)

becomes

\(f(x) = \color{red}{(x^2 - 2)}(x + 2)\)

Ok so far?

Answer 12

ok

Answer 13

Now we use FOIL for the two binomials above:

\(f(x) = (x^2 - 2)(x + 2)\)

\(f(x) = x^3 + 2x^2 - 2x - 4\)

Answer 14

ohhh
ok

Answer 15

so i do like

Answer 16

x^2*x
and so on