Question

Question #4a1e9

Answer 1

`y=-1/272(x+5)(x^2+4x+13)`

Explanation:

All cubic equations can be written in the form `y=a(x-r_1)(x-r_2)(x-r_3)`, where `a` is a constant and `r_1,r_2,r_3` are roots of the equation.

Since we know the roots, we can write the cubic equation as `y=a(x+5)(x+2+3i)(x+2-3i)` (the complex roots of all real polynomials occur in conjugate pairs; thus since one of the roots is `-2-3i`, another root is `-2+3i`).

Simplifying this gives `y=a(x+5)(x^2+4x+13)`.

It is given that `y(3)=-1`, then `-1=a(3+5)((3)^2+4*3+13)=272a`, or `a=-1/272`.

The final answer is `y=-1/272(x+5)(x^2+4x+13)`.

Answer 2

Equation for `f(x)`:
`f(x) = -1/112(x+5)(x^2+5)`

Explanation:

Because one of the zeros is irrational in a way that it includes `i`, there must be another zero that includes the conjugate of that term.

`f(x) = a(x+5)(x-(-3i-2))(x+(-3i-2))`

`f(x) = a(x+5)(x+3i+2)(x-3i-2)`

`f(x) = a(x+5)(x^2 cancel(-3ix)cancel(-2x) cancel(+3ix)+9 cancel(-6i) cancel(+2x) cancel(-6i) -4)`

`f(x) = a(x+5)(x^2+5)`

Now use the given to find `a`: `f(3)=-1`
`-1 = a(3+5)(3^2+5)`

`-1=8*14*a`

`a=-1/(112)`

Equation for `f(x)`:
`f(x) = -1/112(x+5)(x^2+5)`