Question

I am not sure how to do this, if you can help me? Use the given zero to find the remaining zeros of the function.

`h(x)=4x^4+5x^3+94x^2+125x-150`
zero: 5i

Answer 1

The zeroes are `3/4,-2,5i` and `-5i`

Explanation:

We have the polynomial `4x^4+5x^3+94x^2+125x-150=0`.

We know that `5i` is a zero of the polynomial. It follows that `(x-5i)` is a factor of the polynomial.

The Conjugate Pairs Theorem states that if `(a+bi)` is a factor of a polynomial, then `(a-bi)` is also a factor.

Therefore, `(x+5i)` is also a factor of the polynomial.

Since the polynomial is of degree `4`, we have `4` roots. We've found `2`.

Now, we can multiply `(x-5i)` and `(x+5i)`.

`(x-5i)(x+5i)`

`x^2-(5i)^2`

`x^2-(-25)`

`x^2+25`

We can divide `4x^4+5x^3+94x^2+125x-150` by `x^2+25`.

Do not use synthetic division, only long division!

`color(white)(x^2++0x25|i)4x^2+5x-6`
`x^2+0x+25|4x^4+5x^3+94x^2+125x-150`
`color(white)(x^2+0x+25|i)4x^4+0x^3+100x^2`
`color(white)(x^2+0x+25|h)0x^4+5x^3-6x^2+125x`
`color(white)(x^2+0x+25|h)0x^4+5x^3+0x^2+125x`
`color(white)(x^2+0x+25|h)0x^4+0x^3-6x^2+0x-150`
`color(white)(x^2+0x+25|h)0x^4+0x^3-6x^2+0x-150`
`color(white)(x^2+0x+25|h)0`

So we have another factor:

`4x^2+5x-6`

This we can factor again:

We need to multiply the first and last terms: `4*-6=-24`

We need to find two numbers which multiply to give `-24`, and add to give `5`.

`8` and `-3` will work.

`4x^2+8x-3x-6`

`4x(x+2)-3(x+2)`

`(4x-3)(x+2)`

So now, overall, we have `(4x-3)(x+2)(x-5i)(x+5i)=0`

The zeroes are `3/4,-2,5i` and `-5i`