Question

Please help with question 7 I have included the graph for it

Answer 1

The given polynomial is

`x^4-6x^3+25x^2-96x+144`

Using Desmos online graphing calculator, the graph of the function is shown below

From the graph, a zero of the polynomial occurs at the point (3,0)

Hence x = 3 is a zero of the polynomial

Hence x = 3

This implies that x - 3 is a factor of the polynomial

Using synthetic division to verify the zeros

Perform the division

`(x^4-6x^3+25x^2-96x+144)/(x-3)`

The division is shown below

Since the remainder of the polynomial is 0, hence 3 is a zero of the polynomial

Factoring the polynomial gives

Since x - 3 is a factor then

`x^4-6x^3+25x^2-96x+144=\mleft(x-3\mright)(x^4-6x^3+25x^2-96x+144)/(x-3)`

From the synthetic division

`(x^4-6x^3+25x^2-96x+144)/(x-3)=x^3-3x^2+16x-48`

It follows

`x^4-6x^3+25x^2-96x+144=(x-3)(x^3-3x^2+16x-48)`

Factoring

`x^3-3x^2+16x-48`

gives

`\mleft(x-3\mright)\mleft(x^2+16\mright)`

It follows

`x^4-6x^3+25x^2-96x+144=(x-3)(x-3)(x^2+16)`

Hence the imaginary zeros occur at

`(x^2+16)`

Equate to zero and solve

`\begin{gathered} x^2+16=0 \\ \Rightarrow x^2=-16 \\ \Rightarrow x=\pm\sqrt[]{-16}_{} \\ x=\pm4i \end{gathered}`

Therefore, the imaginary zeros are 4i and -4i