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

Please help with question 7 I have included the graph for it-example-1
Please help with question 7 I have included the graph for it-example-1
Please help with question 7 I have included the graph for it-example-2

1 Answer

7 votes

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

Please help with question 7 I have included the graph for it-example-1
Please help with question 7 I have included the graph for it-example-2
User Dmcnelis
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