Question

find all the zerosf(x)=16x^5-72x^4+137x^3+43x^2-244x+120

Answer 1

Plot the function f(x)=16x^5-72x^4+137x^3+43x^2-244x+120 on the graph.

The graph intersect the x-axis at (-1.25,0), (0.75,0) and (1,0). So zeros of the function is x = -1.25, x = 0.75 and x = 1.

The highest degree of function is 5, we obtain three zeros from the graph remaining two zeros not lie on the graph means two zeros are imaginary zeros.

Determine the zeros of the function.

$\begin{gathered} f(x)=16x^5-72x^4+137x^3+43x^2-244x+120 \\ =(4x+5)(x-1)(4x-3)(x^2-4x+8) \end{gathered}$

For zeros,()4x +5 = 0, x - 1 =0, 4x - 3 = 0 and (x^2 -4x +8 ) = 0.

Determine the value of x for equation (x^2 -4x +8 ) = 0.

$\begin{gathered} x=\frac{+4\pm\sqrt[]{(-4)^2-4\cdot1\cdot8}}{2\cdot1} \\ =\frac{4\pm\sqrt[]{16-32}}{2} \\ =2\pm\frac{\sqrt[]{-16}}{2} \\ =2\pm2i \\ =2+2i,2-2i \end{gathered}$

So zeros of the function is,

-1.25, 0.75, 1, 2+2i and 2-2i