Find all the zeros in the polynomial functionf(x)=4x^5-48x^4+169x^3-157x^2-309x-91

[tex]\text{Root =}\frac{Dividers\text{ of coefficient of }last\text{ term}}{Dividers\text{ of coefficient of leading term}}[/tex][tex]\text{Root}=\frac{1,7,13,91}{1,2,4}[/tex]

First root is 7

x = 7 is the zero of the given polnomial

Divide f(x) by (x - 7) by long division method

[tex]\begin{gathered} \frac{f(x)}{x-7}=\frac{4x^5-48x^4_{}+169x^3-157x^2-309x-91}{x-7} \\ \frac{f(x)}{x-7}=4x^4-20x^3+29x^2+46x+13 \end{gathered}[/tex]

Now, factorize the resulting quotient:

[tex]4x^4-20x^3+29x^2+46x+13[/tex]

Again use rational root theorem

fisrt term 4

Factors of 4: 1,2,4

Last term 13

Factors of 13: 1,13

[tex]\text{Root}=\pm\frac{1,13}{1,2,4}[/tex]

root of the function is -1/2

x = -1/2

2x+1 = 0

Divide the polynomial by 2x+1

[tex]\frac{4x^4-20x^3+29x^2+46x+13}{2x+1}=2x^3-11x^2+20x+13[/tex]

Now, factorize the polynomial

[tex]2x^3-11x^2+20x+13[/tex]

Leading term coefficient = 2

Factor of 2: 1,2'

Last term coefficient = 13

Factor of 13 = 1,13

Root = -1/2

x = -1/2

2x+1

Divide the polynomial by 2x+1

[tex]\frac{2x^3-11x^2+20x+13}{2x+1}=x^2-6x+13[/tex]

Factorize:

[tex]x^2-6x_{}+13[/tex]

Appy discriminant rule

[tex]\begin{gathered} x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ x=\frac{6\pm\sqrt{36-4(1)(13}_{}}{2(1)} \\ x=3+2i,3-2i \end{gathered}[/tex]

So, the zeros are

[tex]x=7,-\frac{1}{2},\frac{-1}{2},3+2i,3-2i[/tex]

Answer:

[tex]x=7,-\frac{1}{2},\frac{-1}{2},3+2i,3-2i[/tex]