Question

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

Answer 1

The given polynomial:

`f(x)=4x^5-48x^4_{}+169x^3-157x^2-309x-91`

Use the rational root theorem:

Leading coefficient = 4

The dividers of 4 are : 1,2,4

Last term = 91

The dividers of 91 are: 1,7,13,91

`\text{Root =}\frac{Dividers\text{ of coefficient of }last\text{ term}}{Dividers\text{ of coefficient of leading term}}`
`\text{Root}=(1,7,13,91)/(1,2,4)`

First root is 7

x = 7 is the zero of the given polnomial

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

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

Now, factorize the resulting quotient:

`4x^4-20x^3+29x^2+46x+13`

Again use rational root theorem

fisrt term 4

Factors of 4: 1,2,4

Last term 13

Factors of 13: 1,13

`\text{Root}=\pm(1,13)/(1,2,4)`

root of the function is -1/2

x = -1/2

2x+1 = 0

Divide the polynomial by 2x+1

`(4x^4-20x^3+29x^2+46x+13)/(2x+1)=2x^3-11x^2+20x+13`

Now, factorize the polynomial

`2x^3-11x^2+20x+13`

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

`(2x^3-11x^2+20x+13)/(2x+1)=x^2-6x+13`

Factorize:

`x^2-6x_{}+13`

Appy discriminant rule

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

So, the zeros are

`x=7,-(1)/(2),(-1)/(2),3+2i,3-2i`

Answer:

`x=7,-(1)/(2),(-1)/(2),3+2i,3-2i`