Question

finding all the zeroes of a polynomial factorand please explain1) f(x) =x^4-9x^3+32x^2-10x-522) f(x)=4x^5-48x^4+169x^3-157x^2-309x-91

Answer 1

we have:

`\begin{gathered} f\mleft(x\mright)\: =x^4-9x^3+32x^2-10x-52 \\ x^4-9x^3+32x^2-10x-52=0 \\ then\text{ x intercept }in\text{ (-1,0) and }(2,0) \end{gathered}`

answer: zeros in x = -1 and x = 2

`\begin{gathered} f\mleft(x\mright)=4x^5-48x^4+169x^3-157x^2-309x-91 \\ 4x^5-48x^4+169x^3-157x^2-309x-91=0 \\ then\text{ x intercept in (}7,0)\text{ and (}-(1)/(2),0) \end{gathered}`

answer: zeros in x = 7 and x = -1/2