Step 1
Find the possible zeroes of h(x)
[tex]h(x)-=x^3-5x^2+5x+3[/tex][tex]\begin{gathered} \text{Possible zeroes=}\frac{\pm1}{1},\frac{\pm3_{}}{1} \\ u\sin g\text{ factors of the constant 3 and dividing them by the leading coefficient} \end{gathered}[/tex]
Hence, the first zero will be 3 because
[tex]\begin{gathered} h(3)=3^3-5(3)^2+5(3)+3 \\ h(3)=27-45+15+3=0 \end{gathered}[/tex]
Step 2
Use the factor remainder theorem to get the quadratic equation that will give us the other zeroes.
Note; Since the highest power of the polynomial is 3, the polynomial has 3 roots
Therefore the quotient gotten after the division is;
[tex]x^2-2x-1[/tex]
Step 3
Factorize the quotient using the quadratic formula to the get the other zeroes.
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]
where from the quotient
[tex]\begin{gathered} a=1 \\ b=-2 \\ c=-1 \end{gathered}[/tex][tex]\begin{gathered} x=\frac{-(-2)\pm\sqrt[]{(-2)^2-4(1)(-1)}}{2(1)} \\ x=\frac{2\pm\sqrt[]{4+4}}{2} \\ x=\frac{2\pm\sqrt[]{8}}{2} \\ x=\frac{2+\sqrt[]{8}}{2}\text{ or }\frac{2-\sqrt[]{8}}{2} \\ x=\frac{2+2\sqrt[]{2}}{2}\text{ or }\frac{2-2\sqrt[]{2}}{2} \\ x=1+\sqrt[]{2}\text{ or 1-}\sqrt[]{2} \end{gathered}[/tex]
Hence, the zeroes are;
[tex]x=3,\text{ 1+}\sqrt[]{2},\text{ 1-}\sqrt[]{2}[/tex]
