Question

Please look at the photo! I x = ?, ___ + √2, ___ - √2

Answer 1

Step 1

Find the possible zeroes of h(x)

`h(x)-=x^3-5x^2+5x+3`
`\begin{gathered} \text{Possible zeroes=}(\pm1)/(1),\frac{\pm3_{}}{1} \\ u\sin g\text{ factors of the constant 3 and dividing them by the leading coefficient} \end{gathered}`

Hence, the first zero will be 3 because

`\begin{gathered} h(3)=3^3-5(3)^2+5(3)+3 \\ h(3)=27-45+15+3=0 \end{gathered}`

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;

`x^2-2x-1`

Step 3

Factorize the quotient using the quadratic formula to the get the other zeroes.

`x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}`

where from the quotient

`\begin{gathered} a=1 \\ b=-2 \\ c=-1 \end{gathered}`
`\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}`

Hence, the zeroes are;

`x=3,\text{ 1+}\sqrt[]{2},\text{ 1-}\sqrt[]{2}`