Question

If f(x)=2x^3-6x^2-16x-20 and f(5)=0, then find all of the zeros of f(x) algebraically.

Answer 1

`\begin{gathered} x=5 \\ x=-1+i \\ x=-1-i \end{gathered}`

Step-by-step explanation

Step 1

factorize

`\begin{gathered} f(x)=2x^3-6x^2-16x-20 \\ f(x)=2(x^3-3x^2-8x-10) \\ \end{gathered}`

a)we know that

`\begin{gathered} f(5)=0 \\ \text{hence,}5\text{ is a zero} \end{gathered}`
`\begin{gathered} f(x)=2(x^3-3x^2-8x-10) \\ f(x)=2(x-5)(x^2+2x+2) \end{gathered}`

Step 2

factorize the trinomial, use the quadratic formula

`\begin{gathered} x^2+2x+2=0 \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ x=\frac{-2\pm\sqrt[]{2^2-4\cdot2\cdot1}}{2\cdot1} \\ x=\frac{-2\pm\sqrt[]{-4}}{2} \\ x=\frac{-2\pm\sqrt[]{-4}}{2} \\ x_1=(-2+2i)/(2)=-1+i \\ x_2=(-2-2i)/(2)=-1-i \end{gathered}`

therefore, the zeros are:

`\begin{gathered} x=5 \\ x=-1+i \\ x=-1-i \end{gathered}`

I hope this helps you