Answer:
[tex]\displaystyle x = 1, - 2 + i, -2 -i[/tex]
Step-by-step explanation:
We are given that:
[tex]\displaystyle f(x) = 2x^3 + 6x^2 + 2x - 10 \text{ and } f(1) = 0[/tex]
And we want to find all zeros of f algebraically.
Since f(1) = 0, then (x - 1) is a zero of f.
We can hence factor by synthetic division:
[tex]\begin{table}[]\begin{tabular}{lllll}\multicolumn{1}{l|}{1} & 2 & 6 & 2 & -10 \\\multicolumn{1}{l|}{} & & 6 & 2 & 10 \\ \cline{2-5} & 2 & 8 & 10 & -10 \\ & & & & \end{tabular}\end{table}[/tex][tex]\begin{tabular}{lllll}\multicolumn{1}{l|}{1} & 2 & 6 & 2 & -10 \\\multicolumn{1}{l|}{} & & 2 & 8 & 10 \\ \cline{2-5} & 2 & 8 & 10 & 0 \\ & & & & \end{tabular}[/tex]
Therefore:
[tex]\displaystyle \begin{aligned} f(x) & = 2x^3 + 6x^2 + 2x- 10 \\ \\ & = (x-1)(2x^2 + 8x + 10) \\ \\ &= 2(x-1)(x^2 + 4x + 5)\end{aligned}[/tex]
Therefore, by the Zero Product Property:
[tex]\displaystyle x-1 = 0 \text{ or } x^2 +4x + 5 = 0[/tex]
From the quadratic formula:
[tex]\displaystyle \begin{aligned}x & = \frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ \\ & = \frac{-(4)\pm\sqrt{(4)^2 - 4(1)(5)}}{2(1)} \\ \\ &= \frac{-4\pm\sqrt{-4}}{2} \\ \\ &= \frac{-4\pm 2i}{2} \\ \\ & = -2\pm i \end{aligned}[/tex]
Therefore, all the zeros of f include:
[tex]\displaystyle x = 1, - 2 + i, -2 -i[/tex]
