Question

question: if $f(x)=2x^{3}+6x^{2}+2x - 10$ and $f(1)=0$, then find all of the zeros of $f(x)$ algebraically. score: 0/4 penalty: 1 off synthetic division to find zeros (lev 2)

Explanation:

Step1: Use synthetic division

Since \(f(1) = 0\), we use synthetic - division to divide \(f(x)=2x^{3}+6x^{2}+2x - 10\) by \((x - 1)\). The coefficients are \(2,6,2,-10\).

1	2	6	2	- 10
	2	8	10	0

The quotient is \(2x^{2}+8x + 10\).

Step2: Set the quotient equal to zero

Set \(2x^{2}+8x + 10=0\). Divide through by 2 to simplify: \(x^{2}+4x + 5 = 0\).

Step3: Use the quadratic formula

For a quadratic equation \(ax^{2}+bx + c = 0\) (\(a = 1\), \(b = 4\), \(c = 5\)), the quadratic formula is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).
\[x=\frac{-4\pm\sqrt{4^{2}-4\times1\times5}}{2\times1}=\frac{-4\pm\sqrt{16 - 20}}{2}=\frac{-4\pm\sqrt{-4}}{2}=\frac{-4\pm2i}{2}=-2\pm i\]
We already know \(x = 1\) is a zero.

Answer:

\(x = 1,x=-2 + i,x=-2 - i\)