Question

section 3.5
score: 1/6 answered: 1/6
question 4
find all zeros of f(x) = x³ + 3x² + x - 1. enter the zeros separated by commas. enter exact value, not decimal approximations.
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Explanation:

Step1: Try Rational Roots

By Rational Root Theorem, possible roots are $\pm1$. Test $x = 1$: $f(1)=1 + 3+1 - 1 = 4
eq0$. Test $x=-1$: $f(-1)=-1 + 3 - 1 - 1 = 0$. So $x=-1$ is a root.

Step2: Factor the Polynomial

Divide $f(x)$ by $(x + 1)$ using polynomial long division or synthetic division. Using synthetic division for $x=-1$:
Coefficients: 1 (x³), 3 (x²), 1 (x), -1 (constant)
Bring down 1. Multiply by -1: -1. Add to 3: 2. Multiply by -1: -2. Add to 1: -1. Multiply by -1: 1. Add to -1: 0.
So quotient is $x² + 2x - 1$. Now solve $x² + 2x - 1 = 0$ using quadratic formula $x=\frac{-b\pm\sqrt{b² - 4ac}}{2a}$, where $a = 1$, $b = 2$, $c=-1$.
$x=\frac{-2\pm\sqrt{4 + 4}}{2}=\frac{-2\pm\sqrt{8}}{2}=\frac{-2\pm2\sqrt{2}}{2}=-1\pm\sqrt{2}$.

Step3: List All Zeros

The zeros are $x=-1$, $x=-1+\sqrt{2}$, $x=-1-\sqrt{2}$.

Answer:

-1, -1 + √2, -1 - √2