Question

section 3.8
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question 5
find all zeros of f(x) = x⁵ + 3x³ - 2. enter the zeros separated by commas. enter exact values, not decimal approximations.
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Explanation:

Step1: Set \( f(x) = 0 \)

We have the function \( f(x)=x^{3}+3x^{2}-2 \), so we set \( x^{3}+3x^{2}-2 = 0 \). Let's try to find rational roots first using the Rational Root Theorem. The possible rational roots are factors of the constant term (2) divided by factors of the leading coefficient (1), so possible roots are \( \pm1, \pm2 \).

Step2: Test \( x = 1 \)

Substitute \( x = 1 \) into \( f(x) \): \( f(1)=1^{3}+3(1)^{2}-2=1 + 3-2=2
eq0 \).

Step3: Test \( x=-1 \)

Substitute \( x=-1 \): \( f(-1)=(-1)^{3}+3(-1)^{2}-2=-1 + 3-2=0 \). So \( x=-1 \) is a root.

Step4: Factor the polynomial

Since \( x = -1 \) is a root, \( (x + 1) \) is a factor. We perform polynomial division or use synthetic division to factor \( x^{3}+3x^{2}-2 \) by \( (x + 1) \). Using synthetic division with root -1:

$$
LATEXBLOCK0
$$

So the polynomial factors as \( (x + 1)(x^{2}+2x - 2) \).

Step5: Solve the quadratic equation

Now we solve \( x^{2}+2x - 2=0 \) using the quadratic formula \( x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \), where \( a = 1 \), \( b = 2 \), \( c=-2 \).

$$ x=\frac{-2\pm\sqrt{2^{2}-4(1)(-2)}}{2(1)}=\frac{-2\pm\sqrt{4 + 8}}{2}=\frac{-2\pm\sqrt{12}}{2}=\frac{-2\pm2\sqrt{3}}{2}=-1\pm\sqrt{3} $$

Answer:

\( -1, -1+\sqrt{3}, -1-\sqrt{3} \)