Question

section 4.5 find all the real zeros of

1. ( x^4 - 9x^2 + 4x + 12 = 0 )
2. ( x^3 - 5x^2 - 4x + 20 = 0 )

Explanation:

For Problem 20: \(x^4 - 9x^2 + 4x + 12 = 0\)

Step 1: Try Rational Roots

Use Rational Root Theorem. Possible roots: \(\pm1, \pm2, \pm3, \pm4, \pm6, \pm12\). Test \(x = -1\): \((-1)^4 - 9(-1)^2 + 4(-1) + 12 = 1 - 9 - 4 + 12 = 0\). So \(x + 1\) is a factor.

Step 2: Polynomial Division

Divide \(x^4 - 9x^2 + 4x + 12\) by \(x + 1\) (using synthetic division or long division). Result: \(x^3 - x^2 - 8x + 12\).

Step 3: Factor Cubic

Test \(x = 2\) in \(x^3 - x^2 - 8x + 12\): \(8 - 4 - 16 + 12 = 0\). So \(x - 2\) is a factor. Divide cubic by \(x - 2\): \(x^2 + x - 6\).

Step 4: Factor Quadratic

\(x^2 + x - 6 = (x + 3)(x - 2)\). So the polynomial factors as \((x + 1)(x - 2)^2(x + 3)\).

Step 5: Find Zeros

Set each factor to zero: \(x + 1 = 0 \Rightarrow x = -1\); \(x - 2 = 0 \Rightarrow x = 2\); \(x + 3 = 0 \Rightarrow x = -3\).

For Problem 21: \(x^3 - 5x^2 - 4x + 20 = 0\)

Step 1: Factor by Grouping

Group terms: \((x^3 - 5x^2) + (-4x + 20) = x^2(x - 5) - 4(x - 5) = (x^2 - 4)(x - 5)\).

Step 2: Factor Difference of Squares

\(x^2 - 4 = (x + 2)(x - 2)\). So the polynomial is \((x + 2)(x - 2)(x - 5)\).

Step 3: Find Zeros

Set each factor to zero: \(x + 2 = 0 \Rightarrow x = -2\); \(x - 2 = 0 \Rightarrow x = 2\); \(x - 5 = 0 \Rightarrow x = 5\).

Answer:

s:

1. Real zeros: \(x = -3\), \(x = -1\), \(x = 2\) (with multiplicity 2)
2. Real zeros: \(x = -2\), \(x = 2\), \(x = 5\)