Question

question: f(x) has the given zero. factor fully. f(x) = 3x⁴ + 5x³ + 81x + 135 has a zero at -5/3. fully factor. options: (3x+5)(x+3)(x²-3x+9), -3, (3x+5)(x-3)(x²+3x+9), (x+5/3)(x³+27), (3x+5)(x³-27), all of these options (except none of these options), skip this question, (3x+5)(x³+27), 135, none of these options

Explanation:

Step1: Factor out common terms

Given \( f(x) = 3x^4 + 5x^3 + 81x + 135 \). Group terms: \( (3x^4 + 5x^3) + (81x + 135) \). Factor \( x^3 \) from first group and 27 from second: \( x^3(3x + 5) + 27(3x + 5) \).

Step2: Factor out \( (3x + 5) \)

Now, factor out \( (3x + 5) \) from both terms: \( (3x + 5)(x^3 + 27) \).

Step3: Factor \( x^3 + 27 \) (sum of cubes)

Use sum of cubes formula \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \), where \( a = x \), \( b = 3 \). So \( x^3 + 27 = (x + 3)(x^2 - 3x + 9) \).

Step4: Combine factors

Substitute back: \( f(x) = (3x + 5)(x + 3)(x^2 - 3x + 9) \).

Answer:

\((3x + 5)(x + 3)(x^2 - 3x + 9)\) (corresponding to the option \((3x + 5)(x + 3)(x^{2}-3x + 9)\))