Question

question 8

factor the polynomial function (f(x) = 6x^3 + 17x^2 - 3x - 20).

the zeros are

(f(x) = )

Explanation:

⚡ Using: dividing polynomials · 🆕 New: Factoring Higher-Degree Polynomials
Finding rational roots to break down polynomials

Step 1: Find a rational zero

To find the first zero, we use the Rational Root Theorem. The possible rational roots are factors of the constant term \(-20\) divided by factors of the leading coefficient \(6\).

Let's test small integer values in \(f(x) = 6x^3 + 17x^2 - 3x - 20\):

• Test \(x = 1\):

$$ f(1) = 6(1)^3 + 17(1)^2 - 3(1) - 20 = 6 + 17 - 3 - 20 = 0 $$

Since \(f(1) = 0\), \(x = 1\) is a zero, which means \((x - 1)\) is a factor.

Step 2: Divide the polynomial

We divide \(f(x)\) by \((x - 1)\) using synthetic division to find the remaining quadratic factor:

$$
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$$

The quotient is the quadratic polynomial:

$$ 6x^2 + 23x + 20 $$

Step 3: Factor the quadratic quotient

Now we factor \(6x^2 + 23x + 20\) by grouping. We look for two numbers that multiply to \(6 \times 20 = 120\) and add to \(23\). These numbers are \(8\) and \(15\):

$$ 6x^2 + 8x + 15x + 20 $$

Group the terms:

$$ 2x(3x + 4) + 5(3x + 4) = (2x + 5)(3x + 4) $$

Step 4: Identify all zeros

Set each factor to zero to find the remaining roots:

• \(x - 1 = 0 \implies x = 1\)
• \(2x + 5 = 0 \implies x = -\frac{5}{2}\)
• \(3x + 4 = 0 \implies x = -\frac{4}{3}\)

Answer:

• The zeros are: \(1, -\frac{5}{2}, -\frac{4}{3}\)
• \(f(x) =\) \((x - 1)(2x + 5)(3x + 4)\)