Question

the function $f(x) = 6x^3 - 29x^2 - 6x + 5$ has at least one rational root. use the rational root theorem to find that root, then proceed to find all complex roots. (note: roots may be integer, rational, irrational, and/or complex.)

answer attempt 1 out of 2

there is one root :
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Explanation:

Step1: Apply Rational Root Theorem

The Rational Root Theorem states that possible rational roots are $\pm\frac{p}{q}$, where $p$ is a factor of the constant term and $q$ is a factor of the leading coefficient. For $f(x) = 6x^3 - 29x^2 - 6x + 5$, constant term $p = 5$ (factors: $\pm1, \pm5$) and leading coefficient $q = 6$ (factors: $\pm1, \pm2, \pm3, \pm6$). So possible rational roots are $\pm1, \pm5, \pm\frac{1}{2}, \pm\frac{5}{2}, \pm\frac{1}{3}, \pm\frac{5}{3}, \pm\frac{1}{6}, \pm\frac{5}{6}$.

Test $x = 5$: $f(5) = 6(125) - 29(25) - 6(5) + 5 = 750 - 725 - 30 + 5 = 0$. So $x = 5$ is a root.

Step2: Factor the Polynomial

Since $x = 5$ is a root, $(x - 5)$ is a factor. Use polynomial division or synthetic division. Using synthetic division with root 5:

Coefficients: 6 | -29 | -6 | 5

Bring down 6. Multiply by 5: 30. Add to -29: 1. Multiply by 5: 5. Add to -6: -1. Multiply by 5: -5. Add to 5: 0.

So the quotient is $6x^2 + x - 1$. Factor this: $6x^2 + x - 1 = (3x - 1)(2x + 1)$.

Step3: Find All Roots

Set each factor to zero: $x - 5 = 0 \implies x = 5$; $3x - 1 = 0 \implies x = \frac{1}{3}$; $2x + 1 = 0 \implies x = -\frac{1}{2}$.

Answer:

The rational root found first is $5$, and all complex (including real) roots are $5$, $\frac{1}{3}$, and $-\frac{1}{2}$. (If only the first rational root is needed, the answer is $5$.)