Question

the function $f(x) = 4x^3 - 7x^2 - 34x - 8$ 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 :
\boxed{}

Explanation:

Step1: Apply Rational Root Theorem

The Rational Root Theorem states that possible rational roots of a polynomial \(a_nx^n + a_{n - 1}x^{n-1}+\dots+a_1x + a_0\) are of the form \(\frac{p}{q}\), where \(p\) is a factor of the constant term \(a_0\) and \(q\) is a factor of the leading coefficient \(a_n\). For \(f(x)=4x^3 - 7x^2 - 34x - 8\), \(a_0=-8\) and \(a_n = 4\). Factors of \(p=\pm1,\pm2,\pm4,\pm8\) and factors of \(q=\pm1,\pm2,\pm4\). So possible rational roots are \(\pm1,\pm2,\pm4,\pm8,\pm\frac{1}{2},\pm\frac{1}{4}\).

Step2: Test Possible Roots

Test \(x = 4\): \(f(4)=4(64)-7(16)-34(4)-8=256 - 112-136 - 8=0\). So \(x = 4\) is a root.

Step3: Factor the Polynomial

Since \(x = 4\) is a root, \((x - 4)\) is a factor. Use polynomial division or synthetic division. Using synthetic division with root \(4\):

$$
LATEXBLOCK0
$$

So \(f(x)=(x - 4)(4x^2+9x + 2)\).

Step4: Solve Quadratic Equation

Factor \(4x^2+9x + 2\): \(4x^2+9x + 2=(4x + 1)(x + 2)\). Set to zero: \(4x+1 = 0\Rightarrow x=-\frac{1}{4}\); \(x + 2=0\Rightarrow x=-2\).

Answer:

The roots are \(x = 4\), \(x=-2\), \(x = -\frac{1}{4}\) (for the first root asked in the box, a rational root is \(4\) (or \(-2\) or \(-\frac{1}{4}\), but \(4\) was found first in testing))