Question

the function $f(x) = 2x^4 + x^3 - 4x^2 - 4x - 16$ has at least two rational roots. use the rational root theorem to find those roots, then proceed to find all complex roots. (note: roots may be integer, rational, irrational, and/or complex.)
answer attempt 1 out of 3
there is one root

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 $f(x)=2x^4 + x^3 - 4x^2 - 4x - 16$, $p=\pm1,\pm2,\pm4,\pm8,\pm16$ and $q=\pm1,\pm2$. So possible rational roots are $\pm1,\pm2,\pm4,\pm8,\pm16,\pm\frac{1}{2},\pm\frac{1}{4},\pm\frac{1}{8},\pm\frac{1}{16}$. Test $x = 2$: $f(2)=2(16)+8 - 4(4)-4(2)-16=32 + 8 - 16 - 8 - 16 = 0$. Test $x = -2$: $f(-2)=2(16)-8 - 4(4)-4(-2)-16=32 - 8 - 16 + 8 - 16 = 0$. So $x = 2$ and $x = -2$ are rational roots.

Step2: Factor the Polynomial

Since $x = 2$ and $x = -2$ are roots, $(x - 2)(x + 2)=x^2 - 4$ is a factor. Divide $f(x)$ by $x^2 - 4$:

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

Step3: Solve the Quadratic

Solve $2x^2 + x + 4 = 0$ using quadratic formula $x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}$, where $a = 2$, $b = 1$, $c = 4$. Discriminant $\Delta=1^2 - 4(2)(4)=1 - 32=-31$. So roots are $x=\frac{-1\pm\sqrt{-31}}{4}=\frac{-1\pm i\sqrt{31}}{4}$.

Answer:

The roots of $f(x)$ are $x = 2$, $x = -2$, $x=\frac{-1 + i\sqrt{31}}{4}$, and $x=\frac{-1 - i\sqrt{31}}{4}$.