# Evaluate: x ^ 3+(-3) ^ 4 = 17

## Expression: $$x ^ { 3 } + ( - 3 ) ^ { 4 } = 17$$

Calculate $-3$ to the power of $4$ and get $81$.

$$x^{3}+81=17$$

Subtract $17$ from both sides.

$$x^{3}+81-17=0$$

Subtract $17$ from $81$ to get $64$.

$$x^{3}+64=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form $\frac{p}{q}$, where $p$ divides the constant term $64$ and $q$ divides the leading coefficient $1$. List all candidates $\frac{p}{q}$.

$$±64,±32,±16,±8,±4,±2,±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=-4$$

By Factor theorem, $x-k$ is a factor of the polynomial for each root $k$. Divide $x^{3}+64$ by $x+4$ to get $x^{2}-4x+16$. Solve the equation where the result equals to $0$.

$$x^{2}-4x+16=0$$

All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. Substitute $1$ for $a$, $-4$ for $b$, and $16$ for $c$ in the quadratic formula.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 1\times 16}}{2}$$

Do the calculations.

$$x=\frac{4±\sqrt{-48}}{2}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$x\in \emptyset$$

List all found solutions.

$$x=-4$$

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