Solve a^3+64a-320=0

$$a ^ { 3 } + 64 a - 320 = 0$$

Solve for a

$$a=4$$

Solution Steps

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -320 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±320,±160,±80,±64,±40,±32,±20,±16,±10,±8,±5,±4,±2,±1$$

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.

$$a=4$$

By Factor theorem, a-k is a factor of the polynomial for each root k. Divide a^{3}+64a-320 by a-4 to get a^{2}+4a+80. Solve the equation where the result equals to 0.

$$a^{2}+4a+80=0$$

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 80 for c in the quadratic formula.

$$a=\frac{-4±\sqrt{4^{2}-4\times 1\times 80}}{2}$$

Do the calculations.

$$a=\frac{-4±\sqrt{-304}}{2}$$

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

$$a\in \emptyset$$

List all found solutions.

$$a=4$$