Solve a^4-20a^2+64

$$a ^ { 4 } - 20 a ^ { 2 } + 64$$

Factor

$$\left(a-4\right)\left(a-2\right)\left(a+2\right)\left(a+4\right)$$

Solution Steps

To factor the expression, solve the equation where it equals to 0.

$$a^{4}-20a^{2}+64=0$$

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.

$$a=2$$

By Factor theorem, a-k is a factor of the polynomial for each root k. Divide a^{4}-20a^{2}+64 by a-2 to get a^{3}+2a^{2}-16a-32. To factor the result, solve the equation where it equals to 0.

$$a^{3}+2a^{2}-16a-32=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -32 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±32,±16,±8,±4,±2,±1$$

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.

$$a=-2$$

By Factor theorem, a-k is a factor of the polynomial for each root k. Divide a^{3}+2a^{2}-16a-32 by a+2 to get a^{2}-16. To factor the result, solve the equation where it equals to 0.

$$a^{2}-16=0$$

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, 0 for b, and -16 for c in the quadratic formula.

$$a=\frac{0±\sqrt{0^{2}-4\times 1\left(-16\right)}}{2}$$

Do the calculations.

$$a=\frac{0±8}{2}$$

Solve the equation a^{2}-16=0 when ± is plus and when ± is minus.

$$a=-4 a=4$$

Rewrite the factored expression using the obtained roots.

$$\left(a-4\right)\left(a-2\right)\left(a+2\right)\left(a+4\right)$$

Evaluate

$$a^{4}-20a^{2}+64$$