Solve x^4-41x^2+400=

$${ x }^{ 4 } -41 { x }^{ 2 } +400=$$

Factor

$$\left(x-5\right)\left(x-4\right)\left(x+4\right)\left(x+5\right)$$

Solution Steps

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

$$x^{4}-41x^{2}+400=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 400 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±400,±200,±100,±80,±50,±40,±25,±20,±16,±10,±8,±5,±4,±2,±1$$

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.

$$x=4$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{4}-41x^{2}+400 by x-4 to get x^{3}+4x^{2}-25x-100. To factor the result, solve the equation where it equals to 0.

$$x^{3}+4x^{2}-25x-100=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -100 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±100,±50,±25,±20,±10,±5,±4,±2,±1$$

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.

$$x=-4$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+4x^{2}-25x-100 by x+4 to get x^{2}-25. To factor the result, solve the equation where it equals to 0.

$$x^{2}-25=0$$

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 -25 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 1\left(-25\right)}}{2}$$

Do the calculations.

$$x=\frac{0±10}{2}$$

Solve the equation x^{2}-25=0 when ± is plus and when ± is minus.

$$x=-5 x=5$$

Rewrite the factored expression using the obtained roots.

$$\left(x-5\right)\left(x-4\right)\left(x+4\right)\left(x+5\right)$$

Evaluate

$$x^{4}-41x^{2}+400$$