Solve P(x)=x^4-6x^2+5

$$P ( x ) = x ^ { 4 } - 6 x ^ { 2 } + 5$$

Factor

$$\left(x-1\right)\left(x+1\right)\left(x^{2}-5\right)$$

Solution Steps

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

$$x^{4}-6x^{2}+5=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 5 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±5,±1$$

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.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{4}-6x^{2}+5 by x-1 to get x^{3}+x^{2}-5x-5. To factor the result, solve the equation where it equals to 0.

$$x^{3}+x^{2}-5x-5=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -5 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±5,±1$$

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.

$$x=-1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+x^{2}-5x-5 by x+1 to get x^{2}-5. To factor the result, solve the equation where it equals to 0.

$$x^{2}-5=0$$

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 -5 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 1\left(-5\right)}}{2}$$

Do the calculations.

$$x=\frac{0±2\sqrt{5}}{2}$$

Solve the equation x^{2}-5=0 when ± is plus and when ± is minus.

$$x=-\sqrt{5} x=\sqrt{5}$$

Rewrite the factored expression using the obtained roots. Polynomial x^{2}-5 is not factored since it does not have any rational roots.

$$\left(x-1\right)\left(x+1\right)\left(x^{2}-5\right)$$

Evaluate

$$x^{4}-6x^{2}+5$$