Solve x^4-4x^2+3

$${ x }^{ 4 } -4 { x }^{ 2 } +3$$

Factor

$$\left(x-1\right)\left(x+1\right)\left(x^{2}-3\right)$$

Solution Steps

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

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

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 3 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±3,±1$$

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}-4x^{2}+3 by x-1 to get x^{3}+x^{2}-3x-3. To factor the result, solve the equation where it equals to 0.

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

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -3 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±3,±1$$

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}-3x-3 by x+1 to get x^{2}-3. To factor the result, solve the equation where it equals to 0.

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

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 -3 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 1\left(-3\right)}}{2}$$

Do the calculations.

$$x=\frac{0±2\sqrt{3}}{2}$$

Solve the equation x^{2}-3=0 when ± is plus and when ± is minus.

$$x=-\sqrt{3} x=\sqrt{3}$$

Rewrite the factored expression using the obtained roots. Polynomial x^{2}-3 is not factored since it does not have any rational roots.

$$\left(x-1\right)\left(x+1\right)\left(x^{2}-3\right)$$

Evaluate

$$x^{4}-4x^{2}+3$$