Solve m^4+3m^2-4

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

Factor

$$\left(m-1\right)\left(m+1\right)\left(m^{2}+4\right)$$

Solution Steps

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

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

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -4 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±4,±2,±1$$

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.

$$m=1$$

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

$$m^{3}+m^{2}+4m+4=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 4 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±4,±2,±1$$

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.

$$m=-1$$

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

$$m^{2}+4=0$$

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 4 for c in the quadratic formula.

$$m=\frac{0±\sqrt{0^{2}-4\times 1\times 4}}{2}$$

Do the calculations.

$$m=\frac{0±\sqrt{-16}}{2}$$

Polynomial m^{2}+4 is not factored since it does not have any rational roots.

$$m^{2}+4$$

Rewrite the factored expression using the obtained roots.

$$\left(m-1\right)\left(m+1\right)\left(m^{2}+4\right)$$

Evaluate

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