Solve m^4-5m^2+12m+28=0

$$m ^ { 4 } - 5 m ^ { 2 } + 12 m + 28 = 0$$

Solve for m

$$m=-2$$

Solution Steps

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 28 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±28,±14,±7,±4,±2,±1$$

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.

$$m=-2$$

By Factor theorem, m-k is a factor of the polynomial for each root k. Divide m^{4}-5m^{2}+12m+28 by m+2 to get m^{3}-2m^{2}-m+14. Solve the equation where the result equals to 0.

$$m^{3}-2m^{2}-m+14=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 14 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±14,±7,±2,±1$$

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.

$$m=-2$$

By Factor theorem, m-k is a factor of the polynomial for each root k. Divide m^{3}-2m^{2}-m+14 by m+2 to get m^{2}-4m+7. Solve the equation where the result equals to 0.

$$m^{2}-4m+7=0$$

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, -4 for b, and 7 for c in the quadratic formula.

$$m=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 1\times 7}}{2}$$

Do the calculations.

$$m=\frac{4±\sqrt{-12}}{2}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$m\in \emptyset$$

List all found solutions.

$$m=-2$$