Solve x^4+2x^3-7x^2-8x+12=0

$$x ^ { 4 } + 2 x ^ { 3 } - 7 x ^ { 2 } - 8 x + 12 = 0$$

Solve for x

$$x=1$$

$$x=2$$

$$x=-2$$

$$x=-3$$

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 12 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±12,±6,±4,±3,±2,±1$$

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.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{4}+2x^{3}-7x^{2}-8x+12 by x-1 to get x^{3}+3x^{2}-4x-12. Solve the equation where the result equals to 0.

$$x^{3}+3x^{2}-4x-12=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -12 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±12,±6,±4,±3,±2,±1$$

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.

$$x=2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+3x^{2}-4x-12 by x-2 to get x^{2}+5x+6. Solve the equation where the result equals to 0.

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

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, 5 for b, and 6 for c in the quadratic formula.

$$x=\frac{-5±\sqrt{5^{2}-4\times 1\times 6}}{2}$$

Do the calculations.

$$x=\frac{-5±1}{2}$$

Solve the equation x^{2}+5x+6=0 when ± is plus and when ± is minus.

$$x=-3 x=-2$$

List all found solutions.

$$x=1 x=2 x=-3 x=-2$$