Solve x^5+2x^4-13x^3-26x^2+36x+72

$$x ^ { 5 } + 2 x ^ { 4 } - 13 x ^ { 3 } - 26 x ^ { 2 } + 36 x + 72$$

Factor

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

Solution Steps

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

$$x^{5}+2x^{4}-13x^{3}-26x^{2}+36x+72=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 72 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±72,±36,±24,±18,±12,±9,±8,±6,±4,±3,±2,±1$$

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^{5}+2x^{4}-13x^{3}-26x^{2}+36x+72 by x-2 to get x^{4}+4x^{3}-5x^{2}-36x-36. To factor the result, solve the equation where it equals to 0.

$$x^{4}+4x^{3}-5x^{2}-36x-36=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -36 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±36,±18,±12,±9,±6,±4,±3,±2,±1$$

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^{4}+4x^{3}-5x^{2}-36x-36 by x+2 to get x^{3}+2x^{2}-9x-18. To factor the result, solve the equation where it equals to 0.

$$x^{3}+2x^{2}-9x-18=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -18 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±18,±9,±6,±3,±2,±1$$

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}+2x^{2}-9x-18 by x+2 to get x^{2}-9. To factor the result, solve the equation where it equals to 0.

$$x^{2}-9=0$$

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 -9 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 1\left(-9\right)}}{2}$$

Do the calculations.

$$x=\frac{0±6}{2}$$

Solve the equation x^{2}-9=0 when ± is plus and when ± is minus.

$$x=-3 x=3$$

Rewrite the factored expression using the obtained roots.

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

Evaluate

$$\left(x-2\right)\left(x^{2}-9\right)\left(x+2\right)^{2}$$