Solve 12x^5-8x^4-27x^3+24x^2+5x-6

$$12 x ^ { 5 } - 8 x ^ { 4 } - 27 x ^ { 3 } + 24 x ^ { 2 } + 5 x - 6$$

Factor

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

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 -6 and q divides the leading coefficient 12. One such root is -\frac{1}{2}. Factor the polynomial by dividing it by 2x+1.

$$\left(2x+1\right)\left(6x^{4}-7x^{3}-10x^{2}+17x-6\right)$$

Consider 6x^{4}-7x^{3}-10x^{2}+17x-6. By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -6 and q divides the leading coefficient 6. One such root is 1. Factor the polynomial by dividing it by x-1.

$$\left(x-1\right)\left(6x^{3}-x^{2}-11x+6\right)$$

Consider 6x^{3}-x^{2}-11x+6. By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 6 and q divides the leading coefficient 6. One such root is 1. Factor the polynomial by dividing it by x-1.

$$\left(x-1\right)\left(6x^{2}+5x-6\right)$$

Consider 6x^{2}+5x-6. Factor the expression by grouping. First, the expression needs to be rewritten as 6x^{2}+ax+bx-6. To find a and b, set up a system to be solved.

$$a+b=5 ab=6\left(-6\right)=-36$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -36.

$$-1,36 -2,18 -3,12 -4,9 -6,6$$

Calculate the sum for each pair.

$$-1+36=35 -2+18=16 -3+12=9 -4+9=5 -6+6=0$$

The solution is the pair that gives sum 5.

$$a=-4 b=9$$

Rewrite 6x^{2}+5x-6 as \left(6x^{2}-4x\right)+\left(9x-6\right).

$$\left(6x^{2}-4x\right)+\left(9x-6\right)$$

Factor out 2x in the first and 3 in the second group.

$$2x\left(3x-2\right)+3\left(3x-2\right)$$

Factor out common term 3x-2 by using distributive property.

$$\left(3x-2\right)\left(2x+3\right)$$

Rewrite the complete factored expression.

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

Evaluate

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