Solve 12x^3+16x^2-41x-15

$$12 x ^ { 3 } + 16 x ^ { 2 } - 41 x - 15$$

Factor

$$\left(2x-3\right)\left(2x+5\right)\left(3x+1\right)$$

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 -15 and q divides the leading coefficient 12. One such root is -\frac{1}{3}. Factor the polynomial by dividing it by 3x+1.

$$\left(3x+1\right)\left(4x^{2}+4x-15\right)$$

Consider 4x^{2}+4x-15. Factor the expression by grouping. First, the expression needs to be rewritten as 4x^{2}+ax+bx-15. To find a and b, set up a system to be solved.

$$a+b=4 ab=4\left(-15\right)=-60$$

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 -60.

$$-1,60 -2,30 -3,20 -4,15 -5,12 -6,10$$

Calculate the sum for each pair.

$$-1+60=59 -2+30=28 -3+20=17 -4+15=11 -5+12=7 -6+10=4$$

The solution is the pair that gives sum 4.

$$a=-6 b=10$$

Rewrite 4x^{2}+4x-15 as \left(4x^{2}-6x\right)+\left(10x-15\right).

$$\left(4x^{2}-6x\right)+\left(10x-15\right)$$

Factor out 2x in the first and 5 in the second group.

$$2x\left(2x-3\right)+5\left(2x-3\right)$$

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

$$\left(2x-3\right)\left(2x+5\right)$$

Rewrite the complete factored expression.

$$\left(2x-3\right)\left(3x+1\right)\left(2x+5\right)$$

Evaluate

$$\left(2x-3\right)\left(2x+5\right)\left(3x+1\right)$$