Solve f(1/2)=12x^3+19x^2-13x-6

$$f ( \frac { 1 } { 2 } ) = 12 x ^ { 3 } + 19 x ^ { 2 } - 13 x - 6$$

Factor

$$\left(4x-3\right)\left(x+2\right)\left(3x+1\right)$$

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

$$\left(x+2\right)\left(12x^{2}-5x-3\right)$$

Consider 12x^{2}-5x-3. Factor the expression by grouping. First, the expression needs to be rewritten as 12x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=-5 ab=12\left(-3\right)=-36$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. 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.

$$a=-9 b=4$$

Rewrite 12x^{2}-5x-3 as \left(12x^{2}-9x\right)+\left(4x-3\right).

$$\left(12x^{2}-9x\right)+\left(4x-3\right)$$

Factor out 3x in 12x^{2}-9x.

$$3x\left(4x-3\right)+4x-3$$

Factor out common term 4x-3 by using distributive property.

$$\left(4x-3\right)\left(3x+1\right)$$

Rewrite the complete factored expression.

$$\left(4x-3\right)\left(3x+1\right)\left(x+2\right)$$

Evaluate

$$\left(4x-3\right)\left(x+2\right)\left(3x+1\right)$$