Solve f(x)=48x^3+40x^2-x-3

$$f ( x ) = 48 x ^ { 3 } + 40 x ^ { 2 } - x - 3$$

Factor

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

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 -3 and q divides the leading coefficient 48. One such root is -\frac{3}{4}. Factor the polynomial by dividing it by 4x+3.

$$\left(4x+3\right)\left(12x^{2}+x-1\right)$$

Consider 12x^{2}+x-1. Factor the expression by grouping. First, the expression needs to be rewritten as 12x^{2}+ax+bx-1. To find a and b, set up a system to be solved.

$$a+b=1 ab=12\left(-1\right)=-12$$

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 -12.

$$-1,12 -2,6 -3,4$$

Calculate the sum for each pair.

$$-1+12=11 -2+6=4 -3+4=1$$

The solution is the pair that gives sum 1.

$$a=-3 b=4$$

Rewrite 12x^{2}+x-1 as \left(12x^{2}-3x\right)+\left(4x-1\right).

$$\left(12x^{2}-3x\right)+\left(4x-1\right)$$

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

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

Factor out common term 4x-1 by using distributive property.

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

Rewrite the complete factored expression.

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

Evaluate

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