Solve f(x)=2x^3+10x^2-4x-48

$$f ( x ) = 2 x ^ { 3 } + 10 x ^ { 2 } - 4 x - 48$$

Factor

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

Solution Steps

Factor out 2.

$$2\left(x^{3}+5x^{2}-2x-24\right)$$

Consider x^{3}+5x^{2}-2x-24. By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -24 and q divides the leading coefficient 1. One such root is -4. Factor the polynomial by dividing it by x+4.

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

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

$$a+b=1 ab=1\left(-6\right)=-6$$

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 -6.

$$-1,6 -2,3$$

Calculate the sum for each pair.

$$-1+6=5 -2+3=1$$

The solution is the pair that gives sum 1.

$$a=-2 b=3$$

Rewrite x^{2}+x-6 as \left(x^{2}-2x\right)+\left(3x-6\right).

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

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

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

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

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

Rewrite the complete factored expression.

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

Evaluate

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