Solve 3x^3-15x^2-42x

$$3 x ^ { 3 } - 15 x ^ { 2 } - 42 x$$

Factor

$$3x\left(x-7\right)\left(x+2\right)$$

Solution Steps

Factor out 3.

$$3\left(x^{3}-5x^{2}-14x\right)$$

Consider x^{3}-5x^{2}-14x. Factor out x.

$$x\left(x^{2}-5x-14\right)$$

Consider x^{2}-5x-14. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-14. To find a and b, set up a system to be solved.

$$a+b=-5 ab=1\left(-14\right)=-14$$

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 -14.

$$1,-14 2,-7$$

Calculate the sum for each pair.

$$1-14=-13 2-7=-5$$

The solution is the pair that gives sum -5.

$$a=-7 b=2$$

Rewrite x^{2}-5x-14 as \left(x^{2}-7x\right)+\left(2x-14\right).

$$\left(x^{2}-7x\right)+\left(2x-14\right)$$

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

$$x\left(x-7\right)+2\left(x-7\right)$$

Factor out common term x-7 by using distributive property.

$$\left(x-7\right)\left(x+2\right)$$

Rewrite the complete factored expression.

$$3x\left(x-7\right)\left(x+2\right)$$

Evaluate

$$3x\left(x-7\right)\left(x+2\right)$$