Solve x^3+3x^2-10x

$${ x }^{ 3 } +3 { x }^{ 2 } -10x$$

Factor

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

Solution Steps

Factor out x.

$$x\left(x^{2}+3x-10\right)$$

Consider x^{2}+3x-10. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-10. To find a and b, set up a system to be solved.

$$a+b=3 ab=1\left(-10\right)=-10$$

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 -10.

$$-1,10 -2,5$$

Calculate the sum for each pair.

$$-1+10=9 -2+5=3$$

The solution is the pair that gives sum 3.

$$a=-2 b=5$$

Rewrite x^{2}+3x-10 as \left(x^{2}-2x\right)+\left(5x-10\right).

$$\left(x^{2}-2x\right)+\left(5x-10\right)$$

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

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

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

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

Rewrite the complete factored expression.

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

Evaluate

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