Solve 9x^3+39x^2-30x

$$9 { x }^{ 3 } +39 { x }^{ 2 } -30x$$

Factor

$$3x\left(3x-2\right)\left(x+5\right)$$

Solution Steps

Factor out 3.

$$3\left(3x^{3}+13x^{2}-10x\right)$$

Consider 3x^{3}+13x^{2}-10x. Factor out x.

$$x\left(3x^{2}+13x-10\right)$$

Consider 3x^{2}+13x-10. Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx-10. To find a and b, set up a system to be solved.

$$a+b=13 ab=3\left(-10\right)=-30$$

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 -30.

$$-1,30 -2,15 -3,10 -5,6$$

Calculate the sum for each pair.

$$-1+30=29 -2+15=13 -3+10=7 -5+6=1$$

The solution is the pair that gives sum 13.

$$a=-2 b=15$$

Rewrite 3x^{2}+13x-10 as \left(3x^{2}-2x\right)+\left(15x-10\right).

$$\left(3x^{2}-2x\right)+\left(15x-10\right)$$

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

$$x\left(3x-2\right)+5\left(3x-2\right)$$

Factor out common term 3x-2 by using distributive property.

$$\left(3x-2\right)\left(x+5\right)$$

Rewrite the complete factored expression.

$$3x\left(3x-2\right)\left(x+5\right)$$

Evaluate

$$3x\left(3x-2\right)\left(x+5\right)$$