Solve 15x^3-19x^2-10x

$$15 x ^ { 3 } - 19 x ^ { 2 } - 10 x$$

Factor

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

Solution Steps

Factor out x.

$$x\left(15x^{2}-19x-10\right)$$

Consider 15x^{2}-19x-10. Factor the expression by grouping. First, the expression needs to be rewritten as 15x^{2}+ax+bx-10. To find a and b, set up a system to be solved.

$$a+b=-19 ab=15\left(-10\right)=-150$$

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 -150.

$$1,-150 2,-75 3,-50 5,-30 6,-25 10,-15$$

Calculate the sum for each pair.

$$1-150=-149 2-75=-73 3-50=-47 5-30=-25 6-25=-19 10-15=-5$$

The solution is the pair that gives sum -19.

$$a=-25 b=6$$

Rewrite 15x^{2}-19x-10 as \left(15x^{2}-25x\right)+\left(6x-10\right).

$$\left(15x^{2}-25x\right)+\left(6x-10\right)$$

Factor out 5x in the first and 2 in the second group.

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

Factor out common term 3x-5 by using distributive property.

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

Rewrite the complete factored expression.

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

Evaluate

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