Solve 24y+10xy-6x^2y

$$24 y + 10 x y - 6 x ^ { 2 } y$$

Factor

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

Solution Steps

Factor out 2.

$$2\left(12y+5xy-3x^{2}y\right)$$

Consider 12y+5xy-3x^{2}y. Factor out y.

$$y\left(12+5x-3x^{2}\right)$$

Consider 12+5x-3x^{2}. Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-3x^{2}+5x+12$$

Factor the expression by grouping. First, the expression needs to be rewritten as -3x^{2}+ax+bx+12. To find a and b, set up a system to be solved.

$$a+b=5 ab=-3\times 12=-36$$

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 -36.

$$-1,36 -2,18 -3,12 -4,9 -6,6$$

Calculate the sum for each pair.

$$-1+36=35 -2+18=16 -3+12=9 -4+9=5 -6+6=0$$

The solution is the pair that gives sum 5.

$$a=9 b=-4$$

Rewrite -3x^{2}+5x+12 as \left(-3x^{2}+9x\right)+\left(-4x+12\right).

$$\left(-3x^{2}+9x\right)+\left(-4x+12\right)$$

Factor out 3x in the first and 4 in the second group.

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

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

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

Rewrite the complete factored expression.

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

Evaluate

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