Solve 21x^2y^2+6x^2y^3-15{x}^2{y}^4

$$21 { x }^{ 2 } { y }^{ 2 } +6 { x }^{ 2 } { y }^{ 3 } -15 { x }^{ 2 } { y }^{ 4 }$$

Factor

$$-3\left(5y-7\right)\left(y+1\right)x^{2}y^{2}$$

Solution Steps

Factor out 3.

$$3\left(7x^{2}y^{2}+2x^{2}y^{3}-5x^{2}y^{4}\right)$$

Consider 7x^{2}y^{2}+2x^{2}y^{3}-5x^{2}y^{4}. Factor out x^{2}y^{2}.

$$x^{2}y^{2}\left(7+2y-5y^{2}\right)$$

Consider 7+2y-5y^{2}. Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-5y^{2}+2y+7$$

Factor the expression by grouping. First, the expression needs to be rewritten as -5y^{2}+ay+by+7. To find a and b, set up a system to be solved.

$$a+b=2 ab=-5\times 7=-35$$

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 -35.

$$-1,35 -5,7$$

Calculate the sum for each pair.

$$-1+35=34 -5+7=2$$

The solution is the pair that gives sum 2.

$$a=7 b=-5$$

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

$$\left(-5y^{2}+7y\right)+\left(-5y+7\right)$$

Factor out -y in the first and -1 in the second group.

$$-y\left(5y-7\right)-\left(5y-7\right)$$

Factor out common term 5y-7 by using distributive property.

$$\left(5y-7\right)\left(-y-1\right)$$

Rewrite the complete factored expression.

$$3x^{2}y^{2}\left(5y-7\right)\left(-y-1\right)$$

Evaluate

$$-3\left(5y-7\right)\left(y+1\right)\left(xy\right)^{2}$$