Solve 3x^{3}y^{4}+18x^{2}y^4-6x^2y^4-36xy^4

$$3 x ^ { 3 } y ^ { 4 } + 18 x ^ { 2 } y ^ { 4 } - 6 x ^ { 2 } y ^ { 4 } - 36 x y ^ { 4 }$$

Factor

$$3x\left(x-2\right)\left(x+6\right)y^{4}$$

Solution Steps

Factor out 3.

$$3\left(x^{3}y^{4}+6x^{2}y^{4}-2x^{2}y^{4}-12xy^{4}\right)$$

Consider x^{3}y^{4}+6x^{2}y^{4}-2x^{2}y^{4}-12xy^{4}. Factor out xy^{4}.

$$xy^{4}\left(x^{2}+6x-2x-12\right)$$

Consider x^{2}+6x-2x-12. Multiply and combine like terms.

$$x^{2}+4x-12$$

Consider x^{2}+4x-12. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-12. To find a and b, set up a system to be solved.

$$a+b=4 ab=1\left(-12\right)=-12$$

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 -12.

$$-1,12 -2,6 -3,4$$

Calculate the sum for each pair.

$$-1+12=11 -2+6=4 -3+4=1$$

The solution is the pair that gives sum 4.

$$a=-2 b=6$$

Rewrite x^{2}+4x-12 as \left(x^{2}-2x\right)+\left(6x-12\right).

$$\left(x^{2}-2x\right)+\left(6x-12\right)$$

Factor out x in the first and 6 in the second group.

$$x\left(x-2\right)+6\left(x-2\right)$$

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

$$\left(x-2\right)\left(x+6\right)$$

Rewrite the complete factored expression.

$$3xy^{4}\left(x-2\right)\left(x+6\right)$$

Evaluate

$$3x\left(x-2\right)\left(x+6\right)y^{4}$$