Solve 6y^4+12y^3-48y^2

$$6 y ^ { 4 } + 12 y ^ { 3 } - 48 y ^ { 2 }$$

Factor

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

Solution Steps

Factor out 6.

$$6\left(y^{4}+2y^{3}-8y^{2}\right)$$

Consider y^{4}+2y^{3}-8y^{2}. Factor out y^{2}.

$$y^{2}\left(y^{2}+2y-8\right)$$

Consider y^{2}+2y-8. Factor the expression by grouping. First, the expression needs to be rewritten as y^{2}+ay+by-8. To find a and b, set up a system to be solved.

$$a+b=2 ab=1\left(-8\right)=-8$$

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 -8.

$$-1,8 -2,4$$

Calculate the sum for each pair.

$$-1+8=7 -2+4=2$$

The solution is the pair that gives sum 2.

$$a=-2 b=4$$

Rewrite y^{2}+2y-8 as \left(y^{2}-2y\right)+\left(4y-8\right).

$$\left(y^{2}-2y\right)+\left(4y-8\right)$$

Factor out y in the first and 4 in the second group.

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

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

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

Rewrite the complete factored expression.

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

Evaluate

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