Solve 16xy-81xy^5

$$16 x y - 81 x y ^ { 5 }$$

Factor

$$xy\left(3y-2\right)\left(3y+2\right)\left(-9y^{2}-4\right)$$

Solution Steps

Factor out xy.

$$xy\left(16-81y^{4}\right)$$

Consider 16-81y^{4}. Rewrite 16-81y^{4} as 4^{2}-\left(9y^{2}\right)^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(4-9y^{2}\right)\left(4+9y^{2}\right)$$

Reorder the terms.

$$\left(-9y^{2}+4\right)\left(9y^{2}+4\right)$$

Consider -9y^{2}+4. Rewrite -9y^{2}+4 as 2^{2}-\left(3y\right)^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

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

Reorder the terms.

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

Rewrite the complete factored expression. Polynomial 9y^{2}+4 is not factored since it does not have any rational roots.

$$xy\left(-3y+2\right)\left(3y+2\right)\left(9y^{2}+4\right)$$

Evaluate

$$xy\left(16-81y^{4}\right)$$