Solve 9y^3-57y^2-42y

$$9 y ^ { 3 } - 57 y ^ { 2 } - 42 y$$

Factor

$$3y\left(y-7\right)\left(3y+2\right)$$

Solution Steps

Factor out 3.

$$3\left(3y^{3}-19y^{2}-14y\right)$$

Consider 3y^{3}-19y^{2}-14y. Factor out y.

$$y\left(3y^{2}-19y-14\right)$$

Consider 3y^{2}-19y-14. Factor the expression by grouping. First, the expression needs to be rewritten as 3y^{2}+ay+by-14. To find a and b, set up a system to be solved.

$$a+b=-19 ab=3\left(-14\right)=-42$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -42.

$$1,-42 2,-21 3,-14 6,-7$$

Calculate the sum for each pair.

$$1-42=-41 2-21=-19 3-14=-11 6-7=-1$$

The solution is the pair that gives sum -19.

$$a=-21 b=2$$

Rewrite 3y^{2}-19y-14 as \left(3y^{2}-21y\right)+\left(2y-14\right).

$$\left(3y^{2}-21y\right)+\left(2y-14\right)$$

Factor out 3y in the first and 2 in the second group.

$$3y\left(y-7\right)+2\left(y-7\right)$$

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

$$\left(y-7\right)\left(3y+2\right)$$

Rewrite the complete factored expression.

$$3y\left(y-7\right)\left(3y+2\right)$$

Evaluate

$$3y\left(y-7\right)\left(3y+2\right)$$