Solve 48z^3-102z^2-45z

$$48 z ^ { 3 } - 102 z ^ { 2 } - 45 z$$

Factor

$$3z\left(2z-5\right)\left(8z+3\right)$$

Solution Steps

Factor out 3.

$$3\left(16z^{3}-34z^{2}-15z\right)$$

Consider 16z^{3}-34z^{2}-15z. Factor out z.

$$z\left(16z^{2}-34z-15\right)$$

Consider 16z^{2}-34z-15. Factor the expression by grouping. First, the expression needs to be rewritten as 16z^{2}+az+bz-15. To find a and b, set up a system to be solved.

$$a+b=-34 ab=16\left(-15\right)=-240$$

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 -240.

$$1,-240 2,-120 3,-80 4,-60 5,-48 6,-40 8,-30 10,-24 12,-20 15,-16$$

Calculate the sum for each pair.

$$1-240=-239 2-120=-118 3-80=-77 4-60=-56 5-48=-43 6-40=-34 8-30=-22 10-24=-14 12-20=-8 15-16=-1$$

The solution is the pair that gives sum -34.

$$a=-40 b=6$$

Rewrite 16z^{2}-34z-15 as \left(16z^{2}-40z\right)+\left(6z-15\right).

$$\left(16z^{2}-40z\right)+\left(6z-15\right)$$

Factor out 8z in the first and 3 in the second group.

$$8z\left(2z-5\right)+3\left(2z-5\right)$$

Factor out common term 2z-5 by using distributive property.

$$\left(2z-5\right)\left(8z+3\right)$$

Rewrite the complete factored expression.

$$3z\left(2z-5\right)\left(8z+3\right)$$

Evaluate

$$3z\left(2z-5\right)\left(8z+3\right)$$