Solve 12s^3-13s^2-35s

$$12 s ^ { 3 } - 13 s ^ { 2 } - 35 s$$

Factor

$$s\left(3s-7\right)\left(4s+5\right)$$

Solution Steps

Factor out s.

$$s\left(12s^{2}-13s-35\right)$$

Consider 12s^{2}-13s-35. Factor the expression by grouping. First, the expression needs to be rewritten as 12s^{2}+as+bs-35. To find a and b, set up a system to be solved.

$$a+b=-13 ab=12\left(-35\right)=-420$$

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 -420.

$$1,-420 2,-210 3,-140 4,-105 5,-84 6,-70 7,-60 10,-42 12,-35 14,-30 15,-28 20,-21$$

Calculate the sum for each pair.

$$1-420=-419 2-210=-208 3-140=-137 4-105=-101 5-84=-79 6-70=-64 7-60=-53 10-42=-32 12-35=-23 14-30=-16 15-28=-13 20-21=-1$$

The solution is the pair that gives sum -13.

$$a=-28 b=15$$

Rewrite 12s^{2}-13s-35 as \left(12s^{2}-28s\right)+\left(15s-35\right).

$$\left(12s^{2}-28s\right)+\left(15s-35\right)$$

Factor out 4s in the first and 5 in the second group.

$$4s\left(3s-7\right)+5\left(3s-7\right)$$

Factor out common term 3s-7 by using distributive property.

$$\left(3s-7\right)\left(4s+5\right)$$

Rewrite the complete factored expression.

$$s\left(3s-7\right)\left(4s+5\right)$$

Evaluate

$$s\left(3s-7\right)\left(4s+5\right)$$