Solve 48m^2-40m-18m+15

$$48 m ^ { 2 } - 40 m - 18 m + 15$$

Factor

$$\left(6m-5\right)\left(8m-3\right)$$

Solution Steps

Multiply and combine like terms.

$$48m^{2}-58m+15$$

Factor the expression by grouping. First, the expression needs to be rewritten as 48m^{2}+am+bm+15. To find a and b, set up a system to be solved.

$$a+b=-58 ab=48\times 15=720$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 720.

$$-1,-720 -2,-360 -3,-240 -4,-180 -5,-144 -6,-120 -8,-90 -9,-80 -10,-72 -12,-60 -15,-48 -16,-45 -18,-40 -20,-36 -24,-30$$

Calculate the sum for each pair.

$$-1-720=-721 -2-360=-362 -3-240=-243 -4-180=-184 -5-144=-149 -6-120=-126 -8-90=-98 -9-80=-89 -10-72=-82 -12-60=-72 -15-48=-63 -16-45=-61 -18-40=-58 -20-36=-56 -24-30=-54$$

The solution is the pair that gives sum -58.

$$a=-40 b=-18$$

Rewrite 48m^{2}-58m+15 as \left(48m^{2}-40m\right)+\left(-18m+15\right).

$$\left(48m^{2}-40m\right)+\left(-18m+15\right)$$

Factor out 8m in the first and -3 in the second group.

$$8m\left(6m-5\right)-3\left(6m-5\right)$$

Factor out common term 6m-5 by using distributive property.

$$\left(6m-5\right)\left(8m-3\right)$$

Evaluate

$$\left(6m-5\right)\left(8m-3\right)$$

Solution Steps

Combine -40m and -18m to get -58m.

$$48m^{2}-58m+15$$