Solve 12t^2+15t-4t-5

$$12 t ^ { 2 } + 15 t - 4 t - 5$$

Factor

$$\left(3t-1\right)\left(4t+5\right)$$

Solution Steps

Multiply and combine like terms.

$$12t^{2}+11t-5$$

Factor the expression by grouping. First, the expression needs to be rewritten as 12t^{2}+at+bt-5. To find a and b, set up a system to be solved.

$$a+b=11 ab=12\left(-5\right)=-60$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -60.

$$-1,60 -2,30 -3,20 -4,15 -5,12 -6,10$$

Calculate the sum for each pair.

$$-1+60=59 -2+30=28 -3+20=17 -4+15=11 -5+12=7 -6+10=4$$

The solution is the pair that gives sum 11.

$$a=-4 b=15$$

Rewrite 12t^{2}+11t-5 as \left(12t^{2}-4t\right)+\left(15t-5\right).

$$\left(12t^{2}-4t\right)+\left(15t-5\right)$$

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

$$4t\left(3t-1\right)+5\left(3t-1\right)$$

Factor out common term 3t-1 by using distributive property.

$$\left(3t-1\right)\left(4t+5\right)$$

Evaluate

$$\left(3t-1\right)\left(4t+5\right)$$

Solution Steps

Combine 15t and -4t to get 11t.

$$12t^{2}+11t-5$$