Solve 24t^2+84t+40t+140

$$24 t ^ { 2 } + 84 t + 40 t + 140$$

Factor

$$4\left(2t+7\right)\left(3t+5\right)$$

Solution Steps

Factor out 4.

$$4\left(6t^{2}+21t+10t+35\right)$$

Consider 6t^{2}+21t+10t+35. Multiply and combine like terms.

$$6t^{2}+31t+35$$

Consider 6t^{2}+31t+35. Factor the expression by grouping. First, the expression needs to be rewritten as 6t^{2}+at+bt+35. To find a and b, set up a system to be solved.

$$a+b=31 ab=6\times 35=210$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 210.

$$1,210 2,105 3,70 5,42 6,35 7,30 10,21 14,15$$

Calculate the sum for each pair.

$$1+210=211 2+105=107 3+70=73 5+42=47 6+35=41 7+30=37 10+21=31 14+15=29$$

The solution is the pair that gives sum 31.

$$a=10 b=21$$

Rewrite 6t^{2}+31t+35 as \left(6t^{2}+10t\right)+\left(21t+35\right).

$$\left(6t^{2}+10t\right)+\left(21t+35\right)$$

Factor out 2t in the first and 7 in the second group.

$$2t\left(3t+5\right)+7\left(3t+5\right)$$

Factor out common term 3t+5 by using distributive property.

$$\left(3t+5\right)\left(2t+7\right)$$

Rewrite the complete factored expression.

$$4\left(3t+5\right)\left(2t+7\right)$$

Evaluate

$$24t^{2}+124t+140$$

Solution Steps

Combine 84t and 40t to get 124t.

$$24t^{2}+124t+140$$