Solve (6x-2)+(3x^2+14x-5)

$$( 6 x - 2 ) + ( 3 x ^ { 2 } + 14 x - 5 )$$

Evaluate

$$\left(3x-1\right)\left(x+7\right)$$

Solution Steps

Combine 6x and 14x to get 20x.

$$20x-2+3x^{2}-5$$

Subtract 5 from -2 to get -7.

$$20x-7+3x^{2}$$

Factor

$$\left(3x-1\right)\left(x+7\right)$$

Solution Steps

Multiply and combine like terms.

$$3x^{2}+20x-7$$

Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx-7. To find a and b, set up a system to be solved.

$$a+b=20 ab=3\left(-7\right)=-21$$

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 -21.

$$-1,21 -3,7$$

Calculate the sum for each pair.

$$-1+21=20 -3+7=4$$

The solution is the pair that gives sum 20.

$$a=-1 b=21$$

Rewrite 3x^{2}+20x-7 as \left(3x^{2}-x\right)+\left(21x-7\right).

$$\left(3x^{2}-x\right)+\left(21x-7\right)$$

Factor out x in the first and 7 in the second group.

$$x\left(3x-1\right)+7\left(3x-1\right)$$

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

$$\left(3x-1\right)\left(x+7\right)$$