Solve (2x^2-6x+11)+(7x^2+9x-41)

$$( 2 x ^ { 2 } - 6 x + 11 ) + ( 7 x ^ { 2 } + 9 x - 41 )$$

Evaluate

$$3\left(3x-5\right)\left(x+2\right)$$

Solution Steps

Combine 2x^{2} and 7x^{2} to get 9x^{2}.

$$9x^{2}-6x+11+9x-41$$

Combine -6x and 9x to get 3x.

$$9x^{2}+3x+11-41$$

Subtract 41 from 11 to get -30.

$$9x^{2}+3x-30$$

Factor

$$3\left(3x-5\right)\left(x+2\right)$$

Solution Steps

Multiply and combine like terms.

$$9x^{2}+3x-30$$

Factor out 3.

$$3\left(3x^{2}+x-10\right)$$

Consider 3x^{2}+x-10. Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx-10. To find a and b, set up a system to be solved.

$$a+b=1 ab=3\left(-10\right)=-30$$

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 -30.

$$-1,30 -2,15 -3,10 -5,6$$

Calculate the sum for each pair.

$$-1+30=29 -2+15=13 -3+10=7 -5+6=1$$

The solution is the pair that gives sum 1.

$$a=-5 b=6$$

Rewrite 3x^{2}+x-10 as \left(3x^{2}-5x\right)+\left(6x-10\right).

$$\left(3x^{2}-5x\right)+\left(6x-10\right)$$

Factor out x in the first and 2 in the second group.

$$x\left(3x-5\right)+2\left(3x-5\right)$$

Factor out common term 3x-5 by using distributive property.

$$\left(3x-5\right)\left(x+2\right)$$

Rewrite the complete factored expression.

$$3\left(3x-5\right)\left(x+2\right)$$