Solve (3x^2-6x-45)+(x-5)

$$( 3 x ^ { 2 } - 6 x - 45 ) + ( x - 5 )$$

Evaluate

$$\left(x-5\right)\left(3x+10\right)$$

Solution Steps

Combine -6x and x to get -5x.

$$3x^{2}-5x-45-5$$

Subtract 5 from -45 to get -50.

$$3x^{2}-5x-50$$

Factor

$$\left(x-5\right)\left(3x+10\right)$$

Solution Steps

Multiply and combine like terms.

$$3x^{2}-5x-50$$

Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx-50. To find a and b, set up a system to be solved.

$$a+b=-5 ab=3\left(-50\right)=-150$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -150.

$$1,-150 2,-75 3,-50 5,-30 6,-25 10,-15$$

Calculate the sum for each pair.

$$1-150=-149 2-75=-73 3-50=-47 5-30=-25 6-25=-19 10-15=-5$$

The solution is the pair that gives sum -5.

$$a=-15 b=10$$

Rewrite 3x^{2}-5x-50 as \left(3x^{2}-15x\right)+\left(10x-50\right).

$$\left(3x^{2}-15x\right)+\left(10x-50\right)$$

Factor out 3x in the first and 10 in the second group.

$$3x\left(x-5\right)+10\left(x-5\right)$$

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

$$\left(x-5\right)\left(3x+10\right)$$