Solve x^2+25x+150

$${ x }^{ 2 } +25x+150$$

Factor

$$\left(x+10\right)\left(x+15\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+150. To find a and b, set up a system to be solved.

$$a+b=25 ab=1\times 150=150$$

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 150.

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

Calculate the sum for each pair.

$$1+150=151 2+75=77 3+50=53 5+30=35 6+25=31 10+15=25$$

The solution is the pair that gives sum 25.

$$a=10 b=15$$

Rewrite x^{2}+25x+150 as \left(x^{2}+10x\right)+\left(15x+150\right).

$$\left(x^{2}+10x\right)+\left(15x+150\right)$$

Factor out x in the first and 15 in the second group.

$$x\left(x+10\right)+15\left(x+10\right)$$

Factor out common term x+10 by using distributive property.

$$\left(x+10\right)\left(x+15\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$x^{2}+25x+150=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x=\frac{-25±\sqrt{25^{2}-4\times 150}}{2}$$

Square 25.

$$x=\frac{-25±\sqrt{625-4\times 150}}{2}$$

Multiply -4 times 150.

$$x=\frac{-25±\sqrt{625-600}}{2}$$

Add 625 to -600.

$$x=\frac{-25±\sqrt{25}}{2}$$

Take the square root of 25.

$$x=\frac{-25±5}{2}$$

Now solve the equation x=\frac{-25±5}{2} when ± is plus. Add -25 to 5.

$$x=-\frac{20}{2}$$

Divide -20 by 2.

$$x=-10$$

Now solve the equation x=\frac{-25±5}{2} when ± is minus. Subtract 5 from -25.

$$x=-\frac{30}{2}$$

Divide -30 by 2.

$$x=-15$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -10 for x_{1} and -15 for x_{2}.

$$x^{2}+25x+150=\left(x-\left(-10\right)\right)\left(x-\left(-15\right)\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$x^{2}+25x+150=\left(x+10\right)\left(x+15\right)$$

Evaluate

$$\left(x+10\right)\left(x+15\right)$$