Solve x^2+35x+300

$${ x }^{ 2 } +35x+300$$

Factor

$$\left(x+15\right)\left(x+20\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+300. To find a and b, set up a system to be solved.

$$a+b=35 ab=1\times 300=300$$

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 300.

$$1,300 2,150 3,100 4,75 5,60 6,50 10,30 12,25 15,20$$

Calculate the sum for each pair.

$$1+300=301 2+150=152 3+100=103 4+75=79 5+60=65 6+50=56 10+30=40 12+25=37 15+20=35$$

The solution is the pair that gives sum 35.

$$a=15 b=20$$

Rewrite x^{2}+35x+300 as \left(x^{2}+15x\right)+\left(20x+300\right).

$$\left(x^{2}+15x\right)+\left(20x+300\right)$$

Factor out x in the first and 20 in the second group.

$$x\left(x+15\right)+20\left(x+15\right)$$

Factor out common term x+15 by using distributive property.

$$\left(x+15\right)\left(x+20\right)$$

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.

$$x^{2}+35x+300=0$$

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.

$$x=\frac{-35±\sqrt{35^{2}-4\times 300}}{2}$$

Square 35.

$$x=\frac{-35±\sqrt{1225-4\times 300}}{2}$$

Multiply -4 times 300.

$$x=\frac{-35±\sqrt{1225-1200}}{2}$$

Add 1225 to -1200.

$$x=\frac{-35±\sqrt{25}}{2}$$

Take the square root of 25.

$$x=\frac{-35±5}{2}$$

Now solve the equation x=\frac{-35±5}{2} when ± is plus. Add -35 to 5.

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

Divide -30 by 2.

$$x=-15$$

Now solve the equation x=\frac{-35±5}{2} when ± is minus. Subtract 5 from -35.

$$x=-\frac{40}{2}$$

Divide -40 by 2.

$$x=-20$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -15 for x_{1} and -20 for x_{2}.

$$x^{2}+35x+300=\left(x-\left(-15\right)\right)\left(x-\left(-20\right)\right)$$

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

$$x^{2}+35x+300=\left(x+15\right)\left(x+20\right)$$

Evaluate

$$\left(x+15\right)\left(x+20\right)$$