Solve x^2+20x-300

$$x ^ { 2 } + 20 x - 300$$

Factor

$$\left(x-10\right)\left(x+30\right)$$

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.

$$a+b=20 ab=1\left(-300\right)=-300$$

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 -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.

$$-1+300=299 -2+150=148 -3+100=97 -4+75=71 -5+60=55 -6+50=44 -10+30=20 -12+25=13 -15+20=5$$

The solution is the pair that gives sum 20.

$$a=-10 b=30$$

Rewrite x^{2}+20x-300 as \left(x^{2}-10x\right)+\left(30x-300\right).

$$\left(x^{2}-10x\right)+\left(30x-300\right)$$

Factor out x in the first and 30 in the second group.

$$x\left(x-10\right)+30\left(x-10\right)$$

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

$$\left(x-10\right)\left(x+30\right)$$

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.

$$x^{2}+20x-300=0$$

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.

$$x=\frac{-20±\sqrt{20^{2}-4\left(-300\right)}}{2}$$

Square 20.

$$x=\frac{-20±\sqrt{400-4\left(-300\right)}}{2}$$

Multiply -4 times -300.

$$x=\frac{-20±\sqrt{400+1200}}{2}$$

Add 400 to 1200.

$$x=\frac{-20±\sqrt{1600}}{2}$$

Take the square root of 1600.

$$x=\frac{-20±40}{2}$$

Now solve the equation x=\frac{-20±40}{2} when ± is plus. Add -20 to 40.

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

Divide 20 by 2.

$$x=10$$

Now solve the equation x=\frac{-20±40}{2} when ± is minus. Subtract 40 from -20.

$$x=-\frac{60}{2}$$

Divide -60 by 2.

$$x=-30$$

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 -30 for x_{2}.

$$x^{2}+20x-300=\left(x-10\right)\left(x-\left(-30\right)\right)$$

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

$$x^{2}+20x-300=\left(x-10\right)\left(x+30\right)$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.

$$x ^ 2 +20x -300 = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = -20 rs = -300$$

Two numbers r and s sum up to -20 exactly when the average of the two numbers is \frac{1}{2}*-20 = -10. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = -10 - u s = -10 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -300

$$(-10 - u) (-10 + u) = -300$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$100 - u^2 = -300$$

Simplify the expression by subtracting 100 on both sides

$$-u^2 = -300-100 = -400$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = 400 u = \pm\sqrt{400} = \pm 20$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =-10 - 20 = -30 s = -10 + 20 = 10$$

Evaluate

$$\left(x-10\right)\left(x+30\right)$$