Solve x^2-40x+300

$$x ^ { 2 } - 40 x + 300$$

Factor

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

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.

$$a+b=-40 ab=1\times 300=300$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both 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.

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

The solution is the pair that gives sum -40.

$$a=-30 b=-10$$

Rewrite x^{2}-40x+300 as \left(x^{2}-30x\right)+\left(-10x+300\right).

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

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

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

Factor out common term x-30 by using distributive property.

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

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.

$$x^{2}-40x+300=0$$

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.

$$x=\frac{-\left(-40\right)±\sqrt{\left(-40\right)^{2}-4\times 300}}{2}$$

Square -40.

$$x=\frac{-\left(-40\right)±\sqrt{1600-4\times 300}}{2}$$

Multiply -4 times 300.

$$x=\frac{-\left(-40\right)±\sqrt{1600-1200}}{2}$$

Add 1600 to -1200.

$$x=\frac{-\left(-40\right)±\sqrt{400}}{2}$$

Take the square root of 400.

$$x=\frac{-\left(-40\right)±20}{2}$$

The opposite of -40 is 40.

$$x=\frac{40±20}{2}$$

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

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

Divide 60 by 2.

$$x=30$$

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

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

Divide 20 by 2.

$$x=10$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 30 for x_{1} and 10 for x_{2}.

$$x^{2}-40x+300=\left(x-30\right)\left(x-10\right)$$

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.

$$x ^ 2 -40x +300 = 0$$

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

$$r + s = 40 rs = 300$$

Two numbers r and s sum up to 40 exactly when the average of the two numbers is \frac{1}{2}*40 = 20. 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>

$$r = 20 - u s = 20 + u$$

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

$$(20 - u) (20 + u) = 300$$

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

$$400 - u^2 = 300$$

Simplify the expression by subtracting 400 on both sides

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

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

$$u^2 = 100 u = \pm\sqrt{100} = \pm 10$$

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

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

Evaluate

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