Solve x^2-30x+125

$$x ^ { 2 } - 30 x + 125$$

Factor

$$\left(x-25\right)\left(x-5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+125. To find a and b, set up a system to be solved.

$$a+b=-30 ab=1\times 125=125$$

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 125.

$$-1,-125 -5,-25$$

Calculate the sum for each pair.

$$-1-125=-126 -5-25=-30$$

The solution is the pair that gives sum -30.

$$a=-25 b=-5$$

Rewrite x^{2}-30x+125 as \left(x^{2}-25x\right)+\left(-5x+125\right).

$$\left(x^{2}-25x\right)+\left(-5x+125\right)$$

Factor out x in the first and -5 in the second group.

$$x\left(x-25\right)-5\left(x-25\right)$$

Factor out common term x-25 by using distributive property.

$$\left(x-25\right)\left(x-5\right)$$

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.

$$x^{2}-30x+125=0$$

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.

$$x=\frac{-\left(-30\right)±\sqrt{\left(-30\right)^{2}-4\times 125}}{2}$$

Square -30.

$$x=\frac{-\left(-30\right)±\sqrt{900-4\times 125}}{2}$$

Multiply -4 times 125.

$$x=\frac{-\left(-30\right)±\sqrt{900-500}}{2}$$

Add 900 to -500.

$$x=\frac{-\left(-30\right)±\sqrt{400}}{2}$$

Take the square root of 400.

$$x=\frac{-\left(-30\right)±20}{2}$$

The opposite of -30 is 30.

$$x=\frac{30±20}{2}$$

Now solve the equation x=\frac{30±20}{2} when ± is plus. Add 30 to 20.

$$x=\frac{50}{2}$$

Divide 50 by 2.

$$x=25$$

Now solve the equation x=\frac{30±20}{2} when ± is minus. Subtract 20 from 30.

$$x=\frac{10}{2}$$

Divide 10 by 2.

$$x=5$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 25 for x_{1} and 5 for x_{2}.

$$x^{2}-30x+125=\left(x-25\right)\left(x-5\right)$$

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.

$$x ^ 2 -30x +125 = 0$$

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

$$r + s = 30 rs = 125$$

Two numbers r and s sum up to 30 exactly when the average of the two numbers is \frac{1}{2}*30 = 15. 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>

$$r = 15 - u s = 15 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 125

$$(15 - u) (15 + u) = 125$$

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

$$225 - u^2 = 125$$

Simplify the expression by subtracting 225 on both sides

$$-u^2 = 125-225 = -100$$

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.

$$r =15 - 10 = 5 s = 15 + 10 = 25$$

Evaluate

$$\left(x-25\right)\left(x-5\right)$$