Solve x^2-25x+100

$$x ^ { 2 } - 25 x + 100$$

Factor

$$\left(x-20\right)\left(x-5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+100. To find a and b, set up a system to be solved.

$$a+b=-25 ab=1\times 100=100$$

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 100.

$$-1,-100 -2,-50 -4,-25 -5,-20 -10,-10$$

Calculate the sum for each pair.

$$-1-100=-101 -2-50=-52 -4-25=-29 -5-20=-25 -10-10=-20$$

The solution is the pair that gives sum -25.

$$a=-20 b=-5$$

Rewrite x^{2}-25x+100 as \left(x^{2}-20x\right)+\left(-5x+100\right).

$$\left(x^{2}-20x\right)+\left(-5x+100\right)$$

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

$$x\left(x-20\right)-5\left(x-20\right)$$

Factor out common term x-20 by using distributive property.

$$\left(x-20\right)\left(x-5\right)$$

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.

$$x^{2}-25x+100=0$$

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.

$$x=\frac{-\left(-25\right)±\sqrt{\left(-25\right)^{2}-4\times 100}}{2}$$

Square -25.

$$x=\frac{-\left(-25\right)±\sqrt{625-4\times 100}}{2}$$

Multiply -4 times 100.

$$x=\frac{-\left(-25\right)±\sqrt{625-400}}{2}$$

Add 625 to -400.

$$x=\frac{-\left(-25\right)±\sqrt{225}}{2}$$

Take the square root of 225.

$$x=\frac{-\left(-25\right)±15}{2}$$

The opposite of -25 is 25.

$$x=\frac{25±15}{2}$$

Now solve the equation x=\frac{25±15}{2} when ± is plus. Add 25 to 15.

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

Divide 40 by 2.

$$x=20$$

Now solve the equation x=\frac{25±15}{2} when ± is minus. Subtract 15 from 25.

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

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 20 for x_{1} and 5 for x_{2}.

$$x^{2}-25x+100=\left(x-20\right)\left(x-5\right)$$

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.

$$x ^ 2 -25x +100 = 0$$

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

$$r + s = 25 rs = 100$$

Two numbers r and s sum up to 25 exactly when the average of the two numbers is \frac{1}{2}*25 = \frac{25}{2}. 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>

$$r = \frac{25}{2} - u s = \frac{25}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 100

$$(\frac{25}{2} - u) (\frac{25}{2} + u) = 100$$

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

$$\frac{625}{4} - u^2 = 100$$

Simplify the expression by subtracting \frac{625}{4} on both sides

$$-u^2 = 100-\frac{625}{4} = -\frac{225}{4}$$

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

$$u^2 = \frac{225}{4} u = \pm\sqrt{\frac{225}{4}} = \pm \frac{15}{2}$$

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

$$r =\frac{25}{2} - \frac{15}{2} = 5 s = \frac{25}{2} + \frac{15}{2} = 20$$

Evaluate

$$\left(x-20\right)\left(x-5\right)$$