Solve 5x^2-50x+120

$$5 x ^ { 2 } - 50 x + 120$$

Factor

$$5\left(x-6\right)\left(x-4\right)$$

Solution Steps

Factor out 5.

$$5\left(x^{2}-10x+24\right)$$

Consider x^{2}-10x+24. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+24. To find a and b, set up a system to be solved.

$$a+b=-10 ab=1\times 24=24$$

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 24.

$$-1,-24 -2,-12 -3,-8 -4,-6$$

Calculate the sum for each pair.

$$-1-24=-25 -2-12=-14 -3-8=-11 -4-6=-10$$

The solution is the pair that gives sum -10.

$$a=-6 b=-4$$

Rewrite x^{2}-10x+24 as \left(x^{2}-6x\right)+\left(-4x+24\right).

$$\left(x^{2}-6x\right)+\left(-4x+24\right)$$

Factor out x in the first and -4 in the second group.

$$x\left(x-6\right)-4\left(x-6\right)$$

Factor out common term x-6 by using distributive property.

$$\left(x-6\right)\left(x-4\right)$$

Rewrite the complete factored expression.

$$5\left(x-6\right)\left(x-4\right)$$

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.

$$5x^{2}-50x+120=0$$

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.

$$x=\frac{-\left(-50\right)±\sqrt{\left(-50\right)^{2}-4\times 5\times 120}}{2\times 5}$$

Square -50.

$$x=\frac{-\left(-50\right)±\sqrt{2500-4\times 5\times 120}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-50\right)±\sqrt{2500-20\times 120}}{2\times 5}$$

Multiply -20 times 120.

$$x=\frac{-\left(-50\right)±\sqrt{2500-2400}}{2\times 5}$$

Add 2500 to -2400.

$$x=\frac{-\left(-50\right)±\sqrt{100}}{2\times 5}$$

Take the square root of 100.

$$x=\frac{-\left(-50\right)±10}{2\times 5}$$

The opposite of -50 is 50.

$$x=\frac{50±10}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{50±10}{10}$$

Now solve the equation x=\frac{50±10}{10} when ± is plus. Add 50 to 10.

$$x=\frac{60}{10}$$

Divide 60 by 10.

$$x=6$$

Now solve the equation x=\frac{50±10}{10} when ± is minus. Subtract 10 from 50.

$$x=\frac{40}{10}$$

Divide 40 by 10.

$$x=4$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 6 for x_{1} and 4 for x_{2}.

$$5x^{2}-50x+120=5\left(x-6\right)\left(x-4\right)$$

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.This is achieved by dividing both sides of the equation by 5

$$x ^ 2 -10x +24 = 0$$

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

$$r + s = 10 rs = 24$$

Two numbers r and s sum up to 10 exactly when the average of the two numbers is \frac{1}{2}*10 = 5. 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>

$$r = 5 - u s = 5 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 24

$$(5 - u) (5 + u) = 24$$

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

$$25 - u^2 = 24$$

Simplify the expression by subtracting 25 on both sides

$$-u^2 = 24-25 = -1$$

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

$$u^2 = 1 u = \pm\sqrt{1} = \pm 1$$

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

$$r =5 - 1 = 4 s = 5 + 1 = 6$$

Evaluate

$$5\left(x-6\right)\left(x-4\right)$$