Solve 12x^2-156x+480

$$12 x ^ { 2 } - 156 x + 480$$

Factor

$$12\left(x-8\right)\left(x-5\right)$$

Solution Steps

Factor out 12.

$$12\left(x^{2}-13x+40\right)$$

Consider x^{2}-13x+40. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+40. To find a and b, set up a system to be solved.

$$a+b=-13 ab=1\times 40=40$$

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 40.

$$-1,-40 -2,-20 -4,-10 -5,-8$$

Calculate the sum for each pair.

$$-1-40=-41 -2-20=-22 -4-10=-14 -5-8=-13$$

The solution is the pair that gives sum -13.

$$a=-8 b=-5$$

Rewrite x^{2}-13x+40 as \left(x^{2}-8x\right)+\left(-5x+40\right).

$$\left(x^{2}-8x\right)+\left(-5x+40\right)$$

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

$$x\left(x-8\right)-5\left(x-8\right)$$

Factor out common term x-8 by using distributive property.

$$\left(x-8\right)\left(x-5\right)$$

Rewrite the complete factored expression.

$$12\left(x-8\right)\left(x-5\right)$$

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.

$$12x^{2}-156x+480=0$$

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.

$$x=\frac{-\left(-156\right)±\sqrt{\left(-156\right)^{2}-4\times 12\times 480}}{2\times 12}$$

Square -156.

$$x=\frac{-\left(-156\right)±\sqrt{24336-4\times 12\times 480}}{2\times 12}$$

Multiply -4 times 12.

$$x=\frac{-\left(-156\right)±\sqrt{24336-48\times 480}}{2\times 12}$$

Multiply -48 times 480.

$$x=\frac{-\left(-156\right)±\sqrt{24336-23040}}{2\times 12}$$

Add 24336 to -23040.

$$x=\frac{-\left(-156\right)±\sqrt{1296}}{2\times 12}$$

Take the square root of 1296.

$$x=\frac{-\left(-156\right)±36}{2\times 12}$$

The opposite of -156 is 156.

$$x=\frac{156±36}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{156±36}{24}$$

Now solve the equation x=\frac{156±36}{24} when ± is plus. Add 156 to 36.

$$x=\frac{192}{24}$$

Divide 192 by 24.

$$x=8$$

Now solve the equation x=\frac{156±36}{24} when ± is minus. Subtract 36 from 156.

$$x=\frac{120}{24}$$

Divide 120 by 24.

$$x=5$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 8 for x_{1} and 5 for x_{2}.

$$12x^{2}-156x+480=12\left(x-8\right)\left(x-5\right)$$

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.This is achieved by dividing both sides of the equation by 12

$$x ^ 2 -13x +40 = 0$$

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

$$r + s = 13 rs = 40$$

Two numbers r and s sum up to 13 exactly when the average of the two numbers is \frac{1}{2}*13 = \frac{13}{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>

$$r = \frac{13}{2} - u s = \frac{13}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 40

$$(\frac{13}{2} - u) (\frac{13}{2} + u) = 40$$

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

$$\frac{169}{4} - u^2 = 40$$

Simplify the expression by subtracting \frac{169}{4} on both sides

$$-u^2 = 40-\frac{169}{4} = -\frac{9}{4}$$

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

$$u^2 = \frac{9}{4} u = \pm\sqrt{\frac{9}{4}} = \pm \frac{3}{2}$$

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

$$r =\frac{13}{2} - \frac{3}{2} = 5 s = \frac{13}{2} + \frac{3}{2} = 8$$

Evaluate

$$12\left(x-8\right)\left(x-5\right)$$