Solve 18x^2-60x+48

$$18 x ^ { 2 } - 60 x + 48$$

Factor

$$6\left(x-2\right)\left(3x-4\right)$$

Solution Steps

Factor out 6.

$$6\left(3x^{2}-10x+8\right)$$

Consider 3x^{2}-10x+8. Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx+8. To find a and b, set up a system to be solved.

$$a+b=-10 ab=3\times 8=24$$

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 3x^{2}-10x+8 as \left(3x^{2}-6x\right)+\left(-4x+8\right).

$$\left(3x^{2}-6x\right)+\left(-4x+8\right)$$

Factor out 3x in the first and -4 in the second group.

$$3x\left(x-2\right)-4\left(x-2\right)$$

Factor out common term x-2 by using distributive property.

$$\left(x-2\right)\left(3x-4\right)$$

Rewrite the complete factored expression.

$$6\left(x-2\right)\left(3x-4\right)$$

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.

$$18x^{2}-60x+48=0$$

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.

$$x=\frac{-\left(-60\right)±\sqrt{\left(-60\right)^{2}-4\times 18\times 48}}{2\times 18}$$

Square -60.

$$x=\frac{-\left(-60\right)±\sqrt{3600-4\times 18\times 48}}{2\times 18}$$

Multiply -4 times 18.

$$x=\frac{-\left(-60\right)±\sqrt{3600-72\times 48}}{2\times 18}$$

Multiply -72 times 48.

$$x=\frac{-\left(-60\right)±\sqrt{3600-3456}}{2\times 18}$$

Add 3600 to -3456.

$$x=\frac{-\left(-60\right)±\sqrt{144}}{2\times 18}$$

Take the square root of 144.

$$x=\frac{-\left(-60\right)±12}{2\times 18}$$

The opposite of -60 is 60.

$$x=\frac{60±12}{2\times 18}$$

Multiply 2 times 18.

$$x=\frac{60±12}{36}$$

Now solve the equation x=\frac{60±12}{36} when ± is plus. Add 60 to 12.

$$x=\frac{72}{36}$$

Divide 72 by 36.

$$x=2$$

Now solve the equation x=\frac{60±12}{36} when ± is minus. Subtract 12 from 60.

$$x=\frac{48}{36}$$

Reduce the fraction \frac{48}{36} to lowest terms by extracting and canceling out 12.

$$x=\frac{4}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 2 for x_{1} and \frac{4}{3} for x_{2}.

$$18x^{2}-60x+48=18\left(x-2\right)\left(x-\frac{4}{3}\right)$$

Subtract \frac{4}{3} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$18x^{2}-60x+48=18\left(x-2\right)\times \frac{3x-4}{3}$$

Cancel out 3, the greatest common factor in 18 and 3.

$$18x^{2}-60x+48=6\left(x-2\right)\left(3x-4\right)$$

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 18

$$x ^ 2 -\frac{10}{3}x +\frac{8}{3} = 0$$

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

$$r + s = \frac{10}{3} rs = \frac{8}{3}$$

Two numbers r and s sum up to \frac{10}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{10}{3} = \frac{5}{3}. 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>

$$r = \frac{5}{3} - u s = \frac{5}{3} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{8}{3}

$$(\frac{5}{3} - u) (\frac{5}{3} + u) = \frac{8}{3}$$

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

$$\frac{25}{9} - u^2 = \frac{8}{3}$$

Simplify the expression by subtracting \frac{25}{9} on both sides

$$-u^2 = \frac{8}{3}-\frac{25}{9} = -\frac{1}{9}$$

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

$$u^2 = \frac{1}{9} u = \pm\sqrt{\frac{1}{9}} = \pm \frac{1}{3}$$

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

$$r =\frac{5}{3} - \frac{1}{3} = 1.333 s = \frac{5}{3} + \frac{1}{3} = 2.000$$

Evaluate

$$6\left(x-2\right)\left(3x-4\right)$$