Solve 48x^2-60x+18

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

Factor

$$6\left(2x-1\right)\left(4x-3\right)$$

Solution Steps

Factor out 6.

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

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

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

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

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

Factor out 2x in the first and -1 in the second group.

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

Factor out common term 4x-3 by using distributive property.

$$\left(4x-3\right)\left(2x-1\right)$$

Rewrite the complete factored expression.

$$6\left(4x-3\right)\left(2x-1\right)$$

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.

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

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.

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

Square -60.

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

Multiply -4 times 48.

$$x=\frac{-\left(-60\right)±\sqrt{3600-192\times 18}}{2\times 48}$$

Multiply -192 times 18.

$$x=\frac{-\left(-60\right)±\sqrt{3600-3456}}{2\times 48}$$

Add 3600 to -3456.

$$x=\frac{-\left(-60\right)±\sqrt{144}}{2\times 48}$$

Take the square root of 144.

$$x=\frac{-\left(-60\right)±12}{2\times 48}$$

The opposite of -60 is 60.

$$x=\frac{60±12}{2\times 48}$$

Multiply 2 times 48.

$$x=\frac{60±12}{96}$$

Now solve the equation x=\frac{60±12}{96} when ± is plus. Add 60 to 12.

$$x=\frac{72}{96}$$

Reduce the fraction \frac{72}{96} to lowest terms by extracting and canceling out 24.

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

Now solve the equation x=\frac{60±12}{96} when ± is minus. Subtract 12 from 60.

$$x=\frac{48}{96}$$

Reduce the fraction \frac{48}{96} to lowest terms by extracting and canceling out 48.

$$x=\frac{1}{2}$$

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

$$48x^{2}-60x+18=48\left(x-\frac{3}{4}\right)\left(x-\frac{1}{2}\right)$$

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

$$48x^{2}-60x+18=48\times \frac{4x-3}{4}\left(x-\frac{1}{2}\right)$$

Subtract \frac{1}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$48x^{2}-60x+18=48\times \frac{4x-3}{4}\times \frac{2x-1}{2}$$

Multiply \frac{4x-3}{4} times \frac{2x-1}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$48x^{2}-60x+18=48\times \frac{\left(4x-3\right)\left(2x-1\right)}{4\times 2}$$

Multiply 4 times 2.

$$48x^{2}-60x+18=48\times \frac{\left(4x-3\right)\left(2x-1\right)}{8}$$

Cancel out 8, the greatest common factor in 48 and 8.

$$48x^{2}-60x+18=6\left(4x-3\right)\left(2x-1\right)$$

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 48

$$x ^ 2 -\frac{5}{4}x +\frac{3}{8} = 0$$

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

$$r + s = \frac{5}{4} rs = \frac{3}{8}$$

Two numbers r and s sum up to \frac{5}{4} exactly when the average of the two numbers is \frac{1}{2}*\frac{5}{4} = \frac{5}{8}. 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>

$$r = \frac{5}{8} - u s = \frac{5}{8} + u$$

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

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

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

$$\frac{25}{64} - u^2 = \frac{3}{8}$$

Simplify the expression by subtracting \frac{25}{64} on both sides

$$-u^2 = \frac{3}{8}-\frac{25}{64} = -\frac{1}{64}$$

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

$$u^2 = \frac{1}{64} u = \pm\sqrt{\frac{1}{64}} = \pm \frac{1}{8}$$

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

$$r =\frac{5}{8} - \frac{1}{8} = 0.500 s = \frac{5}{8} + \frac{1}{8} = 0.750$$

Evaluate

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