Solve 12x^2-16x+5=

$$12 x ^ { 2 } - 16 x + 5 =$$

Factor

$$\left(2x-1\right)\left(6x-5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 12x^{2}+ax+bx+5. To find a and b, set up a system to be solved.

$$a+b=-16 ab=12\times 5=60$$

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 60.

$$-1,-60 -2,-30 -3,-20 -4,-15 -5,-12 -6,-10$$

Calculate the sum for each pair.

$$-1-60=-61 -2-30=-32 -3-20=-23 -4-15=-19 -5-12=-17 -6-10=-16$$

The solution is the pair that gives sum -16.

$$a=-10 b=-6$$

Rewrite 12x^{2}-16x+5 as \left(12x^{2}-10x\right)+\left(-6x+5\right).

$$\left(12x^{2}-10x\right)+\left(-6x+5\right)$$

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

$$2x\left(6x-5\right)-\left(6x-5\right)$$

Factor out common term 6x-5 by using distributive property.

$$\left(6x-5\right)\left(2x-1\right)$$

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.

$$12x^{2}-16x+5=0$$

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.

$$x=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 12\times 5}}{2\times 12}$$

Square -16.

$$x=\frac{-\left(-16\right)±\sqrt{256-4\times 12\times 5}}{2\times 12}$$

Multiply -4 times 12.

$$x=\frac{-\left(-16\right)±\sqrt{256-48\times 5}}{2\times 12}$$

Multiply -48 times 5.

$$x=\frac{-\left(-16\right)±\sqrt{256-240}}{2\times 12}$$

Add 256 to -240.

$$x=\frac{-\left(-16\right)±\sqrt{16}}{2\times 12}$$

Take the square root of 16.

$$x=\frac{-\left(-16\right)±4}{2\times 12}$$

The opposite of -16 is 16.

$$x=\frac{16±4}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{16±4}{24}$$

Now solve the equation x=\frac{16±4}{24} when ± is plus. Add 16 to 4.

$$x=\frac{20}{24}$$

Reduce the fraction \frac{20}{24} to lowest terms by extracting and canceling out 4.

$$x=\frac{5}{6}$$

Now solve the equation x=\frac{16±4}{24} when ± is minus. Subtract 4 from 16.

$$x=\frac{12}{24}$$

Reduce the fraction \frac{12}{24} to lowest terms by extracting and canceling out 12.

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5}{6} for x_{1} and \frac{1}{2} for x_{2}.

$$12x^{2}-16x+5=12\left(x-\frac{5}{6}\right)\left(x-\frac{1}{2}\right)$$

Subtract \frac{5}{6} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$12x^{2}-16x+5=12\times \frac{6x-5}{6}\left(x-\frac{1}{2}\right)$$

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

$$12x^{2}-16x+5=12\times \frac{6x-5}{6}\times \frac{2x-1}{2}$$

Multiply \frac{6x-5}{6} times \frac{2x-1}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$12x^{2}-16x+5=12\times \frac{\left(6x-5\right)\left(2x-1\right)}{6\times 2}$$

Multiply 6 times 2.

$$12x^{2}-16x+5=12\times \frac{\left(6x-5\right)\left(2x-1\right)}{12}$$

Cancel out 12, the greatest common factor in 12 and 12.

$$12x^{2}-16x+5=\left(6x-5\right)\left(2x-1\right)$$

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 12

$$x ^ 2 -\frac{4}{3}x +\frac{5}{12} = 0$$

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

$$r + s = \frac{4}{3} rs = \frac{5}{12}$$

Two numbers r and s sum up to \frac{4}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{4}{3} = \frac{2}{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>

$$r = \frac{2}{3} - u s = \frac{2}{3} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{5}{12}

$$(\frac{2}{3} - u) (\frac{2}{3} + u) = \frac{5}{12}$$

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

$$\frac{4}{9} - u^2 = \frac{5}{12}$$

Simplify the expression by subtracting \frac{4}{9} on both sides

$$-u^2 = \frac{5}{12}-\frac{4}{9} = -\frac{1}{36}$$

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

$$u^2 = \frac{1}{36} u = \pm\sqrt{\frac{1}{36}} = \pm \frac{1}{6}$$

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

$$r =\frac{2}{3} - \frac{1}{6} = 0.500 s = \frac{2}{3} + \frac{1}{6} = 0.833$$

Evaluate

$$\left(2x-1\right)\left(6x-5\right)$$