Solve 12x^2-10x+2

$$12 x ^ { 2 } - 10 x + 2$$

Factor

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

Solution Steps

Factor out 2.

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

Consider 6x^{2}-5x+1. Factor the expression by grouping. First, the expression needs to be rewritten as 6x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=-5 ab=6\times 1=6$$

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 6.

$$-1,-6 -2,-3$$

Calculate the sum for each pair.

$$-1-6=-7 -2-3=-5$$

The solution is the pair that gives sum -5.

$$a=-3 b=-2$$

Rewrite 6x^{2}-5x+1 as \left(6x^{2}-3x\right)+\left(-2x+1\right).

$$\left(6x^{2}-3x\right)+\left(-2x+1\right)$$

Factor out 3x in the first and -1 in the second group.

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

Factor out common term 2x-1 by using distributive property.

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

Rewrite the complete factored expression.

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

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.

$$12x^{2}-10x+2=0$$

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.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 12\times 2}}{2\times 12}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 12\times 2}}{2\times 12}$$

Multiply -4 times 12.

$$x=\frac{-\left(-10\right)±\sqrt{100-48\times 2}}{2\times 12}$$

Multiply -48 times 2.

$$x=\frac{-\left(-10\right)±\sqrt{100-96}}{2\times 12}$$

Add 100 to -96.

$$x=\frac{-\left(-10\right)±\sqrt{4}}{2\times 12}$$

Take the square root of 4.

$$x=\frac{-\left(-10\right)±2}{2\times 12}$$

The opposite of -10 is 10.

$$x=\frac{10±2}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{10±2}{24}$$

Now solve the equation x=\frac{10±2}{24} when ± is plus. Add 10 to 2.

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

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

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

Now solve the equation x=\frac{10±2}{24} when ± is minus. Subtract 2 from 10.

$$x=\frac{8}{24}$$

Reduce the fraction \frac{8}{24} to lowest terms by extracting and canceling out 8.

$$x=\frac{1}{3}$$

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

$$12x^{2}-10x+2=12\left(x-\frac{1}{2}\right)\left(x-\frac{1}{3}\right)$$

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

$$12x^{2}-10x+2=12\times \frac{2x-1}{2}\left(x-\frac{1}{3}\right)$$

Subtract \frac{1}{3} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$12x^{2}-10x+2=12\times \frac{2x-1}{2}\times \frac{3x-1}{3}$$

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

$$12x^{2}-10x+2=12\times \frac{\left(2x-1\right)\left(3x-1\right)}{2\times 3}$$

Multiply 2 times 3.

$$12x^{2}-10x+2=12\times \frac{\left(2x-1\right)\left(3x-1\right)}{6}$$

Cancel out 6, the greatest common factor in 12 and 6.

$$12x^{2}-10x+2=2\left(2x-1\right)\left(3x-1\right)$$

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

$$x ^ 2 -\frac{5}{6}x +\frac{1}{6} = 0$$

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

$$r + s = \frac{5}{6} rs = \frac{1}{6}$$

Two numbers r and s sum up to \frac{5}{6} exactly when the average of the two numbers is \frac{1}{2}*\frac{5}{6} = \frac{5}{12}. 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>

$$r = \frac{5}{12} - u s = \frac{5}{12} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{1}{6}

$$(\frac{5}{12} - u) (\frac{5}{12} + u) = \frac{1}{6}$$

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

$$\frac{25}{144} - u^2 = \frac{1}{6}$$

Simplify the expression by subtracting \frac{25}{144} on both sides

$$-u^2 = \frac{1}{6}-\frac{25}{144} = -\frac{1}{144}$$

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

$$u^2 = \frac{1}{144} u = \pm\sqrt{\frac{1}{144}} = \pm \frac{1}{12}$$

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

$$r =\frac{5}{12} - \frac{1}{12} = 0.333 s = \frac{5}{12} + \frac{1}{12} = 0.500$$

Evaluate

$$12x^{2}-10x+2$$