Solve 112x^2-62x+6=

$$112 x ^ { 2 } - 62 x + 6 =$$

Factor

$$2\left(7x-3\right)\left(8x-1\right)$$

Solution Steps

Factor out 2.

$$2\left(56x^{2}-31x+3\right)$$

Consider 56x^{2}-31x+3. Factor the expression by grouping. First, the expression needs to be rewritten as 56x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

$$a+b=-31 ab=56\times 3=168$$

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 168.

$$-1,-168 -2,-84 -3,-56 -4,-42 -6,-28 -7,-24 -8,-21 -12,-14$$

Calculate the sum for each pair.

$$-1-168=-169 -2-84=-86 -3-56=-59 -4-42=-46 -6-28=-34 -7-24=-31 -8-21=-29 -12-14=-26$$

The solution is the pair that gives sum -31.

$$a=-24 b=-7$$

Rewrite 56x^{2}-31x+3 as \left(56x^{2}-24x\right)+\left(-7x+3\right).

$$\left(56x^{2}-24x\right)+\left(-7x+3\right)$$

Factor out 8x in the first and -1 in the second group.

$$8x\left(7x-3\right)-\left(7x-3\right)$$

Factor out common term 7x-3 by using distributive property.

$$\left(7x-3\right)\left(8x-1\right)$$

Rewrite the complete factored expression.

$$2\left(7x-3\right)\left(8x-1\right)$$

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.

$$112x^{2}-62x+6=0$$

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.

$$x=\frac{-\left(-62\right)±\sqrt{\left(-62\right)^{2}-4\times 112\times 6}}{2\times 112}$$

Square -62.

$$x=\frac{-\left(-62\right)±\sqrt{3844-4\times 112\times 6}}{2\times 112}$$

Multiply -4 times 112.

$$x=\frac{-\left(-62\right)±\sqrt{3844-448\times 6}}{2\times 112}$$

Multiply -448 times 6.

$$x=\frac{-\left(-62\right)±\sqrt{3844-2688}}{2\times 112}$$

Add 3844 to -2688.

$$x=\frac{-\left(-62\right)±\sqrt{1156}}{2\times 112}$$

Take the square root of 1156.

$$x=\frac{-\left(-62\right)±34}{2\times 112}$$

The opposite of -62 is 62.

$$x=\frac{62±34}{2\times 112}$$

Multiply 2 times 112.

$$x=\frac{62±34}{224}$$

Now solve the equation x=\frac{62±34}{224} when ± is plus. Add 62 to 34.

$$x=\frac{96}{224}$$

Reduce the fraction \frac{96}{224} to lowest terms by extracting and canceling out 32.

$$x=\frac{3}{7}$$

Now solve the equation x=\frac{62±34}{224} when ± is minus. Subtract 34 from 62.

$$x=\frac{28}{224}$$

Reduce the fraction \frac{28}{224} to lowest terms by extracting and canceling out 28.

$$x=\frac{1}{8}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{3}{7} for x_{1} and \frac{1}{8} for x_{2}.

$$112x^{2}-62x+6=112\left(x-\frac{3}{7}\right)\left(x-\frac{1}{8}\right)$$

Subtract \frac{3}{7} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$112x^{2}-62x+6=112\times \frac{7x-3}{7}\left(x-\frac{1}{8}\right)$$

Subtract \frac{1}{8} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$112x^{2}-62x+6=112\times \frac{7x-3}{7}\times \frac{8x-1}{8}$$

Multiply \frac{7x-3}{7} times \frac{8x-1}{8} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$112x^{2}-62x+6=112\times \frac{\left(7x-3\right)\left(8x-1\right)}{7\times 8}$$

Multiply 7 times 8.

$$112x^{2}-62x+6=112\times \frac{\left(7x-3\right)\left(8x-1\right)}{56}$$

Cancel out 56, the greatest common factor in 112 and 56.

$$112x^{2}-62x+6=2\left(7x-3\right)\left(8x-1\right)$$

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 112

$$x ^ 2 -\frac{31}{56}x +\frac{3}{56} = 0$$

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

$$r + s = \frac{31}{56} rs = \frac{3}{56}$$

Two numbers r and s sum up to \frac{31}{56} exactly when the average of the two numbers is \frac{1}{2}*\frac{31}{56} = \frac{31}{112}. 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>

$$r = \frac{31}{112} - u s = \frac{31}{112} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{3}{56}

$$(\frac{31}{112} - u) (\frac{31}{112} + u) = \frac{3}{56}$$

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

$$\frac{961}{12544} - u^2 = \frac{3}{56}$$

Simplify the expression by subtracting \frac{961}{12544} on both sides

$$-u^2 = \frac{3}{56}-\frac{961}{12544} = -\frac{289}{12544}$$

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

$$u^2 = \frac{289}{12544} u = \pm\sqrt{\frac{289}{12544}} = \pm \frac{17}{112}$$

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

$$r =\frac{31}{112} - \frac{17}{112} = 0.125 s = \frac{31}{112} + \frac{17}{112} = 0.429$$

Evaluate

$$112x^{2}-62x+6$$