Solve 12x^2-31x+7

$$12 x ^ { 2 } - 31 x + 7$$

Factor

$$\left(3x-7\right)\left(4x-1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 12x^{2}+ax+bx+7. To find a and b, set up a system to be solved.

$$a+b=-31 ab=12\times 7=84$$

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 84.

$$-1,-84 -2,-42 -3,-28 -4,-21 -6,-14 -7,-12$$

Calculate the sum for each pair.

$$-1-84=-85 -2-42=-44 -3-28=-31 -4-21=-25 -6-14=-20 -7-12=-19$$

The solution is the pair that gives sum -31.

$$a=-28 b=-3$$

Rewrite 12x^{2}-31x+7 as \left(12x^{2}-28x\right)+\left(-3x+7\right).

$$\left(12x^{2}-28x\right)+\left(-3x+7\right)$$

Factor out 4x in the first and -1 in the second group.

$$4x\left(3x-7\right)-\left(3x-7\right)$$

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

$$\left(3x-7\right)\left(4x-1\right)$$

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.

$$12x^{2}-31x+7=0$$

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.

$$x=\frac{-\left(-31\right)±\sqrt{\left(-31\right)^{2}-4\times 12\times 7}}{2\times 12}$$

Square -31.

$$x=\frac{-\left(-31\right)±\sqrt{961-4\times 12\times 7}}{2\times 12}$$

Multiply -4 times 12.

$$x=\frac{-\left(-31\right)±\sqrt{961-48\times 7}}{2\times 12}$$

Multiply -48 times 7.

$$x=\frac{-\left(-31\right)±\sqrt{961-336}}{2\times 12}$$

Add 961 to -336.

$$x=\frac{-\left(-31\right)±\sqrt{625}}{2\times 12}$$

Take the square root of 625.

$$x=\frac{-\left(-31\right)±25}{2\times 12}$$

The opposite of -31 is 31.

$$x=\frac{31±25}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{31±25}{24}$$

Now solve the equation x=\frac{31±25}{24} when ± is plus. Add 31 to 25.

$$x=\frac{56}{24}$$

Reduce the fraction \frac{56}{24} to lowest terms by extracting and canceling out 8.

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

Now solve the equation x=\frac{31±25}{24} when ± is minus. Subtract 25 from 31.

$$x=\frac{6}{24}$$

Reduce the fraction \frac{6}{24} to lowest terms by extracting and canceling out 6.

$$x=\frac{1}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{7}{3} for x_{1} and \frac{1}{4} for x_{2}.

$$12x^{2}-31x+7=12\left(x-\frac{7}{3}\right)\left(x-\frac{1}{4}\right)$$

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

$$12x^{2}-31x+7=12\times \frac{3x-7}{3}\left(x-\frac{1}{4}\right)$$

Subtract \frac{1}{4} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$12x^{2}-31x+7=12\times \frac{3x-7}{3}\times \frac{4x-1}{4}$$

Multiply \frac{3x-7}{3} times \frac{4x-1}{4} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$12x^{2}-31x+7=12\times \frac{\left(3x-7\right)\left(4x-1\right)}{3\times 4}$$

Multiply 3 times 4.

$$12x^{2}-31x+7=12\times \frac{\left(3x-7\right)\left(4x-1\right)}{12}$$

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

$$12x^{2}-31x+7=\left(3x-7\right)\left(4x-1\right)$$

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

$$x ^ 2 -\frac{31}{12}x +\frac{7}{12} = 0$$

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

$$r + s = \frac{31}{12} rs = \frac{7}{12}$$

Two numbers r and s sum up to \frac{31}{12} exactly when the average of the two numbers is \frac{1}{2}*\frac{31}{12} = \frac{31}{24}. 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>

$$r = \frac{31}{24} - u s = \frac{31}{24} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{7}{12}

$$(\frac{31}{24} - u) (\frac{31}{24} + u) = \frac{7}{12}$$

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

$$\frac{961}{576} - u^2 = \frac{7}{12}$$

Simplify the expression by subtracting \frac{961}{576} on both sides

$$-u^2 = \frac{7}{12}-\frac{961}{576} = -\frac{625}{576}$$

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

$$u^2 = \frac{625}{576} u = \pm\sqrt{\frac{625}{576}} = \pm \frac{25}{24}$$

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

$$r =\frac{31}{24} - \frac{25}{24} = 0.250 s = \frac{31}{24} + \frac{25}{24} = 2.333$$

Evaluate

$$\left(3x-7\right)\left(4x-1\right)$$