Solve 56c^2+31c+3

$$56 c ^ { 2 } + 31 c + 3$$

Factor

$$\left(7c+3\right)\left(8c+1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 56c^{2}+ac+bc+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 positive, a and b are both positive. 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.

$$a=7 b=24$$

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

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

Factor out 7c in the first and 3 in the second group.

$$7c\left(8c+1\right)+3\left(8c+1\right)$$

Factor out common term 8c+1 by using distributive property.

$$\left(8c+1\right)\left(7c+3\right)$$

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.

$$56c^{2}+31c+3=0$$

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.

$$c=\frac{-31±\sqrt{31^{2}-4\times 56\times 3}}{2\times 56}$$

Square 31.

$$c=\frac{-31±\sqrt{961-4\times 56\times 3}}{2\times 56}$$

Multiply -4 times 56.

$$c=\frac{-31±\sqrt{961-224\times 3}}{2\times 56}$$

Multiply -224 times 3.

$$c=\frac{-31±\sqrt{961-672}}{2\times 56}$$

Add 961 to -672.

$$c=\frac{-31±\sqrt{289}}{2\times 56}$$

Take the square root of 289.

$$c=\frac{-31±17}{2\times 56}$$

Multiply 2 times 56.

$$c=\frac{-31±17}{112}$$

Now solve the equation c=\frac{-31±17}{112} when ± is plus. Add -31 to 17.

$$c=-\frac{14}{112}$$

Reduce the fraction \frac{-14}{112} to lowest terms by extracting and canceling out 14.

$$c=-\frac{1}{8}$$

Now solve the equation c=\frac{-31±17}{112} when ± is minus. Subtract 17 from -31.

$$c=-\frac{48}{112}$$

Reduce the fraction \frac{-48}{112} to lowest terms by extracting and canceling out 16.

$$c=-\frac{3}{7}$$

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

$$56c^{2}+31c+3=56\left(c-\left(-\frac{1}{8}\right)\right)\left(c-\left(-\frac{3}{7}\right)\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$56c^{2}+31c+3=56\left(c+\frac{1}{8}\right)\left(c+\frac{3}{7}\right)$$

Add \frac{1}{8} to c by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$56c^{2}+31c+3=56\times \frac{8c+1}{8}\left(c+\frac{3}{7}\right)$$

Add \frac{3}{7} to c by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$56c^{2}+31c+3=56\times \frac{8c+1}{8}\times \frac{7c+3}{7}$$

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

$$56c^{2}+31c+3=56\times \frac{\left(8c+1\right)\left(7c+3\right)}{8\times 7}$$

Multiply 8 times 7.

$$56c^{2}+31c+3=56\times \frac{\left(8c+1\right)\left(7c+3\right)}{56}$$

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

$$56c^{2}+31c+3=\left(8c+1\right)\left(7c+3\right)$$

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 56

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

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.

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

Evaluate

$$\left(7c+3\right)\left(8c+1\right)$$