Solve 18d^2-96d+78

$$18 d ^ { 2 } - 96 d + 78$$

Factor

$$6\left(d-1\right)\left(3d-13\right)$$

Solution Steps

Factor out 6.

$$6\left(3d^{2}-16d+13\right)$$

Consider 3d^{2}-16d+13. Factor the expression by grouping. First, the expression needs to be rewritten as 3d^{2}+ad+bd+13. To find a and b, set up a system to be solved.

$$a+b=-16 ab=3\times 13=39$$

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 39.

$$-1,-39 -3,-13$$

Calculate the sum for each pair.

$$-1-39=-40 -3-13=-16$$

The solution is the pair that gives sum -16.

$$a=-13 b=-3$$

Rewrite 3d^{2}-16d+13 as \left(3d^{2}-13d\right)+\left(-3d+13\right).

$$\left(3d^{2}-13d\right)+\left(-3d+13\right)$$

Factor out d in the first and -1 in the second group.

$$d\left(3d-13\right)-\left(3d-13\right)$$

Factor out common term 3d-13 by using distributive property.

$$\left(3d-13\right)\left(d-1\right)$$

Rewrite the complete factored expression.

$$6\left(3d-13\right)\left(d-1\right)$$

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.

$$18d^{2}-96d+78=0$$

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.

$$d=\frac{-\left(-96\right)±\sqrt{\left(-96\right)^{2}-4\times 18\times 78}}{2\times 18}$$

Square -96.

$$d=\frac{-\left(-96\right)±\sqrt{9216-4\times 18\times 78}}{2\times 18}$$

Multiply -4 times 18.

$$d=\frac{-\left(-96\right)±\sqrt{9216-72\times 78}}{2\times 18}$$

Multiply -72 times 78.

$$d=\frac{-\left(-96\right)±\sqrt{9216-5616}}{2\times 18}$$

Add 9216 to -5616.

$$d=\frac{-\left(-96\right)±\sqrt{3600}}{2\times 18}$$

Take the square root of 3600.

$$d=\frac{-\left(-96\right)±60}{2\times 18}$$

The opposite of -96 is 96.

$$d=\frac{96±60}{2\times 18}$$

Multiply 2 times 18.

$$d=\frac{96±60}{36}$$

Now solve the equation d=\frac{96±60}{36} when ± is plus. Add 96 to 60.

$$d=\frac{156}{36}$$

Reduce the fraction \frac{156}{36} to lowest terms by extracting and canceling out 12.

$$d=\frac{13}{3}$$

Now solve the equation d=\frac{96±60}{36} when ± is minus. Subtract 60 from 96.

$$d=\frac{36}{36}$$

Divide 36 by 36.

$$d=1$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{13}{3} for x_{1} and 1 for x_{2}.

$$18d^{2}-96d+78=18\left(d-\frac{13}{3}\right)\left(d-1\right)$$

Subtract \frac{13}{3} from d by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$18d^{2}-96d+78=18\times \frac{3d-13}{3}\left(d-1\right)$$

Cancel out 3, the greatest common factor in 18 and 3.

$$18d^{2}-96d+78=6\left(3d-13\right)\left(d-1\right)$$

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 18

$$x ^ 2 -\frac{16}{3}x +\frac{13}{3} = 0$$

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

$$r + s = \frac{16}{3} rs = \frac{13}{3}$$

Two numbers r and s sum up to \frac{16}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{16}{3} = \frac{8}{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>

$$r = \frac{8}{3} - u s = \frac{8}{3} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{13}{3}

$$(\frac{8}{3} - u) (\frac{8}{3} + u) = \frac{13}{3}$$

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

$$\frac{64}{9} - u^2 = \frac{13}{3}$$

Simplify the expression by subtracting \frac{64}{9} on both sides

$$-u^2 = \frac{13}{3}-\frac{64}{9} = -\frac{25}{9}$$

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

$$u^2 = \frac{25}{9} u = \pm\sqrt{\frac{25}{9}} = \pm \frac{5}{3}$$

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

$$r =\frac{8}{3} - \frac{5}{3} = 1.000 s = \frac{8}{3} + \frac{5}{3} = 4.333$$

Evaluate

$$6\left(d-1\right)\left(3d-13\right)$$