Solve 6g^2-39g+18

$$6 g ^ { 2 } - 39 g + 18$$

Factor

$$3\left(g-6\right)\left(2g-1\right)$$

Solution Steps

Factor out 3.

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

Consider 2g^{2}-13g+6. Factor the expression by grouping. First, the expression needs to be rewritten as 2g^{2}+ag+bg+6. To find a and b, set up a system to be solved.

$$a+b=-13 ab=2\times 6=12$$

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 12.

$$-1,-12 -2,-6 -3,-4$$

Calculate the sum for each pair.

$$-1-12=-13 -2-6=-8 -3-4=-7$$

The solution is the pair that gives sum -13.

$$a=-12 b=-1$$

Rewrite 2g^{2}-13g+6 as \left(2g^{2}-12g\right)+\left(-g+6\right).

$$\left(2g^{2}-12g\right)+\left(-g+6\right)$$

Factor out 2g in the first and -1 in the second group.

$$2g\left(g-6\right)-\left(g-6\right)$$

Factor out common term g-6 by using distributive property.

$$\left(g-6\right)\left(2g-1\right)$$

Rewrite the complete factored expression.

$$3\left(g-6\right)\left(2g-1\right)$$

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.

$$6g^{2}-39g+18=0$$

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.

$$g=\frac{-\left(-39\right)±\sqrt{\left(-39\right)^{2}-4\times 6\times 18}}{2\times 6}$$

Square -39.

$$g=\frac{-\left(-39\right)±\sqrt{1521-4\times 6\times 18}}{2\times 6}$$

Multiply -4 times 6.

$$g=\frac{-\left(-39\right)±\sqrt{1521-24\times 18}}{2\times 6}$$

Multiply -24 times 18.

$$g=\frac{-\left(-39\right)±\sqrt{1521-432}}{2\times 6}$$

Add 1521 to -432.

$$g=\frac{-\left(-39\right)±\sqrt{1089}}{2\times 6}$$

Take the square root of 1089.

$$g=\frac{-\left(-39\right)±33}{2\times 6}$$

The opposite of -39 is 39.

$$g=\frac{39±33}{2\times 6}$$

Multiply 2 times 6.

$$g=\frac{39±33}{12}$$

Now solve the equation g=\frac{39±33}{12} when ± is plus. Add 39 to 33.

$$g=\frac{72}{12}$$

Divide 72 by 12.

$$g=6$$

Now solve the equation g=\frac{39±33}{12} when ± is minus. Subtract 33 from 39.

$$g=\frac{6}{12}$$

Reduce the fraction \frac{6}{12} to lowest terms by extracting and canceling out 6.

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 6 for x_{1} and \frac{1}{2} for x_{2}.

$$6g^{2}-39g+18=6\left(g-6\right)\left(g-\frac{1}{2}\right)$$

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

$$6g^{2}-39g+18=6\left(g-6\right)\times \frac{2g-1}{2}$$

Cancel out 2, the greatest common factor in 6 and 2.

$$6g^{2}-39g+18=3\left(g-6\right)\left(2g-1\right)$$

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 6

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

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

$$r + s = \frac{13}{2} rs = 3$$

Two numbers r and s sum up to \frac{13}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{13}{2} = \frac{13}{4}. 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>

$$r = \frac{13}{4} - u s = \frac{13}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 3

$$(\frac{13}{4} - u) (\frac{13}{4} + u) = 3$$

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

$$\frac{169}{16} - u^2 = 3$$

Simplify the expression by subtracting \frac{169}{16} on both sides

$$-u^2 = 3-\frac{169}{16} = -\frac{121}{16}$$

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

$$u^2 = \frac{121}{16} u = \pm\sqrt{\frac{121}{16}} = \pm \frac{11}{4}$$

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

$$r =\frac{13}{4} - \frac{11}{4} = 0.500 s = \frac{13}{4} + \frac{11}{4} = 6$$

Evaluate

$$3\left(g-6\right)\left(2g-1\right)$$