Solve 3c^2-11c+6

$$3 c ^ { 2 } - 11 c + 6$$

Factor

$$\left(c-3\right)\left(3c-2\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 3c^{2}+ac+bc+6. To find a and b, set up a system to be solved.

$$a+b=-11 ab=3\times 6=18$$

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 18.

$$-1,-18 -2,-9 -3,-6$$

Calculate the sum for each pair.

$$-1-18=-19 -2-9=-11 -3-6=-9$$

The solution is the pair that gives sum -11.

$$a=-9 b=-2$$

Rewrite 3c^{2}-11c+6 as \left(3c^{2}-9c\right)+\left(-2c+6\right).

$$\left(3c^{2}-9c\right)+\left(-2c+6\right)$$

Factor out 3c in the first and -2 in the second group.

$$3c\left(c-3\right)-2\left(c-3\right)$$

Factor out common term c-3 by using distributive property.

$$\left(c-3\right)\left(3c-2\right)$$

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.

$$3c^{2}-11c+6=0$$

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.

$$c=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 3\times 6}}{2\times 3}$$

Square -11.

$$c=\frac{-\left(-11\right)±\sqrt{121-4\times 3\times 6}}{2\times 3}$$

Multiply -4 times 3.

$$c=\frac{-\left(-11\right)±\sqrt{121-12\times 6}}{2\times 3}$$

Multiply -12 times 6.

$$c=\frac{-\left(-11\right)±\sqrt{121-72}}{2\times 3}$$

Add 121 to -72.

$$c=\frac{-\left(-11\right)±\sqrt{49}}{2\times 3}$$

Take the square root of 49.

$$c=\frac{-\left(-11\right)±7}{2\times 3}$$

The opposite of -11 is 11.

$$c=\frac{11±7}{2\times 3}$$

Multiply 2 times 3.

$$c=\frac{11±7}{6}$$

Now solve the equation c=\frac{11±7}{6} when ± is plus. Add 11 to 7.

$$c=\frac{18}{6}$$

Divide 18 by 6.

$$c=3$$

Now solve the equation c=\frac{11±7}{6} when ± is minus. Subtract 7 from 11.

$$c=\frac{4}{6}$$

Reduce the fraction \frac{4}{6} to lowest terms by extracting and canceling out 2.

$$c=\frac{2}{3}$$

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

$$3c^{2}-11c+6=3\left(c-3\right)\left(c-\frac{2}{3}\right)$$

Subtract \frac{2}{3} from c by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$3c^{2}-11c+6=3\left(c-3\right)\times \frac{3c-2}{3}$$

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

$$3c^{2}-11c+6=\left(c-3\right)\left(3c-2\right)$$

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 3

$$x ^ 2 -\frac{11}{3}x +2 = 0$$

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

$$r + s = \frac{11}{3} rs = 2$$

Two numbers r and s sum up to \frac{11}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{11}{3} = \frac{11}{6}. 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>

$$r = \frac{11}{6} - u s = \frac{11}{6} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 2

$$(\frac{11}{6} - u) (\frac{11}{6} + u) = 2$$

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

$$\frac{121}{36} - u^2 = 2$$

Simplify the expression by subtracting \frac{121}{36} on both sides

$$-u^2 = 2-\frac{121}{36} = -\frac{49}{36}$$

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

$$u^2 = \frac{49}{36} u = \pm\sqrt{\frac{49}{36}} = \pm \frac{7}{6}$$

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

$$r =\frac{11}{6} - \frac{7}{6} = 0.667 s = \frac{11}{6} + \frac{7}{6} = 3$$

Evaluate

$$\left(c-3\right)\left(3c-2\right)$$