Solve 3c^2-20c+32

$$3 c ^ { 2 } - 20 c + 32$$

Factor

$$\left(c-4\right)\left(3c-8\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 3c^{2}+ac+bc+32. To find a and b, set up a system to be solved.

$$a+b=-20 ab=3\times 32=96$$

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 96.

$$-1,-96 -2,-48 -3,-32 -4,-24 -6,-16 -8,-12$$

Calculate the sum for each pair.

$$-1-96=-97 -2-48=-50 -3-32=-35 -4-24=-28 -6-16=-22 -8-12=-20$$

The solution is the pair that gives sum -20.

$$a=-12 b=-8$$

Rewrite 3c^{2}-20c+32 as \left(3c^{2}-12c\right)+\left(-8c+32\right).

$$\left(3c^{2}-12c\right)+\left(-8c+32\right)$$

Factor out 3c in the first and -8 in the second group.

$$3c\left(c-4\right)-8\left(c-4\right)$$

Factor out common term c-4 by using distributive property.

$$\left(c-4\right)\left(3c-8\right)$$

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.

$$3c^{2}-20c+32=0$$

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.

$$c=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 3\times 32}}{2\times 3}$$

Square -20.

$$c=\frac{-\left(-20\right)±\sqrt{400-4\times 3\times 32}}{2\times 3}$$

Multiply -4 times 3.

$$c=\frac{-\left(-20\right)±\sqrt{400-12\times 32}}{2\times 3}$$

Multiply -12 times 32.

$$c=\frac{-\left(-20\right)±\sqrt{400-384}}{2\times 3}$$

Add 400 to -384.

$$c=\frac{-\left(-20\right)±\sqrt{16}}{2\times 3}$$

Take the square root of 16.

$$c=\frac{-\left(-20\right)±4}{2\times 3}$$

The opposite of -20 is 20.

$$c=\frac{20±4}{2\times 3}$$

Multiply 2 times 3.

$$c=\frac{20±4}{6}$$

Now solve the equation c=\frac{20±4}{6} when ± is plus. Add 20 to 4.

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

Divide 24 by 6.

$$c=4$$

Now solve the equation c=\frac{20±4}{6} when ± is minus. Subtract 4 from 20.

$$c=\frac{16}{6}$$

Reduce the fraction \frac{16}{6} to lowest terms by extracting and canceling out 2.

$$c=\frac{8}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 4 for x_{1} and \frac{8}{3} for x_{2}.

$$3c^{2}-20c+32=3\left(c-4\right)\left(c-\frac{8}{3}\right)$$

Subtract \frac{8}{3} from c by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$3c^{2}-20c+32=3\left(c-4\right)\times \frac{3c-8}{3}$$

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

$$3c^{2}-20c+32=\left(c-4\right)\left(3c-8\right)$$

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

$$x ^ 2 -\frac{20}{3}x +\frac{32}{3} = 0$$

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

$$r + s = \frac{20}{3} rs = \frac{32}{3}$$

Two numbers r and s sum up to \frac{20}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{20}{3} = \frac{10}{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>

$$r = \frac{10}{3} - u s = \frac{10}{3} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{32}{3}

$$(\frac{10}{3} - u) (\frac{10}{3} + u) = \frac{32}{3}$$

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

$$\frac{100}{9} - u^2 = \frac{32}{3}$$

Simplify the expression by subtracting \frac{100}{9} on both sides

$$-u^2 = \frac{32}{3}-\frac{100}{9} = -\frac{4}{9}$$

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

$$u^2 = \frac{4}{9} u = \pm\sqrt{\frac{4}{9}} = \pm \frac{2}{3}$$

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

$$r =\frac{10}{3} - \frac{2}{3} = 2.667 s = \frac{10}{3} + \frac{2}{3} = 4.000$$

Evaluate

$$\left(c-4\right)\left(3c-8\right)$$