Solve 3m^2-32m+20

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

Factor

$$\left(m-10\right)\left(3m-2\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 3m^{2}+am+bm+20. To find a and b, set up a system to be solved.

$$a+b=-32 ab=3\times 20=60$$

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 60.

$$-1,-60 -2,-30 -3,-20 -4,-15 -5,-12 -6,-10$$

Calculate the sum for each pair.

$$-1-60=-61 -2-30=-32 -3-20=-23 -4-15=-19 -5-12=-17 -6-10=-16$$

The solution is the pair that gives sum -32.

$$a=-30 b=-2$$

Rewrite 3m^{2}-32m+20 as \left(3m^{2}-30m\right)+\left(-2m+20\right).

$$\left(3m^{2}-30m\right)+\left(-2m+20\right)$$

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

$$3m\left(m-10\right)-2\left(m-10\right)$$

Factor out common term m-10 by using distributive property.

$$\left(m-10\right)\left(3m-2\right)$$

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.

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

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.

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

Square -32.

$$m=\frac{-\left(-32\right)±\sqrt{1024-4\times 3\times 20}}{2\times 3}$$

Multiply -4 times 3.

$$m=\frac{-\left(-32\right)±\sqrt{1024-12\times 20}}{2\times 3}$$

Multiply -12 times 20.

$$m=\frac{-\left(-32\right)±\sqrt{1024-240}}{2\times 3}$$

Add 1024 to -240.

$$m=\frac{-\left(-32\right)±\sqrt{784}}{2\times 3}$$

Take the square root of 784.

$$m=\frac{-\left(-32\right)±28}{2\times 3}$$

The opposite of -32 is 32.

$$m=\frac{32±28}{2\times 3}$$

Multiply 2 times 3.

$$m=\frac{32±28}{6}$$

Now solve the equation m=\frac{32±28}{6} when ± is plus. Add 32 to 28.

$$m=\frac{60}{6}$$

Divide 60 by 6.

$$m=10$$

Now solve the equation m=\frac{32±28}{6} when ± is minus. Subtract 28 from 32.

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

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

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 10 for x_{1} and \frac{2}{3} for x_{2}.

$$3m^{2}-32m+20=3\left(m-10\right)\left(m-\frac{2}{3}\right)$$

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

$$3m^{2}-32m+20=3\left(m-10\right)\times \frac{3m-2}{3}$$

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

$$3m^{2}-32m+20=\left(m-10\right)\left(3m-2\right)$$

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

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

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

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

Two numbers r and s sum up to \frac{32}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{32}{3} = \frac{16}{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>

$$r = \frac{16}{3} - u s = \frac{16}{3} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{20}{3}

$$(\frac{16}{3} - u) (\frac{16}{3} + u) = \frac{20}{3}$$

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

$$\frac{256}{9} - u^2 = \frac{20}{3}$$

Simplify the expression by subtracting \frac{256}{9} on both sides

$$-u^2 = \frac{20}{3}-\frac{256}{9} = -\frac{196}{9}$$

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

$$u^2 = \frac{196}{9} u = \pm\sqrt{\frac{196}{9}} = \pm \frac{14}{3}$$

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

$$r =\frac{16}{3} - \frac{14}{3} = 0.667 s = \frac{16}{3} + \frac{14}{3} = 10$$

Evaluate

$$\left(m-10\right)\left(3m-2\right)$$