# The Algebra Announcement

## Harvard Mathematicians Discover "Factoring"

Mathematicians at Harvard University have made history in the math world today, with new, innovative means of solving equations. They call the process "factoring". Factoring an equation shows many a new way to consider algebraic equations. It is especially helpful, Mathematicians claim, in finding the x-intercepts of the equation.

## Factoring with the X-Box

Factoring using the X-box method is probably the most complicated and involves the most steps.

1. Begin with the quadratic, -3x² +4x +4 = 0. Determine the a, b, and c coefficients.

2. Draw an X. In the top of the X, put the product of the a and c coefficients. In the bottom of the X, put the b coefficient.

(a)(c) = top

(-3)(4) = -12.

3. Determine the factors which when multiplied, equal the product of a and c, and when added, equal b. Put the factors on the sides of the X.

(6)(-2) = -12.

6 + -2 = 4.

4. Next, draw a box next to the X. Draw a 2 by 2 inside the square so there are 4 smaller spaces inside the box.

5. Inside the top left space, put the a term from the original quadratic. In the top right, put the first factor from the X. Add an x to the factor. In the bottom left, put the other factor from the X, with an x at the end. In the bottom right, put the c term from the original quadratic.

6. Now you must determine the factors which will be written on the top columns and right side of the Box. Each factor must be a factor for both terms in the row or column. When multiplied together, the product must equal the corresponding term in the space.

(3x)(-x) = -3x².

(3x)(2) = 6x.

(2)(-x) = -2x.

(2)(2) = 4.

7. Now you will set up a new equation using the factors on the outside of the Box. Take the factors on the top of each column and put them in parentheses. Multiply them by the factors on the right side of the box, in parentheses. Set the whole equation equal to zero. The quadratic is now in factored form.

(3x + 2)(-x + 2) = 0.

8. According to the zero product property, each parentheses can equal zero. We can use this to find the solution. Set each parentheses equal to zero and solve for x.

3x +2 = 0.

-x + 2 = 0.

9. Examine the solutions. Real number solutions are where the graph of the quadratic crosses the x-axis. Imaginary solutions are solutions to the quadratic but not x-intercepts of the graph.

## Factoring using the Quadratic Formula

The quadratic formula is very simple to use and works every time. The formula is: x = (-b ± √(b)² - (4)(a)(c)) / (2(a)). a, b, and c terms correspond to the standard form of a quadratic, ax2 +bx + c.

1. Take the quadratic 5x² -6x -11 = 0.

2. Identify the a, b, and c coefficients.

a = 5 b = -6 c = -11

3. Plug the a, b, and c coefficients into the equation.

x = (6 ± √(-6)² - (4)(5)(-11)) / (2(5))

4. Solve the equation.

x = (3 ± √46) / 5.

5. Examine the solutions. Real number solutions are where the graph of the quadratic crosses the x-axis. Imaginary solutions are solutions to the quadratic but not x-intercepts of the graph.

## Factoring by Completing the Square

Completing the square forces the quadratic to be a perfect square trinomial. The perfect square trinomial can be rewritten as the square of a binomial. After square rooting both sides, the equation can be easily solved.

1. Begin with the quadratic 2x² - 4x -8 = 0.

2. Set the a and b terms equal to the c term. Add or subtract the c value on both sides of the equation.

2x² - 4x = 8.

3. Divide by the equation by the a coefficient so it is equal to one.

x² -2x = 4.

4. Add a blank space to both sides of the equation. This helps to keep the equation organized.

x² - 2x + = 4 + .

5. In each blank space put 1/2 of the b term squared. It is helpful to write it as (1/2b)2 unsimplified.

x² - 2x + (1)² = 4 + (1)².

6. Simplify the other side of the equation.

x² -2x + (1)² = 5.

7. You have now made a perfect square trinomial. Next, re-write as the square of a binomial. Put the square root of a term, x, the sign of the b term, and the square root of the c term in parentheses. Square the parentheses.

(x - 1)² = 5.

8. Now you can solve for x. Square root both sides. Simplify.

x = 1 ± √5.

9. Examine the solutions. Real number solutions are where the graph of the quadratic crosses the x-axis. Imaginary solutions are solutions to the quadratic but not x-intercepts of the graph.