# Math: Slide and Divide

*Explaining the Mystery*

As an engineer, I have had numerous high school and college math classes:

- Algebra I
- Algebra II
- Geometry
- Trigonometry
- Calculus I
- Calculus II
- Calculus III
- Differential Equations
- Numerical Methods

Yet, I never cease to be amazed by the so-called "new math". One of the things that threw me for a loop the other day is when my kid asked me for an explanation for this factoring of a binomial. He had written down every step the teacher had written on the board, but was completely befuddled by the what he had copied verbatim. And quite honestly, I did not understand it either because it seemingly violates the laws of algebra, yet allegedly produces the correct answer:

**Line 1:** $$ 3x^2 - 17x + 20 $$

**Line 2:** $$ x^2 - 17x + 60 $$

**Line 3:** $$ (x - 5)(x - 12) $$

**Line 4:** $$ \left( x - \frac{5}{3} \right) \left( x - \frac{12}{3} \right) $$

**Line 5:** $$ (3x - 5)(x - 4) $$

Some of that makes sense. But, the reason I say it "seemingly violates the laws of algebra" is because **there is not a way using algebra to get from Line 1 to Line 2. Similarly, there is no way to get from Line 3 to Line 4 using algebra**. Based on the *distributive property* ...

- If Line 1 is divided by 3 to remove the coefficient 3 from the first addend 3x
^{2}then the second and third addends -17x and 20 must also be divided by 3, but they are not. - If Line 1 is multiplied by 3 to triple the third addend 20 then the first and second addends 3x
^{2}and -17x must also be multiplied by 3, but they are not. - If Line 3 is divided by 3
^{2}such that (x - 5) and (x - 12) are each divided by 3, then the x variables should also be divided by 3, but they are not.

As it turns out, the steps in the example above are not actually algebraic steps. Rather, **they are steps of an algorithm**. If all of the steps of the algorithm are executed correctly, then the algorithm yields the correct result. This algorithm goes by several different names, with the more common names shown below:

- Slide and Divide
- Swing and Divide
- Slip and Slide

After some deep thought on this apparent conundrum, I concluded that **the variable x that exists on Lines 1, 4 & 5 is not actually the same variable x that exists on Line 2 & 3**. That is why there is not an algebraic way to move from Line 1 to Line 2, or from Line 3 to Line 4! But, the steps of the algorithm can be reconciled with algebra by changing the x variable on Lines 2 & 3 to a new variable, named T for "temporary":

**Line 1:** Original Binomial

$$ 3x^2 - 17x + 20 $$

**Line 2:** Slide the 3 and Change the Variable Name

$$ T^2 - 17T + 60 $$

**Line 3:** Factor the Binomial

$$ (T - 5)(T - 12) $$

**Line 4:** Divide by 3 and Restore the Variable Name

$$ \left( x - \frac{5}{3} \right) \left( x - \frac{12}{3} \right) $$

**Line 5:** Simplify to Standard Form

$$ (3x - 5)(x - 4) $$

Once the temporary variable T is used, the lines above are actually algebra with some of the intermediate steps removed. The rest of this post is dedicated to "show all the work" so people can understand why it works instead of just blindly trusting in a confusing magic sequence of steps.

In order to factor the following binomial (which is assumed to equal zero)

Line 1:$$ 3x^2 - 17x + 20 $$

**First**, divide by 3 to remove the coefficient of the first addend, and simplify

$$ \frac{3x^2 - 17x + 20}{3} $$

$$ \frac{3x^2}{3} - \frac{17x}{3} + \frac{20}{3} $$

$$ x^2 - \frac{17}{3}x + \frac{20}{3} $$

**Second**, define variable x in terms of a new, temporary variable T divided by the original coefficient of the first addend.

$$ x = \frac{T}{3} $$

**Third**, substitute the new definition of x back into the equation, and simplify

$$ x^2 - \frac{17}{3}x + \frac{20}{3} $$

$$ \left ( \frac{T}{3} \right )^2 - \left( \frac{17}{3} \right) \left(\frac{T}{3} \right) + \frac{20}{3} $$

$$ \frac{T^2}{3^2} - \frac{17T}{3^2} + \frac{20}{3} $$

**Fourth**, multiple by 3^{2} to remove the fractions on the first and second addends in the expression above, and simplify.

$$ 3^2 \left( \frac{T^2}{3^2} - \frac{17T}{3^2} + \frac{20}{3} \right) $$

$$ 3^2 \left( \frac{T^2}{3^2} \right) - 3^2 \left( \frac{17T}{3^2} \right) + 3^2 \left( \frac{20}{3} \right) $$

$$ T^2 - 17T + 3 \bullet 20 $$

Line 2:$$ T^2 - 17T + 60 $$

There it is: **the slide**:

- The original coefficient, 3, of the first addend has been removed from the first addend
- Nothing has been done to the original coefficient, -17, of the second addend
- The original coefficient, 3, of the first addend has been multiplied by the original coefficient, 20, of the third addend

**Fifth**, factor the new binomial that is written in terms of T.

$$ T^2 - 17T + 60 $$

Line 3:$$ (T - 5)(T - 12) $$

**Sixth**, find the definition of T, and substitute it back into the equation above:

$$ x = \frac{T}{3} $$

$$ T = 3x $$

$$ (T - 5)(T - 12) $$

$$ (3x - 5)(3x - 12) $$

**Seventh**, divide the expression above by 3^{2} to remove the coefficient, 3, from both terms, and simplify.

$$ (3x - 5)(3x - 12) $$

$$ \frac{(3x - 5)(3x - 12)}{3^2} $$

$$ \left( \frac{3x - 5}{3} \right) \left( \frac{ 3x - 12}{3} \right) $$

Line 4:$$ \left( x - \frac{5}{3} \right) \left( x - \frac{12}{3} \right) $$

There it is: **the divide**: The factors obtained after doing the slide, 5 and 12, have been divided by the original coefficient, 3, of the first addend.

**Finally**, the factorization can be simplified into common form:

Line 5:$$ (3x - 5)(x - 4) $$

Programmer, Engineer