## Saturday, October 19, 2019

### Re-learning and Enjoying Polynomial Division with students.

They had just learned how to multiply two linear binomial expressions like
(x + 2).(x - 5) , (3x - 2).(5 - 4x) , etc. in two ways -  pictorially as well as symbolically (i.e. by expanding).

I now wanted to see their approach for the division problems, like for problems of the type:
(x^2 + 5x + 6) / (x + 2)

Polynomial division was not yet taught to them in their school. So I should have first given them a simpler problem like the example above (all +ve terms), however for some reason I directly pushed them into the challenging zone this time. This was the problem I gave;

(x^2 + 3x - 10) / (x - 2)

I would suggest you to pause and think for a while - to assure yourself as to why this could be a little difficult problem, esp for those who do not know the procedure to solve this... See if you can solve this problem using a way which was not taught to you :)

-----

After some time, one of them came to me and showed me his final result. He said it is 3 wholes and (x^2 - 4) / (x-2) . I must confess at this moment that I was completely surprised by such an answer. And I wonder if you too would expect or have seen the quotient of polynomial division in such a form.

I suggested him to go and explain his approach on the board for others to know. And this is what he did -

Note his diagrammatic representation for the expression (x^2 + 3x - 10) . The ten small shaded circles represent negative ten. Then he explained -

We need to divide  (x^2 + 3x - 10)   by  the expression  (x - 2)  , which means we need to find out how many (x-2) are there in (x^2 + 3x - 10).

Looking at the terms 3x and -6, we can see that (x - 2)  is present 3 whole number of times in this expression. So what is left now is (x^2 - 4),  which when divided by (x - 2) will give us  (x^2 - 4) / (x - 2).

When one of the students did not understand this, he gave an explanation using a numerical example of 4 / 3 (he did not fully write the long division process till the end (remainder=0), but he explained the process well verbally).

Everyone agreed with this result. I asked them if we could verify this, to which one of them said - Yes, we can multiply and check. And this is what he did.

It was an Aha moment for me!  What about you?  :-)

Now, there was a student who said that he had got a different answer. Others were surprised by this remark. He was asked to come forward and explain his approach.

This is his work:

- We need to find out (x-2) multiplied by what will give us (x^2 + 3x - 10)

- Since there is one x^2 term, it means that the multiplier of (x-2) should have at least one 'x' term..[ while saying this, he wrote 'x' next to (x-2) ]

- Now, multiplying this 'x' by (x-2)  gives us x^2 - 2x

- But we need +3x  and not -2x.... So we need to add +5 to this multiplier 'x' so that this +5 after multiplying by 'x' of the expression (x - 2) gives +5x , which after combining with the -2x we have, will effectively yield the desired +3x  [while saying this, he wrote +5 next to 'x']

- Further, this +5 and -2 will multiply to give -10 too...So the answer is (x+5)

"So is the movie over?", I asked them with the hope that they should loudly say - NO.

And yes, they did not disappoint me :)

We now need to prove that both these answers are same.

"Oh is it? Why can't these 2 answers be different ? We have seen problems having multiple correct answers", I continue probing them as if I was unaware of whats going on in their mind.

Sir, how can division of the same set of 2 numbers give different answers?, argued one of them.

"Hmm... But these 2 expressions look completely different,", - my counter.

Yes, but then they should be equivalent....

"Have we seen such cases earlier ?"

Yes, many times.., came their quick reply.

I felt a sense of accomplishment with this conversation... So now their goal was -

I doubt if we have seen a T.P.T. algebraic statement of this form in any textbook  :)
Thanks to my students, they keep offering me numerous wonderful learning opportunities !

Do not hesitate in pausing for a while, trying to prove this on your own first.

-----

People with some algebraic knowledge will mostly and quickly factorize the expression in Numerator as (x+2)(x-2) and 'cancel off' one of the factors viz. (x+2) with the expression in Denominator...  But would this exercise leave us with (x+5) as desired  ?

And before you say yes, let me just tell you that these students are Not even aware of this identity of difference of squares [ a^ - b^2 = (a+b)(a+b) ], which we could instantly see and use.

So then how would they go ahead??

Yes, that's the interesting part which even keeps me on toes while I work with them.

I saw that they were just staring at these expressions for some time. I thought that I should intervene and offer them some clue and I did that. But soon I realized that I was wrong in doing so.

I asked them "what does the expression on the left look like?"

They said, it resembles a Fraction.

"Yes, and what about the one on right side"

They said, it looks like a Whole number.

When asked for the reason, they said - RHS expression (x+5) has the denominator=1.

"So what do you think, what should happen in the left side expression?", I ask them.

One of them quickly said, (x-2) should be factor of (x^2 - 4)...

And why so ?

Only then its denominator will go away.

While I was about to relish with this thought process, meanwhile one of them was already scribbling something on the board and he interrupted us ...

Sir, it is proved.... both expressions are equal !... he exclaimed in delight.

Oh wow !!  I felt a bit tempted to correct his vocab.... but that was not so important now....

Just study the left side of his work above. What he saw and said is -

We need to prove the left expression to be equal to x+5.....Now, there is already a 3 in the left side expression.... So this means that the fractional part of this left expression should get simplified to x+2,  so that this when added to 3 wholes, will give us x+5 as desired.

He further continued....

So then I checked whether the fractional part is really equal to x+2 or not.

'How?'

I multiplied (x-2) and (x+2).... and we get x^2 - 4. .. After factorizing the Nr., we can divide both Nr. and Dr. by the term (x-2)  and then we will be left with (x+2),  which when added to 3, gives us (x+5) as desired.

------------

To this, another student joined us, saying even he has proved both the expressions equal, but has used a little different method to get x+2.

I thought whether the Dr. is equal to the square of Nr. i.e. whether (x^2 - 4) = (x-2)^2  ?
But then I noticed that after multiplying (x-2) by (x-2), I will get  +4  and not -4 as desired in the expression (x^2 - 4) ..... So I changed the sign of 2 of one of the x-2, making it x+2..... And then when I multiplied these two terms (x-2) and (x+2), I got  (x^2 - 4)......  And then I did same as what he has done (above method) to get x+5.

-------------------

I could clearly see that the sense of wonder and accomplishment of proving the equivalence of two resulting expressions, obtained from their two different approaches - gave them more joy, satisfaction and confidence than the answer of the main (division) problem...

What do you think, would have happened if I had directly showed them or given them the procedure of dividing the polynomials as given in textbooks?  How would your students see / solve this problem if you don't give them the ready-made recipe ?

Will be eager to know your views and comments on this piece...

Thanks and Regards

PS: Students belong to class-7 marathi medium government school in navi-mumbai. I work with them voluntarily after their school hours as a part of maths enrichment program. www.supportmentor.weebly.com