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)

I looked at everyone and they were already with him. Before I could ask him about verification, he had already begun -

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

And yes, they did not disappoint me :)

"Why not?", I ask them.

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

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