[S1_20160227FZCK] Factorisation by Chunking

Problem / Question

 

Solution

Commentary

     Here I illustrate the usefulness of chunking to factorise (AmE: factor) an algebraic expression.  Observe that  3a – 2b  is a repeated part of the expression.  I call it a “chunk”.  To make it clear, I rewrite  (3a – 2b)²  as   (3a – 2b)(3a – 2b)  so that you can see it as two copies of the same chunk.  I highlight in yellow one copy of  (3a – 2b)  from each of 
(3a – 2b) (3a – 2b)   and  -3(3a – 2b).  The remaining stuff are highlighted in blue and green.  Take out the yellow chunk as common factor by writing it out on the left in the third line, shown in yellow.  You can pull out the common factor by writing it out to the right if you want, but here I chose to put it on the left.  The result would be equivalent anyway.  Once you have written out the common factor,  you write out the other stuff (shown highlighted in blue and green) into another other bracket.

     Once you understand how it works, you can actually do the second line mentally and write down the answer straightaway.  Chunking is a very useful technique in mathematics.  Here are some more examples of the technique of chunking: (1), (2), (3).

Heuristics Used

H04. Look for pattern(s)

H10. Simplify the problem

H11. Solve part of the problem

Suitable Levels

* Lower Secondary Mathematics (Sec 1 ~ grade 7)

* GCE ‘O’ Level “Elementary” Mathematics

* other syllabuses that involve algebra and factorisation (factoring)

* any learner who is interested

[AM_20151127DF2D] Differentiation with Chunking and Elimination

Question

Introduction

     Although this looks like a differential equation question, the student is not required to solve the differential equation.  The requirement is just to derive the equation.  This would be a challenging question for secondary 4 (~ grade 10) students taking Additional Mathematics or their counterparts in Integrated Programme schools.

Strategy

     One way to do this is to differentiate the given equation once and again and just verify the equation by substitution.  The problem is that when we repeatedly apply the Product rule

the terms tend to sprawl.  A way to keep things neat is to try to recognise chunks and also use elimination.

Solution

Remarks

     After differentiating once, we notice that  10xe^2x   is twice of  5xe^2x,  and this allows the simplification in [1].  The second differentiation yields  10e^2x   which, we notice, is twice of  5e^2x.  We can get rid of that term.   Multiplying equation [1] by 2 gives  10e^2x  in equation [3],  which matches nicely with the same term in  [2].  So we can eliminate that term via elimination.  After that, we just need to rearrange things to get the final equation.

Heuristics Used

H04. Look for pattern(s)

H05. Work backwards

H10. Simplify the problem

H11. Solve part of the problem

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* GCE ‘O’ Level Additional Mathematics, “Integrated Programme Mathematics”

* GCE ‘A’ Levels H2 Mathematics (revision)

* AP Calculus AB / BC (revision)

* University / College calculus (revision)

* other syllabuses that involve differentiation

* any learner interested in calculus

[AM_20151119ISSD] Differences of Squares Hiding under Square Roots

Question

Introduction

     Here is another question involving surds.  As we know, surds are literally absurd, because they are irrational.  How to we do this one?

Strategy

Solution

Remark

     Reflect: What did you learn from solving this question?

     For another example of using the difference of squares formula, please look at this article.

Heuristics Used

H04. Look for pattern(s)

H09. Restate the problem in another way

H10. Simplify the problem

H11. Solve part of the problem

Suitable Levels

* GCE ‘O’ Level Additional Mathematics (indices / surds)

* challenge for GCE ‘O’ Level “Elementary” Mathematics (indices)

* revision for IB Mathematics HL / SL

* other syllabuses that involve indices and/or surds
* any precocious or independent learner who is interested

[AM_20151117ISCP] Estimating a Crazy “Prosperous” surd to the Nearest Integer

Question

Introduction

     To the Chinese, the number  8  (八)  is considered to be auspicious, because it sounds like “prosper” (发) in the various Chinese languages/dialects.  But the LHS expression featured above seems too prosperous for comfort.  There is an explosion of  8s  coupled with eighth roots.  How to even handle that?

Strategy

     As usual, often one good tactic is to look for patterns or chunks.  [H04]  Can you see the sub-expressions containing the eighth roots (highlighted in green and pink)?

Do you notice any similarities between the two chunks?  Do you notice any difference(s)?

If we call the green chunk  a  and the pink chunk  b,  we can remove the roots by taking the eighth powers.  Then we get whole numbers, which are less complicated.

Solution

Remark

     The crux of the problem is the factorisation of  a⁸ – b⁸.  It is based on repeated application of the difference of squares    X² – Y² = (X + Y) (X – Y)  formula which schools expect students to know.
     For another example of using the difference of squares formula, please look at thisarticle.

Heuristics Used

H04. Look for pattern(s)

H09. Restate the problem in another way

H10. Simplify the problem

H11. Solve part of the problem

H13* Use Equation / write a Mathematical Sentence

Suitable Levels
* “IP Mathematics” so called
* challenge for students taking GCE ‘O’ Level Additional Mathematics
* other syllabuses that involve surds
* precocious kids who like to test themselves

[S2_20151107SLFC] Simultaneous Linear Equations with Fractional Coefficients

Question 

Introduction

     Somebody said, “Dear Algebra, please stop asking me about your  x.  She is not coming back.  Don’t ask me  y!”

     Here we have here linear equations that appear to be more tricky than the usual fanfare.  This is because the coefficients of the unknowns  x  and  y  are fractions.  In school, students learn to solve these using the method of substitution and the method of elimination.  They tend to prefer to do it by the former method, because psychologically it seems easier to accept.  However, the latter matter is generally more effective and yields a shorter solution.  This question is like a fly trap for those who prefer the method of substitution.  You make either  x  or  y  the subject from one of the equations, and then substitute that into the other equation.  This yields a complicated algebraic fractions within algebraic fractions.  The more complicated your solution is, the higher chance there is for making careless mistakes.  It is a good idea that students get out of their comfort zones and adopt a new skill.  So how do we do it?

The smart tactic

     First, we need to clear away the fractions by multiplying through with the LCM of the denominators appearing in each equation.  For the first equation, LCM(4, 8) = 8.  Since  4  is a factor of  8,  8  is like a giant that absorbs the number  4.  OK, so we multiply the first equation through by  8.  For the second equation, we multiply every term by LCM (3, 2) = 6

Solution

Remarks

     As you can see, equations [3] and [4] both contain – 3y  and this can be eliminated via subtraction.  So the value of  x  comes out easily.  Now once  x  is known, one can find the  y!

For another example of simultaneous equations, please refer to this article.

Heuristics Used

H04. Look for pattern(s)

H05. Work backwards

H10. Simplify the problem

H11. Solve part of the problem

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* Lower Secondary Mathematics (Secondary 2)

* GCE ‘O’ Level “Elementary” Mathematics (revision)

* other syllabuses that involve algebra, linear simultaneous equations

[S2_20150520SERC] Simultaneous Equations with Reciprocals

Question

Introduction

     This is a secondary 2 (» grade 8) problem that involves linear simultaneous equations in two unknown.  It looks non-linear and indeed if you try to solve for  x  and  y  directly,  you might get into a big mess with extraneous solutions to boot.  What should our approach be?

Observations

     It is always a good idea to stare at the question for a little while longer before jumping in to try to solve it.  Let us (mentally) reformat the equations a little bit.

     Do you notice any repeated chunks?  Chunking is very useful.  Let use substitutions for those chunks to simplify the equations.

Solution

Final Remarks

     To recap: It is a good learn to make observations first before attempt to solve the problem.  If you see any repeated chunks, it is a good idea to use substitution to simplify the problem.  Once the reciprocals have been substituted, we try to use elimination (which is usually the more effective method).  Also, try as far as possible to avoid fractions.  By multiplying equation [2] by 5,  we get – 30Y, which can be cancelled if we multiply  15Y by 2.  Thus  Y  is eliminated.  The problem becomes easy from this point onwards.
     For another example of simultaneous equations, please refer to thisarticle.

HeuristicsUsed

H04. Look for pattern(s)

H09. Restate the problem in another way

H10. Simplify the problem

H11. Solve part of the problem

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* Lower Secondary Mathematics (Secondary 1)

* GCE ‘O’ Level “Elementary” Mathematics (revision)

* other syllabuses that involve algebra and simultaneous linear equations

[S2_20150519XFCH] Chunking for Factorisation

Question

Introduction

     This is a question on factorisation (Am: factoring).  The key to doing this is chunking.  Can you see any chunk (maybe hidden) that is being repeated?  Take this out as the common factor and simplify.  Remember the square-of-differenceidentity  (a – b)² = a² – 2ab + b² .

Solution

Summary

     As in this question,  chunking is a very powerful tactic.  When you attempt to solve mathematics problems, open your eyes wide!

Heuristics Used

H04. Look for pattern(s)

H10. Simplify the problem

H11. Solve part of the problem

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* Lower Secondary Mathematics (usually secondary 2 ~ grade 8 approx.)

* other syllabuses that involve algebraic factorisation and expansion

[S2_20150513IXS] The Power of Repeated Chunks

Question

Introduction

     This is a question on indices (taught in secondary 2 or 3) that looks daunting.  There is a way to unravel it.  Look at the equation carefully.  Can you see anything that looks like it is repeated?  What about that small little “+1”?  How do you deal with it?

Reminders and Tips

    

Solution

Summary

     The key to solving this question is to apply the laws of indices and recognising repeated chunks.  This allows one to see the overall structure of the expression and this often leads to a simplification of the expression involved.  It is good to recognise the powers of small prime numbers.  Like atoms, prime numbers are the building blocks of all whole numbers.  Using prime bases allows us to compare indices more easily.

Heuristics Used

H04. Look for pattern(s)   e.g. repeated chunks

H09. Restate the problem in another way

H10. Simplify the problem

H11. Solve part of the problem

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* Lower Secondary Mathematics (usually Sec 2)

* “Elementary” Mathematics or “E Maths”

* Additional Mathematics (revision)

* other syllabuses that involve indices