# Using a normal calculator to do Algebraic Expansion

### By Owen Cheong

(Math, Physics tutor at The Edge Learning Center)

When you are asked to expand an expression like (3z+5)(2z+7), we usually apply the FOIL method: (Source)

Now, I would like to introduce another way to expand brackets, all you need is a normal calculator. To expand the algebraic expression, we just need to substitute x = 100 and plug it into a calculator. Let’s begin with a question from an IGCSE Edexcel paper as an example:

[2014 June Paper 3HR Q5b]

Expand and simplify (x+2)(x+5)

If we make x=100, then the above expression will become 102×105, which is equal to 10710. Now, we just need to identify the terms based on the following rules:

Counting from RIGHT TO LEFT:

• The first two digits represent the constant term (term without x)
• The next two digits represent the coefficient of x
• The following two digits represent the coefficient of x², and so on So the solution for the problem is 1x²+7x+10, but we usually write x²+7x+10

This method is extremely efficient when it comes to binomial expansion. For example, if we want to expand (2x+1) , we can just substitute x=100 to the expression and figure out the result in the calculator: By applying the rules as discussed above, we can immediately conclude that  We need to do some adjustment if some expected terms are negative numbers. Let’s take a look at the following example:

[2017 May Paper 3H Q1d]

Expand and simplify (x-3)(x+7)

If we substitute x=100 in the above expression, we get

(100-3)(100+7)=97×107=10379

According to the rules we have discussed, the constant term is 79. However this is false because when we use FOIL to expand (x-3)(x+7), the constant term should be equal to -3×7=-21. Therefore, when we expect some term(s) to be negative, we need to apply the following adjustment rules:

• Subtract 100 from the ‘two-digit pair’ if the required term is expected to be negative
• When the previous term of the pattern is negative, then the following ‘two-digit pair’ needs to be increased by 1 (For example, if the x term is negative, then the two-digit number that represents x² term needs to be increased by 1) In this example, the constant term is expected to be negative, so we do 79 – 100= -21. Therefore, the constant term is equal to -21. Since the constant term is negative, the two-digit number that represents the x term needs to be increased by 1, so ‘03’ plus 1 becomes ‘04’. This means that the x term is equal to +4. As a result, (x-3)(x+7)=x²+4x-21

In fact, this method can also be applied to other types of questions. It is also helpful when doing polynomial long division:

[Cambridge IGCSE Additional Math 0606 June 2014 Paper 22 Q3ii]

Show that 3x³-14x²-7x + 10 can be written in the form (x+1)(ax²+bx+c), where a, b and c are constants to be found.

Let’s focus on finding the quadratic expression in the above question. The straightforward way to do this is to apply polynomial long division, since Now, let’s try to apply the calculator method by substituting x=100 into the expression on the right-hand side of the above equation: Using the adjustment rules for the x and x² term, we get:

ax²+bx+c=3x²-17x+10 Although it is a fancy way to do algebraic manipulation, there are some limitations to using this method. This method cannot be applied when the expected coefficient is greater than 100. For example, if we need to expand (50x-62)(20x+61), unfortunately this method won’t work in this scenario.

Therefore, in general, the FOIL method is still recommended when you need to show your work on your exam. However, the calculator method provides an excellent way to double check your answers on your exam.

Let’s try the following problems by first using FOIL and then checking the answer by using the calculator method.

[IGCSE Edexcel 2017 Jan Paper 3H Q9]

FOIL Method:

(2x+3)²-(2x+3)(x-5)=(4x²+6x+6x+9)-(2x²-10x+3x-15)

= (4x²+12x+9)-(2x²-7x-15)

= 2x²+19x+24

(2×100+3)²-(2×100+3)(100-5)=203²-203×95=21924

Obviously, the result 21924 yields the answer 2x²+19x+24

Read more from Owen in his previous blog “ Winning a Nobel Prize by Solving ‘1+1’ ”

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