Five Techniques all Math Students Should Know

To be a successful student in mathematics there are several techniques that should be mastered.

Splitting up Multiplication

This technique is extremely useful. I use it almost every day and I don’t have to pull out my calculator to figure out moderately difficult multiplication.

Any complicated multiplication problem can be broken up. Here’s a simple example: $25*5= (20+5)*5= 20*5+5*5= 100+25= 125$ $25*5= (15+10)*5= 75+50= 125$ $25*5= (16+9)*5= 80+45= 125$

The trick is to break it up by place. By this, I mean break it up into the 1’s, 10’s, 100’s, 1000’s place and so on. $125*3= (100+20+5)*3= 300+60+15= 375$

If I were to do this problem in my head, I would start with the largest number: the 1 in the hundred’s spot. I would get 300 and I would hold that number in my mind while I did the second largest spot. I would get 60 and then add them together. I would then hold 360 in my mind and do the last multiplication. This way I only need to keep track of one number in addition to the multiplication I’m doing at the time.

If you can master this technique, math will go so much faster and you will perform a lot better in your classes. It takes practice though, and I would recommend double checking with the calculator at first.

It might sound crazy, but it makes sense right? If we add by zero nothing changes, but then what could we possibly accomplish? The key lies in the fact that we add by the equivalent of zero.

Let’s go back to the basics. We’ll start with the number 0. We’ll add 1 to it and then we subtract 1 from it. We’ll end up with 0 again! That looks like: $0+1-1= 0$

That’s exactly what we want. We can even do $0+2-2= 0$

As long as we add and subtract the same number, the original number doesn’t change. This can come in handy for situations that arise when completing the square. For example: $x^2+8x+2= 0$

If we wanted to factor this normally, we would have trouble; however, we can add by zero (In this case 0=16-16) to create a perfect square, which is easy to factor: $x^2+8x+(16-16)+2= 0$ $(x^2+8x+16)-16+2= 0$ $(x+4)^2= 14$ $\sqrt{(x+4)^2}= \sqrt{14}$ $x+4=\pm\sqrt{14}$ $x=-4\pm\sqrt{14}$

Notice how adding by zero is the same as adding to both sides: $0= 0$ $0+5= 0+5$ $0+5-5= 0$

Multiplying by One

This is probably the most important technique any student could use. The first time it appears is when students need to get a common denominator when adding fractions: $\frac{2}{3} + \frac{4}{5}$ $(1)*\frac{2}{3} + (1)*\frac{4}{5}$ $\frac{5}{5}*\frac{2}{3}+\frac{3}{3}*\frac{4}{5}$ $\frac{10}{15}+\frac{12}{15}= \frac{22}{15}$

The next time it comes up is in algebra when students need to get a radical out of the denominator: $\dfrac{3}{\sqrt{3}}*(1)$ $\dfrac{3}{\sqrt{3}}*(\dfrac{\sqrt{3}}{\sqrt{3}})= \dfrac{3*\sqrt{3}}{3}$

Multiplying by the Conjugate

This is actually a variation of multiplying by one. In this situation there is a binomial in the denominator with an imaginary term in it: $\dfrac{4}{3+i}$

To get rid of the i in the denominator we have to take advantage of two things.

1. $i*i= -1$, which is a real number.
2. When we multiply two binomials together we normally get a middle term, but if we multiply by the conjugate (fancy term for replace the plus sign with a minus or vice versa) we eliminate the middle term.

Example of getting a middle term: $(x+3)*(x+5)$ $x^2+3x+5x+15$ $x^2+8x+15$

Multiplying by the conjugate to eliminate middle term: $(x+3)*(x-3)$ $x^2+3x-3x-9$ $x^2-9$

So, back to our original example. Now that our binomial is in the denominator of our fraction, we can’t just multiply by $(3-i)$. We have to multiply by $\dfrac{3-i}{3-i}$, which equals one. Let’s see what happens: $\dfrac{4}{3+i}*\dfrac{3-i}{3-i}$

The numerator simplifies to $4*(3-i)= 12-4i$, which is expected. We can’t totally eliminate the “i”, but we can get it into the numerator.

The denominator is then multiplied out to get: $(3+i)*(3-i)= 9+3i-3i-i^2= 9-i^2= 9-(-1)= 9+1= 10$

So if we put everything together we’ll get: $\dfrac{12-4i}{10}= \dfrac{2(6-2i)}{10}= \dfrac{6-2i}{5}$

Mental Percents

Finally, the last technique, at least for now, is being able to do percents quickly. I see students struggle with percents all the time and it doesn’t need to be like that. To figure out any percent I always begin by moving the decimal one spot to the left. This gets me 10%. $10\%$ of $2345.0= 234.5$ $10\%$ of $834503485 = 83450348.5$ $10\%$ of $0.46= 0.046$

Then I can multiply it by 2 or 3 to get in the ballpark of most percentages. If I need to get 60% I’ll just multiply it again. If I need 70% I can just subtract my 30% value from the 100% value. $40\%= 4*10\%$ $40\%$ of $2345= 4*234.5= 938$

If you’re faced with a really ugly number with a lot of digits like 834503485 and you only need to get close, then round it to something nicer like 835000000. $20\%$ of $834503485\approx 2* 83500000\approx 167000000$

If the percent I’m trying to find ends in a 5, I’ll halve my 10% value and then add it back to my ballpark estimate. $25\%$ of $2345= 2*10\% + 5\%= 2*234.5+\dfrac{234.5}{2}= 469+117.25= 586.25$

If the percentage ends in something other than 5 or 0 I can just move the decimal spot once more to the left and get 1%. $21\%$ of $2345= 2*10\% + \dfrac{10\%}{10}= 2*234.5+\dfrac{234.5}{10}= 469+23.45= 492.45$

Using this method you can figure out almost any percent by only multiplying or dividing by 2, 3, or 10 (moving the decimal place).

I certainly hope this post has been helpful! If you thought it was educational, please share it with others who might benefit from it. If you have any questions, feel free to leave a comment or contact me!