The 17th Math Special for Christmas is about the Vedic Math sub-Sutra, “By Alternate Elimination and Retention”. To simplify solution of some problems, we sometimes ignore or temporarily eliminate some variables and retain only the others. One of the most common applications of this sutra is in finding the x and y intercepts of a straight line like 2y + 3x = 18. We eliminate y by equating it to 0 and solve for x. We can also eliminate x to solve for y.
In combination with other Sutras like “By Addition and By subtraction” and “Proportionately”, we can use it to find the Highest Common Factor (or Greatest Common Divisor) of numbers and polynomials by addition or subtraction of multiples of the numbers. This method is similar but more flexible than the Euclidian algorithm used to solve this type of problem.
Proportionately will be discussed by Mr. Rolito Asombra, coach of the San Jose National High School in the elementary group while “by Addition and by Subtraction” will be used by Engr. Emmanuel Nadela, former chairman of the Department of Mechanical Engineering at the Cebu Institute of Technology – University in the high school group on Dec 5, 2021 in the Inspirational Maths from India (Year 4). Just click on this link: https://bit.ly/IMI4Reg to register for the event.
If we want to find the GCD of 589 and 437, we can first find their difference: 589 – 437 = 152. Now we can do one of two things: 1) we can subtract twice 152 from 437, 437 – 2 x 152 = 437 – 304 = 133 and then perform this subtraction: 152 – 133 = 19 or 2) subtract 437 from 3 times 152: 3 x 152 – 437 = 456 – 437 = 19. In both cases we obtained 19 which we can verify as the HCF of the two numbers.
We can also create zeroes at the end of the numbers if we add 3 times 437 to 589, we will get 3 x 437 + 589 = 1311 + 589 = 1900. Obviously 100 is not a factor of the 2 numbers but 19 can be.
In our featured example we alternately eliminated the highest and lowest terms of the polynomials to get their Highest Common Factor, as shown in our infographics.
If we subtract the second polynomial from twice the first, we will eliminate the x cubed terms and get (33x2 + 55x + 22) which can be factored into 11(3x2 + 5x + 2)
2(3x3 + 17x2 + 22x + 8) = (6x3 + 34x2 + 44x + 16)
(6x3 + 34x2 + 44x + 16) – (6x3 + x2 – 11x – 6) = 33x2 + 55x + 22 = 11(3x2 + 5x + 2)
And when we add thrice the first polynomial and four times the second, we will eliminate the constant terms and get (33x3 + 55x2 + 22x) which can be factored into 11x(3x2 + 5x + 2)
3(3x3 + 17x2 + 22x + 8) = (9x3 + 51x2 + 66x + 24)
4(6x3 + x2 – 11x – 6) = (24x3 + 4x2 – 44x – 24)
(9x3 + 51x2 + 66x + 24) +(24x3 + 4x2 – 44x – 24) = (33x3 + 55x2 + 22x)
= 11x(3x2 + 5x + 2)
The GCF then is (3x2 + 5x + 2).