MSC #19: The Product of the Sums is the Sum of the Product
“Given that 3x^2 +3x −5 is a factor of 6x^4 − 9x^3 − 7x^2 + 43x − 30, which of the following is another factor? A) 2x^2 − 6x + 6; B) 2x^2 − 7x + 6; C) 2x^2 + 7x + 6; D) 2x^2 + 5x + 6; E) 2x^2 − 5x + 6
We used this question which was given in the senior and open levels of the 2nd International Vedic Mathematics Olympiad (IVMO 2022) to illustrate the how our previous post, MSC #18: The First by the First and the Last by the Last (FFLL), could be applied to solve this type of problem.
However, all the given choices conform with the results of FFLL. We then need this 19th of our 30 MATH-Inic Specials for Christmas series to choose the correct answer.
Our Math Special for today is one of my favorites because of its numerous applications. I first wrote about using the “Product of the Sums is the Sum of the Product (PSSP)” in my book “Algebra Made Easy as Arithmetic” which was published in January 2017 as a means of checking the results of polynomial multiplication and together with another Vedic Math sub-Sutra, The First by the First and the Last by the Last in factoring polynomials.
Because of this, my daughter Veronica and I were invited by the Institute for the Advancement of Vedic Mathematics (IAVM) to conduct a two-part online webinar to explain our technique.
I also wrote an article which was published in the Vedic Mathematics newsletter of the Vedic Mathematics Academy (UK) about the use to the two sub-Sutras in extracting the square root of a perfect square polynomial.( https://vedicmaths.org/2018/issue-119-vedic-math-in-the-philippines).
Although I first learned about the use of digit sums during my elementary grades in the early 1960s to check the answers to arithmetic operations, it was only recently, when I began to study Vedic Mathematics (VM) that I learned about the PSSP sub-Sutra and that it can be applied to coefficients of polynomials.
But what excited me more is when I used PSSP in factoring polynomials. I came across a YouTube video explaining the use of the “AC Method” of factoring using the trinomial 6x2 + 59x + 144 as an example. Of course, we know that it will take some time before the factor pair of 864 (6 x 144) which would add up to 59 is identified.
Because of my VM training, I quickly noticed that 1) the digit sum of the coefficients is [(144+ 6) + 59] = 209 by “completing the whole”; 2) 209 is divisible by 11 (2 – 0 + 9 = 11) and 3) 209 = 19 x 11 using “the First by the First and the Last by the Last”.
209 is the digit sum of the product, so it must be equal to the product of the digit sums of the two factors. 209 = 209 x 1 = 11 x 19. We can quickly discard the 209 x 1 combination because even the biggest factors of 6 and 144 will not add up to 209. So, the sums of the coefficients of the factors must be 11 and 19.
6x2 can be factored into 6x and x, or 2x and 3x.
Now if we consider 6x as one of the factors, it needs a constant term of 13 so that their digit sum will be 19 or 5 to have a sum of 11. Obviously 13 and 5 are not factors of 144, so we need to rule out the 6x and x combination as a solution.
It leaves us the 2x by 3x option.
2x needs a 17 to have their digit sum be equal to 19. But 17 is also not a factor of 144.
To have a digit sum of 11, 2x needs to have a constant term of 9 and 9 is a factor of 144. We also note that 3x needs a 16 so that their sum of 19. So, the factors of the polynomial (6x2 + 59x + 144) would be (2x + 9) and (3x + 16)
In our featured example, we want to find the factors 15x2 + 14x – 8. The sum of the coefficients is 21which can be factored as 1 x 21 or 3 x 7. We can quickly discard the 1 x 21 factor pair, because to make a sum of 21, we need 15, the biggest factor of the coefficient of x2 , + 6 which is not a factor of the constant -8 or 8 + 13, which is not a factor of 15.
That leaves us the pair, 3 x 7. Now the coefficient of x2, 15 can be factored as (1 x 15) or (3 x5). But as we have explained in the previous paragraph, we cannot use 15.
Now, how can we make (3x + a) and (5x + b) have their sum of coefficients equal to 3 and 7 so as to have a product of 21? Obviously the coefficients of (3x + a) can not be equal to 3, because a must be 0. (3 + a) then must be equal to 7 and a must be 4. (5 + b) must then be equal to 3 and b = – 2.
Therefore, 15x2 + 14x – 8 = (3x + 4)(5x – 2)
In the IVMO 2022 problem, since the application of the First by the First and the Last by the Last resulted in the other factor having the form (2x2 + Bx + 6), which is shared by all the choices, we need to use PSSP.
The sum of the coefficients of (6x^4 − 9x^3 − 7x^2 + 43x – 30) is (6 – 9 – 7 + 43 -30 = 3) while that of (3x^2 +3x −5) is (3 + 3 – 5 = 1). The sum of coefficients of the other factor must be 3/1 = 3.
If the sum of the coefficients of (2x2 + Bx + 6) is 3, then B must be – 5. Therefore, the correct answer is E) 2x^2 − 5x + 6.
