# 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 **2 ^{nd} 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 19^{th} 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 **6x ^{2} + 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.

**6x ^{2}**

^{ }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 **(6x ^{2} + 59x + 144)** would be

**(2x + 9)**and

**(3x + 16)**

In our featured example, we want to find the factors **15x ^{2}**

**+ 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

**x**, + 6 which is not a factor of the constant

^{2 }**-8**or

**8 + 13**, which is not a factor of

**15.**

That leaves us the pair, **3 x 7.** Now the coefficient of **x ^{2}**,

**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, **15x ^{2}**

**+ 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 **(2x**^{2}** + 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 **(2x**^{2}** + Bx + 6) **is** 3, **then **B **must be **– 5. **Therefore, the correct answer is E) 2x^2 − 5x + 6.