# MSC #19 – The Product of the Sums is the Sum of the Product

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 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.

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 sum of the coefficients** A** and **B** given that **(2x ^{2} + Ax – 4) (3x^{2} -2x + B) = 6x^{4} + 11x^{3} – 16x^{2} + 23x – 12**.

Using The technique we discussed yesterday, we can use “**the Last by the Last**” to find B. We multiply the last terms of the trinomial and equate the product to the last term of the 4^{th} degree polynomial: **– 4 x B = -12.** **B**, then is **-12/ -4 = 3**.

Next, we apply PSSP. Using only the coefficients and substituting **4** for **B**, we have:

**(2 + A – 4)(3 – 2 + 3) = (6 + 11 – 16 + 23 – 12)**

**(A – 2)(4) = 12**

**4A – 8 = 12**

**4A = 20; A = 5**

**A + B = 5 + 3 = 8 **