“The area of a rectangle is 2x^4 – 3x^3 + x^2 – 3x + 12. If its width is x^2 -3x + 3, What is its length?
a) 2x^2 + 4
b) 2x^2 + x + 4
c) 2x^2 + 2x + 4
d) 2x^2 + 3x + 4
e) 2x^2 + 4x + 4”
Less than 40% of the more than 1,400 participants in the intermediate group of the 4th Philippine National Vedic Mathematics Olympiad held last October, 2023 was able to get the correct answer to this algebraic division question.
The conventional solution is to use synthetic division. But some students may not have experienced using it with quadratic divisors.
We can also use the “Transpose and Apply” technique in Vedic Mathematics, where we will transpose (change the sign) of all the terms of the divisor.
But the easiest solution to his problem is to use the combination of two Vedic Math Sutras, The First by the First and the Last by the Last(FFLL) and The Product of the Sums is the Sum of the Product(PSSP).
We obtain the first term of the quotient by dividing the first term of the dividend by the first term of the divisor(FF): 2x^4/x^2 = 2x^2.
Likewise, we get the last term of the quotient by dividing the last term of the dividend by the last term of the divisor(LL): 12/3 = 4.
Since all the choices have 2x^2 and 4 as their first and last terms, we have to use the second sutra, PSSP. The product of the sums of the coefficients of the factors is equal to the sum of the coefficient of the product. If x is the sum of the coefficients of the algebraic expression for the length, then
x ( 1 – 3 + 3) = (2 – 3 + 1 – 3 + 12)
x(1) = 9
x = 9
Only choice d, 2x^2 + 3x + 4 has a sum of coefficients equal to 9.