# Coefficients of a perfect square

**(4x^4 – 4x^3 + Ax^2 – 8x + 16) is a perfect square. What is the value of **

**the coefficient, A, of the x^2 term?**

Only 161 out of 1,413 who participated in the Intermediate category of the 4^{th} Philippine National Vedic Mathematics Olympiad were able to get the correct answer to this question.

Since it was given that this 4^{th} degree polynomial is a perfect square, its square root will have the form (ax^2 + bx + c).

Using **the First by the First**, we have (ax^2)^2 = a^2x^4 = 4x^4. Therefore a = 2.

Next, we have, [2(2x^2)(bx)] = 4bx^3 = -4x^3. b, then must be -1.

The second to the last term, -8x is obtained from [2(bx)(c)] or 2(-x)(c) = -8x. c must be 4.

The square root must be (2x^2 – x + 4).

Ax^2 = [2(2x^2)4 + (-x)^2] = [16x^2 + x^2] = 17x^2, **A = 17**.

We can check our answer by using the sutra, **the Product of the Sums is the Sum of the Product**:

(2x^2 – x + 4)^2 = (4x^4 – 4x^3 + Ax^2 – 8x + 16).

Using the digit sums of the coefficients, we have

(2 – 1 + 4)^2 ?= (4 – 4 + 17 – 8 + 16)

or (5)^2 ?= 25; **25 = 25**