Find the values of m and n such that x^4+6x^3+13x^2+mx+n , is a perfect square

Answer 1

we must have squared an expression such as
(x^2 + bx + c)

let's see what we get:
(x^2 + bx + c)^2
= (x^2 + bx + c)(x^2 + bx + c)
= x^4 + bx^3 + cx^2 + bx^3 + b^2x^2 + bcx +cx^2 + 2bcx + c^2
= x^4 + 2bx^3 + x^2(c+b^2+c) + 2bcx + c^2
comparing this with
x^4+6x^3+13x^2+mx+n , we have
2b=6 , ...... 2c +b^2 = 13 ..... 2bc = m ... and c^2 = n
b = 3 ............ c = 2 ........... m=12 ........ n=4

(I started with b=3 and the then substituted across the line

So if m=12 and n=4
we should end up with
(x^2 + 3x + 2)^2
= x^4+6x^3+13x^2+12x+n4

as seen by

http://www.wolframalpha.com/input/?i=expand+%28x%5E2+%2B+3x+%2B+2%29%5E2