If the polynomial 4x^4+12x^3+17x^2+mx+4 be a perfect square, then find the value of m
Answers
Answer:
If the polynomial
16x4−24x3+41x2−mx+16
is the square of something, then that something has to be a polynomial of degree 2: ax2+bx+c . When you square this, the coefficient of x4 is a2 , so this needs to be 16 and a is ±4 . Similarly, c is also ±4 .
Since we are only interested in the square of this quadratic, we might as well assume that the leading coefficient is positive (otherwise negate the whole thing; the square won't change).
Our quadratic polynomial is now known to be 4x2+bx±4 .
Next, let's compare the coefficient of x3 . On the one hand it's −24 , and on the other hand it's 8b (this is what you get when you expand the square). So b has to be −3 .
Our quadratic is now known almost entirely; there's only some ambiguity of sign:
4x2−3x±4 .
When you square this, the coefficient of x2 comes to ±32+9 . For this to be 41 we need the 32 to come positive, so the constant coefficient needs to be positive as well, and our quadratic is:
4x2−3x+4 .
Finally, the coefficient of the linear term of the square is now seen to be -24, so your m must be 24.