Respuesta :
The answers are 4, 64, 324, and 1024. A proof is below.
Let the trinomial factors be [tex]ax^2 + bx + c[/tex] and [tex]dx^2 + ex + f[/tex]. Then, their product is [tex](ad)x^4 + (ae+bd)x^3 + (fa+be+cd)x^2 + (bf+ce)x + cf[/tex]. We'll solve this problem by analyzing each of these coefficients.
Since the [tex]x^4[/tex] coefficient is [tex]ad[/tex] and all of the variables are integers, we have either [tex]a = d = 1[/tex] or [tex]a = d = -1[/tex]. Without loss of generality, let [tex]a=d=1[/tex]. (In the -1 case, we could simply multiply all the variables by -1 to find an analogous 1 case.)
So, we're back to [tex]x^2 + bx + c[/tex] and [tex]x^2 + ex + f[/tex]. We have that the [tex]x^3[/tex] coefficient is [tex]b+e[/tex], and since that term is zero in this polynomial, we have [tex]b+e=0[/tex]. This means that [tex]e=-b[/tex].
Our polynomials are now [tex]x^2 + bx + c[/tex] and [tex]x^2 - bx + f[/tex]. Let's consider the [tex]x[/tex] coefficient. It shows that we must have [tex]bf - bc = 0[/tex], so either [tex]b = 0[/tex] or [tex]f - c = 0[/tex].
Let's look at the case with [tex]b=0[/tex]. In this case, our polynomials become [tex]x^2 + c[/tex] and [tex]x^2 + f[/tex]. Then, we must have, for the [tex]x^2[/tex] coefficient to be zero, that [tex]c+f=0[/tex], so [tex]c = -f[/tex]. But then, considering the constant term, we'd need [tex]-c^2 = k[/tex]. However, we are given that k is positive, and we know that [tex]-c^2[/tex] is negative, so this case doesn't work. Hence, we must have that [tex]f - c = 0[/tex], so [tex]f = c[/tex].
Our polynomials are thus [tex]x^2 + bx + c[/tex] and [tex]x^2 - bx + c[/tex]. Hence, [tex]k = c^2[/tex], and since [tex]c[/tex] is an integer, [tex]k[/tex] must be a perfect square. Let's look at the [tex]x^2[/tex] coefficient again. We must have that [tex]2c - b^2 = 0[/tex], so [tex]b^2 = 2c[/tex]. Hence, [tex]c[/tex] must itself be half of a perfect square, since [tex]b[/tex] is an integer.
Let's consider values of [tex]b[/tex]. Since [tex]b^2 = 2c[/tex], [tex]b^2[/tex] is even, so [tex]b[/tex] is even. If [tex]b = 2[/tex], [tex]c = 2[/tex], and [tex]k = c^2 = 4[/tex]. If [tex]b = 4[/tex], [tex]c = 8[/tex], and [tex]k = 64[/tex]. If [tex]b = 6[/tex], [tex]c = 18[/tex], so [tex]k = 324[/tex]. If [tex]b = 8[/tex], [tex]c = 32[/tex], and [tex]k = 1024[/tex]. If [tex]b = 10[/tex], [tex]c = 50[/tex], so [tex]k = 2500[/tex]: but that doesn't work, because [tex]2500 > 2000[/tex]. So, our working values for [tex]k[/tex] are 4, 64, 324, and 1024.
This was a very drawn-out problem, so if you have any questions about how I worked through any of the steps, please feel free to ask!
