Question

f(x) is a polynomial with positive integral coefficients such that f(1) = 12, f(12) = 2080 then find f(-1).

Answer 1

0

Answer 2

Let the polynomial be $f(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$. Given that the coefficients $a_i$ are positive integers, which means $a_i \ge 1$ for all $i = 0, 1, \dots, n$.

1. Use the condition $f(1) = 12$: Substituting $x=1$ into the polynomial: $f(1) = a_n (1)^n + a_{n-1} (1)^{n-1} + \dots + a_1 (1) + a_0$ $f(1) = a_n + a_{n-1} + \dots + a_1 + a_0 = 12$. This means the sum of all coefficients is 12. Since each $a_i \ge 1$, the maximum number of terms (and thus the maximum degree $n$) is 12. If there are $k$ terms, then $k \le 12$. The degree $n = k-1$. So $n \le 11$.

2. Use the condition $f(12) = 2080$ to determine the degree of the polynomial: $f(12) = a_n (12)^n + a_{n-1} (12)^{n-1} + \dots + a_1 (12) + a_0 = 2080$.

Let's test possible degrees for $n$:

• Case 1: $n=0$ $f(x) = a_0$. From $f(1)=12$, we have $a_0=12$. Then $f(12) = 12$. This contradicts $f(12)=2080$. So, $n \ne 0$.

• Case 2: $n=1$ $f(x) = a_1 x + a_0$. From $f(1)=12$: $a_1 + a_0 = 12$ (Equation 1) From $f(12)=2080$: $12a_1 + a_0 = 2080$ (Equation 2) Subtract (Equation 1) from (Equation 2): $(12a_1 + a_0) - (a_1 + a_0) = 2080 - 12$ $11a_1 = 2068$ $a_1 = \frac{2068}{11} = 188$. Substitute $a_1=188$ into (Equation 1): $188 + a_0 = 12 \implies a_0 = 12 - 188 = -176$. Since $a_0 = -176$ is not a positive integer, $n \ne 1$.

• Case 3: $n=2$ $f(x) = a_2 x^2 + a_1 x + a_0$. From $f(1)=12$: $a_2 + a_1 + a_0 = 12$ (Equation 3) From $f(12)=2080$: $144a_2 + 12a_1 + a_0 = 2080$ (Equation 4) Subtract (Equation 3) from (Equation 4): $(144a_2 + 12a_1 + a_0) - (a_2 + a_1 + a_0) = 2080 - 12$ $143a_2 + 11a_1 = 2068$. Divide by 11: $13a_2 + a_1 = 188$. Since $a_1 \ge 1$, we have $13a_2 \le 187 \implies a_2 \le \frac{187}{13} \approx 14.38$. Also, since $a_0 \ge 1$, from (Equation 3), $a_2 + a_1 \le 11$. Substitute $a_1 = 188 - 13a_2$ into $a_2 + a_1 \le 11$: $a_2 + (188 - 13a_2) \le 11$ $188 - 12a_2 \le 11$ $177 \le 12a_2$ $a_2 \ge \frac{177}{12} = 14.75$. This contradicts $a_2 \le 14.38$. So, $n \ne 2$.

• Case 4: $n=3$ $f(x) = a_3 x^3 + a_2 x^2 + a_1 x + a_0$. From $f(1)=12$: $a_3 + a_2 + a_1 + a_0 = 12$ (Equation 5) From $f(12)=2080$: $12^3 a_3 + 12^2 a_2 + 12a_1 + a_0 = 2080$ $1728a_3 + 144a_2 + 12a_1 + a_0 = 2080$ (Equation 6) Since $a_3 \ge 1$, $1728a_3 \le 2080$. This implies $a_3$ must be 1. Substitute $a_3=1$ into (Equation 6): $1728(1) + 144a_2 + 12a_1 + a_0 = 2080$ $144a_2 + 12a_1 + a_0 = 2080 - 1728 = 352$ (Equation 7) Substitute $a_3=1$ into (Equation 5): $1 + a_2 + a_1 + a_0 = 12 \implies a_2 + a_1 + a_0 = 11$ (Equation 8) Subtract (Equation 8) from (Equation 7): $(144a_2 + 12a_1 + a_0) - (a_2 + a_1 + a_0) = 352 - 11$ $143a_2 + 11a_1 = 341$. Divide by 11: $13a_2 + a_1 = 31$. Since $a_1 \ge 1$, $13a_2 \le 30 \implies a_2 \le \frac{30}{13} \approx 2.3$. So $a_2$ can be 1 or 2.

  • If $a_2 = 1$: $13(1) + a_1 = 31 \implies a_1 = 18$. Substitute $a_2=1, a_1=18$ into (Equation 8): $1 + 18 + a_0 = 11 \implies 19 + a_0 = 11 \implies a_0 = -8$. Not a positive integer.
  • If $a_2 = 2$: $13(2) + a_1 = 31 \implies 26 + a_1 = 31 \implies a_1 = 5$. Substitute $a_2=2, a_1=5$ into (Equation 8): $2 + 5 + a_0 = 11 \implies 7 + a_0 = 11 \implies a_0 = 4$. This is a positive integer. Thus, we found the coefficients: $a_3=1, a_2=2, a_1=5, a_0=4$. All are positive integers. The polynomial is $f(x) = x^3 + 2x^2 + 5x + 4$.

• Case 5: $n \ge 4$ If $n=4$, the term $a_4 x^4$ would be $a_4 (12)^4 = a_4 \times 20736$. Even if $a_4=1$, $20736 > 2080$. So, the degree cannot be 4 or higher.

Therefore, the unique polynomial satisfying the given conditions is $f(x) = x^3 + 2x^2 + 5x + 4$.

3. Calculate $f(-1)$: Substitute $x=-1$ into the polynomial: $f(-1) = (-1)^3 + 2(-1)^2 + 5(-1) + 4$ $f(-1) = -1 + 2(1) - 5 + 4$ $f(-1) = -1 + 2 - 5 + 4$ $f(-1) = 1 - 5 + 4$ $f(-1) = -4 + 4$ $f(-1) = 0$.

The final answer is $\boxed{0}$.