Questão 10

**Evaluating mathematical expressions** I’m trying to analyze the expression more clearly. It seems like I need to directly compare coefficients for n and distinct values for n, which can't be zero, especially considering when n equals 1. I should clarify that the maximums I’m calculating result in three distinct values that yield squares. I’m aiming to produce a structured response, approximately four paragraphs long, discussing the math with around 600 tokens in total. Let's see how to present this efficiently!

Since p has real coefficients and 1+i is a root, 1-i must be too, so x²-2x+2 divides p. I'm comparing degrees: p(p(x)) has degree n² while p(x³) has degree 3n, and their difference must have degree 8, so I'm testing values of n to see when these match up or when the leading terms cancel appropriately. For n=3 the leading coefficients cancel when a=1, forcing p(x) = (x²-2x+2)(x-r). Using p(1)=-1 gives r=2, so p(x) = x³-4x²+6x-4, and I verify the x⁸ coefficient in p(p(x))-p(x³) is nonzero, confirming degree 8 works. Computing p(-1) = -1-4-6-4 = -15, matching answer E.

Since p has real coefficients and p(1+i)=0, I know p(1-i)=0 too, so x²-2x+2 divides p(x), meaning degree n≥2. Comparing degrees of p(p(x)) and p(x³), I need n²=3n since otherwise the degrees can't match the required difference, forcing n=3 and leading coefficient a=1, so p(x)=(x²-2x+2)(x+...). Using p(1)=-1 I solve for the constant, getting p(x)=x³-4x²+6x-4. I check the degree-8 term in p(p(x))-p(x³) is nonzero, confirming consistency, then compute p(-1) = (1+2+2)(-3) = -15.

Since p has real coefficients and p(1+i)=0, I know p(1-i)=0 too, so (x²-2x+2) must divide p. Now I'm comparing degrees: p(p(x))-p(x³) has degree 8, so with p of degree n and leading coefficient a, I need n² vs 3n — checking n=3 gives matching degree 9... wait, that's inconsistent with the target degree 8, so I need to reconsider which case actually works. For n=3 with leading coefficient 1, I write p=(x²-2x+2)(x-r), use p(1)=-1 to solve r=2, giving p(x)=x³-4x²+6x-4. Expanding p(p(x)) and p(x³) confirms the degree-8 difference has a nonzero coefficient (-12), so this works, and evaluating p(-1) gives -15.

The problem is in Portuguese. I will translate it. The polynomial p(x) has real coefficients, so its complex roots come in conjugate pairs. Since p(1+i)=0, p(1−i)=0 as well. The quadratic factor (x−1)²+1 divides p(x). This gives the divisor x²−2x+2. Let p(x) = q(x)·r(x) with q(x) = x²−2x+2 and r(1) = −1. The degree of p(p(x)) is n² and of p(x³) is 3n. Since the difference has degree 8, n² must equal 8, which is impossible for integer n. This contradiction suggests the leading terms may cancel. The only remaining possibility is n=3. For that degree, both p(p(x)) and p(x³) have degree 9, so their difference could drop to exactly 8 if the leading coefficients cancel. The assumption that n=3 is the only viable degree is being reconsidered. If n were greater than 3, p(p(x)) would have degree n² and p(x³) would have degree 3n; since these are different, their leading terms cannot cancel to produce a polynomial of degree 8. The earlier conclusion that n must equal 3 appears inconsistent with the requirement. The assumption that p has degree 3 is now being verified. With q(x) = x²−2x+2 as the divisor, p(x) = (x²−2x+2)(ax+b) where a ≠ 0. The condition p(1) = −1 yields b = −1−a. The leading term of p(p(x)) is a⁴x⁹. The leading coefficient of p(p(x)) is a⁴x⁹ and of p(x³) is ax⁹. For the difference to have degree 8, the leading terms must cancel, so a⁴ − a = 0. Since a ≠ 0, a³ = 1 and a = 1. Then b = −2, giving p(x) = (x²−2x+2)(x−2). Expanding yields p(x) = x³−4x²+6x−4. The next step is verifying that p(p(x))−p(x³) has degree 8. The leading terms canceling to degree 8 is not enough. The degree must be exactly 8, so the coefficient of x⁸ must be nonzero. I checked other degrees: n=2 gives degree 6, n=1 gives degree 3, and n≥4 gives degree n², none of which equal 8. This confirms n=3 is the only possibility. p(-1) = (5)(-3) = -15, which is option E. The next step is to verify that the coefficient of x⁸ in p(p(x))−p(x³) is nonzero, confirming the degree is exactly 8. The difference p(p(x)) − p(x³) is rewritten as u³ − x⁹ + b(u² − x⁶) + c(u − x³), where u = p(x). Factoring u³ − x⁹ gives (u − x³)(u² + ux³ + x⁶). Since u − x³ = b x² + c x + d, the leading term of the difference is b x⁸. Because b = −4, this term is −4x⁸, confirming the degree is exactly 8.

We need answer question. Need solve. Let's restate. Problem: Seja p(x) um polinômio com coeficientes reais tal que p(1+i)=0 e p(1)=-1. Sabendo que p(p(x)) - p(x^3) é um polinômio de grau 8, o valor de p(-1) é? Alternativas A -50, B -5, C -1, D -3, E -15. Need solve. We need infer polynomial p real coefficients, p(1+i)=0 -> since real coefficients, 1-i also root. p(1)=-1. p(p(x)) - p(x^3) degree 8. Need p(-1). Let's analyze. Let deg p = n. p has roots 1±i maybe multiple? Since p(1±i)=0. p(p(x)) - p(x^3). Degree generally n^2 due p(p(x)), p(x^3) degree 3n maybe. But leading cancellations? Need degree exactly 8. If n? p degree n. p(p(x)) degree n^2. p(x^3) degree 3n. The difference degree max(n^2,3n) unless cancellation. Given degree 8. Possible n? n^2=8 => n not integer. 3n=8 => n noninteger. Difference degree maybe less via cancellation. n? Let's explore. If n=2? p(p) degree 4, p(x^3) degree 6, max 6, not 8. n=3? n^2=9, 3n=9, difference maybe degree 8 or less, possible. n=4? n^2=16, 3n=12, difference likely degree 16 unless leading cancel? But p(p(x)) leading coefficient a_n (p(x))^n? Actually p(p(x)) leading term a_n (p(x))^n = a_n (a_n x^n + ...)^n = a_n^{n+1} x^{n^2}. p(x^3) leading a_n x^{3n}. For n=4, n^2=16 >12, degree 16 unless a_n=0 no. So n=3 possible? Let's investigate. n=3: p(x)=a(x - (1+i))(x - (1-i))(x - r) = a[(x-1)^2+1](x-r) = a(x^2 -2x +2)(x-r). p(1)=a(1-2+2)(1-r)=a(1)(1-r)=a(1-r)=-1 => a(1-r)=-1. Compute p(p(x)) - p(x^3) degree. Let's do. For cubic p(x)=a x^3 + b x^2 + c x + d. Roots 1±i imply sum roots=2+r= -b/a => b = -a(2+r). Pair sums: (1+i)(1-i)+(1+i)r+(1-i)r =2 +2r? Actually (1+i)(1-i)=2; r[(1+i)+(1-i)]=2r => total 2+2r = c/a => c=a(2+2r). Product roots=(1+i)(1-i)r=2r = -d/a => d=-2ar. So p(x)=a[x^3 -(2+r)x^2 +2(1+r)x -2r]. Check p(1)=a[1-(2+r)+2(1+r)-2r]=a[1-2-r+2+2r-2r]=a[1? Wait 1-2+2=1; -r+2r-2r = -r? Actually -r+2r-2r = -r; constant -2r? Let's recompute: p(1)=a[1 - (2+r) + 2(1+r) - 2r] = a[1 -2 - r +2 +2r -2r] = a[1 - r]. Yes = -1 => a= -1/(1-r)=1/(r-1). Good. Now p(p(x)) - p(x^3). Since p degree 3, p(p) degree 9? Wait n^2=9, p(x^3) degree 9. Leading terms: p(p(x)) leading coefficient: p(x) leading a x^3, then p(y) leading a y^3 -> a (a x^3)^3 = a^4 x^9. So leading coeff a^4. p(x^3) leading coeff a (x^3)^3 = a x^9. Difference leading coeff (a^4 - a) x^9. If a≠1 and a≠0 (a? a=1? If a=1 then p(1)=-1 => a(1-r)=-1 => 1-r=-1 => r=2; maybe). For degree 8, leading coeff must vanish: a^4 - a =0 => a(a^3 -1)=0 => a=1? a=0 impossible? Or complex? a real. a=0 or a=1 or a^3=1 real => a=1. So a=1 possible. Then p(x)= ... with a=1, r? p(1)=1-r=-1 => r=2. So p(x)=(x^2-2x+2)(x-2)= x^3 -4x^2 +6x -4? Let's compute: (x^2-2x+2)(x-2)= x^3 -2x^2 -2x^2 +4x +2x -4 = x^3 -4x^2 +6x -4. p(1)=1-4+6-4=-1 ok. p(1+i)=? ( (1+i)^2 -2(1+i)+2)= (1+2i-1 -2 -2i +2)=0? yes 0? 1+2i-1=2i -2 -2i = -2 +2=0. Then p(x^3)? p(x^3)=x^9 -4x^6 +6x^3 -4. p(p(x)) degree? p(p(x)) = (p(x))^3 -4(p(x))^2 +6p(x) -4; degree 9? p(x)^3 leading x^9, coefficient 1; -4p(x)^2 leading -4 x^6 smaller; so p(p)-p(x^3) leading x^9 - x^9 =0; next degree? Compute exactly maybe degree? Let's compute to see degree maybe 7? But problem says degree 8. Maybe a not 1? Wait if a=1, difference degree maybe not 8? Need compute. Let's compute p(p(x))-p(x^3) for p=x^3 -4x^2+6x-4. p(x)=a=1. Need p(p(x)) degree? p(p)= p(x)^3 -4 p(x)^2 +6 p(x) -4. p(x)^3 leading x^9, coefficient 1. p(x^3)=x^9 -4x^6+6x^3 -4. Difference: (p(x)^3 - x^9) + [-4p(x)^2 +4x^6] + [6p(x)-6x^3] + [ -4 +4] maybe. p(x)^3 - x^9: Let p=x^3 + q, q=-4x^2+6x-4. (x^3+q)^3 - x^9 = 3x^6 q + 3x^3 q^2 + q^3. q degree2, so 3x^6 q degree8 coefficient 3*(-4)= -12 x^8. 3x^3 q^2 degree7? q^2 leading 16x^4 => 3x^3*16x^4=48x^7. q^3 degree6. Next -4p^2 +4x^6: -4[(x^3+q)^2] +4x^6 = -4(2x^3 q + q^2) = -8x^3 q -4q^2 = -8x^3(-4x^2+6x-4) -4(16x^4 -48x^3+...?) degree5? -8x^3 q =32x^5 -48x^4 +32x^3; -4q^2 = -4(16x^4 -48x^3+...)= -64x^4 +192x^3... So degree5. Then 6p-6x^3 = 6q = -24x^2+36x-24 degree2. So difference degree 8 with coefficient -12 x^8? Then degree 8 indeed. Value p(-1)=? p(-1)= -1 -4 -6 -4? Wait p(x)=x^3 -4x^2 +6x -4. p(-1)= -1 -4 -6 -4 = -15? That's option E. But wait check p(-1) = (-1)^3 -4(-1)^2 +6(-1)-4 = -1 -4 -6 -4 = -15. Yes option E -15. But need ensure no other cases? The problem likely answer E? But let's verify if a=1 unique due degree 8? We saw degree could be 8 if a=1? But maybe other a (a^3=1 real only 1). But if a? p real coefficients, leading coeff a real. a^4 - a=0 => a=0 or a=1. a=0 impossible degree 3? Actually polynomial degree 3, so a≠0, hence a=1. Yes. Then p determined by r? p(1)=-1 gives r=2. So p unique. Then p(-1)=-15. But need be careful: Could p degree lower? If p constant? But p(1+i)=0 impossible? No. What if p degree >3? But degree difference 8 maybe possible with cancellation if n> ? Let's analyze thoroughly because problem statement might have multiple possibilities? Maybe n=4? Let's see. Let deg p = n. p(p) degree n^2, p(x^3) degree 3n. Difference degree 8. If n>3, n^2 >3n for n>3? n=4:16>12; n=5:25>15 etc. Difference degree n^2 unless leading coefficient of p(p) vanishes? Leading coefficient of p(p) = a^{n+1} (where a leading coeff of p). p(x^3) leading = a. Difference leading = a^{n+1} - a if n^2=3n? Actually if n^2=3n i.e. n=3, compare; if n>3, n^2>3n, so p(p) dominates unless a^{n+1}=0? a≠0, so degree n^2, cannot be 8. If n<3? n=0,1,2: p(p) degree n^2 < or =4, p(x^3) degree 3n; difference degree max(3n,n^2)=3n for n=1,2; could be 3 or 6, not 8. So n must be 3. Good. Then leading cancellation possible as above. So unique. But let's double-check: For n=3, difference leading coefficient indeed a^4 - a. If a=1 degree maybe 8; if a? Real a from p(1)=-1: a(1-r)=-1. If a=0 impossible; a^3=1 => a=1 only real. Then r=2. Yes. Answer E -15. Need provide explanation in Portuguese. Final line Resposta: E. Ensure no issue. But let's consider if p could have degree 0? No p(1+i)=0 impossible since constant not zero? p(1)=-1 => p= -1 constant, p(1+i)=-1 not 0. Degree 1? p(x)=ax+b, real coefficients. p(1+i)=a(1+i)+b=0 => a=0 and b=0? Because real and imaginary parts: a=0, b=0 => p=0, but p(1)=-1 contradiction. So deg≥2. But real coefficients with non-real root => degree at least 2. Degree 2? Then p(p) degree 4, p(x^3) degree 6, difference degree 6 not 8. So n=3. Good. We should present solution elegantly. Potential pitfalls: The polynomial p(x) could have degree 3 but leading coefficient maybe not cancel? We proved cancellation necessary for degree 8 because if not, degree 9. But let's compute: If n=3, degree of p(p) is 9, p(x^3) degree 9, difference could be 9 unless leading coefficient equal. So a^4 = a => a=1. Then p(1+i)=0 and p(1)=-1 determine p. Actually p(1+i)=0 gives roots 1±i -> factor (x-1)^2+1 = x^2-2x+2. So p(x)= (x^2-2x+2)(ax+b)? Wait if a=1 leading coefficient 1, then p(x)= (x^2-2x+2)(x - r) as above. Then p(1)=-1 gives (1)(1-r)= -1 => r=2. So p(x)=x^3-4x^2+6x-4. Then p(-1)=-15. Need maybe show degree 8 statement actually satisfied; p(p)-p(x^3) degree 8 automatically? Not necessary but maybe to confirm no contradiction. Let's compute leading term to ensure degree exactly 8: coefficient of x^8 = -12 ≠0. So degree 8 indeed. But problem states it is degree 8; we used to infer a=1. Good. Let's craft final response with steps: 1. Seja p(x)=a_n x^n+... . Since p(1+i)=0 e coeficientes reais, 1-i também raiz. Grau? If n≠3 impossible? Justify by degrees: deg p(p)=n^2, deg p(x^3)=3n. Para que diferença tenha grau 8, n deve ser 3? Actually need careful: if n=3, possible; if n>3, degree n^2 >3n >8? For n>3, n^2≥16, so degree ≥16 (unless leading coeff zero impossible). If n<3, degree max(3n,n^2) ≤6 <8. So n=3. 2. Então p(x)=a(x-(1+i))(x-(1-i))(x-r)=a[(x-1)^2+1](x-r)=a(x^2-2x+2)(x-r). p(1)=-1 => a(1-r)=-1. Leading coefficient a. 3. Compare leading terms in p(p(x))-p(x^3): p(p) leading coefficient a^4, p(x^3) leading a. Para grau 8, coeficiente de x^9 deve anular: a^4 - a =0 => a(a^3-1)=0. Como a≠0 (grau 3), a^3=1 => a=1 (real). 4. Então 1-r=-1 => r=2. Logo p(x)=x^3-4x^2+6x-4. 5. p(-1)= -1 -4 -6 -4? Wait compute: (-1)^3 = -1; -4(-1)^2=-4; 6(-1)=-6; -4; total -15. Yes. Resposta E. Need maybe mention "grau 8" after cancellation exactly; but enough. Let's final in Portuguese.

**Exploring Polynomial Properties** I'm currently analyzing a polynomial with real coefficients, given $p(1+i) = 0$ and $p(1) = -1$. My focus is on understanding the implications of $p(p(x)) - p(x^3)$ being a degree 8 polynomial, specifically to determine $p(-1)$. The presence of a complex root suggests its conjugate must also be a root. **Determining Polynomial Degree** My current focus is on deducing the degree of the polynomial. The condition that $p(p(x)) - p(x^3)$ is degree 8 strongly implies that the leading terms of $p(p(x))$ and $p(x^3)$ must cancel. This cancellation occurs precisely when $n^2 = 3n$, leading to $n=3$. Thoughts: $p(x)$ has real coefficients. Given $p(1+i) = 0$. Since coefficients are real, if $1+i$ is a root, then its conjugate $1-i$ must also be a root. So, $(x - (1+i))$ and $(x - (1-i))$ are factors of $p(x)$. Their product is $((x-1)-i)((x-1)+i) = (x-1)^2 - i^2 = (x-1)^2 + 1 = x^2 - 2x + 1 + 1 = x^2 - 2x + 2$. This quadratic is a factor of $p(x)$. We deduced that $p(x)$ has degree $n=3$. So $p(x)$ must be of the form $p(x) = (x^2 - 2x + 2)(ax+b)$ for some real coefficients $a$ and $b$, with $a \neq 0$. We are also given $p(1) = -1$. Let's substitute $x=1$ into $p(x)$: $p(1) = (1^2 - 2(1) + 2)(a(1)+b)$ $p(1) = (1 - 2 + 2)(a+b)$ $p(1) = (1)(a+b)$ $p(1) = a+b$ Since $p(1) = -1$, we have $a+b = -1$. Now let's go back to the degree 8 condition. $p(x) = ax^3 + (b-2a)x^2 + (2a-2b)x + 2b$. Let $p(x) = Ax^3 + Bx^2 + Cx + D$, where $A=a$, $B=b-2a$, $C=2a-2b$, $D=2b$. $p(p(x)) = A(p(x))^3 + B(p(x))^2 + C(p(x)) + D$. The highest degree term is $A(Ax^3)^3 = A^4 x^9$. $p(x^3) = A(x^3)^3 + B(x^3)^2 + C(x^3) + D = Ax^9 + Bx^6 + Cx^3 + D$. For the degree of $p(p(x)) - p(x^3)$ to be 8, the $x^9$ terms must cancel. So $A^4 x^9$ from $p(p(x))$ must cancel with $Ax^9$ from $p(x^3)$. This means $A^4 = A$. Since $A = a \neq 0$, we can divide by $A$. $A^3 = 1$. So $A=1$. Since $A=a$, we have $a=1$. Now we use $a+b = -1$ and $a=1$: $1+b = -1$ $b = -2$. So, $p(x) = (x^2 - 2x + 2)(x-2)$. Let's expand this: $p(x) = x^3 - 2x^2 + 2x - 2x^2 + 4x - 4$ $p(x) = x^3 - 4x^2 + 6x - 4$. We need to find $p(-1)$. $p(-1) = (-1)^3 - 4(-1)^2 + 6(-1) - 4$ $p(-1) = -1 - 4(1) - 6 - 4$ $p(-1) = -1 - 4 - 6 - 4$ $p(-1) = -15$. This matches option D. Final check of the problem statement: $p(1+i) = 0$. $p(x) = (x^2 - 2x + 2)(x-2)$. $1+i$ is a root of $x^2 - 2x + 2 = 0$. So $p(1+i)=0$. Correct. $p(1) = 1^3 - 4(1)^2 + 6(1) - 4 = 1 - 4 + 6 - 4 = -1$. Correct. Now consider $p(p(x)) - p(x^3)$. $p(x) = x^3 - 4x^2 + 6x - 4$. $a=1$, $b=-2$. $A=1$, $B=b-2a = -2-2(1) = -4$. $C=2a-2b = 2(1)-2(-2) = 2+4=6$. $D=2b = 2(-2) = -4$. $p(p(x)) = (p(x))^3 - 4(p(x))^2 + 6(p(x)) - 4$. The highest term is $(x^3)^3 = x^9$. The coefficient is $1^4 = 1$. $p(x^3) = (x^3)^3 - 4(x^3)^2 + 6(x^3) - 4 = x^9 - 4x^6 + 6x^3 - 4$. The coefficient of $x^9$ in $p(p(x))$ is $A^4 = 1^4 = 1$. The coefficient of $x^9$ in $p(x^3)$ is $A=1$. These cancel. What is the next highest degree term? $p(p(x))$: $A(p(x))^3 = 1(x^3 - 4x^2 + 6x - 4)^3$. The next term after $x^9$ in $(x^3 - 4x^2 + \dots)^3$ will come from terms like $3(x^3)^2(-4x^2) = -12x^8$. So the $x^8$ term from $A(p(x))^3$ is $1 \cdot (-12x^8) = -12x^8$. Now consider $B(p(x))^2 = -4(x^3 - 4x^2 + \dots)^2$. The highest degree term is $-4(x^3)^2 = -4x^6$. There is no $x^8$ term here. So the $x^8$ term in $p(p(x))$ comes solely from $A(p(x))^3$, and it is $-12x^8$. The polynomial $p(x^3)$ has no $x^8$ term. So $p(p(x)) - p(x^3)$ has an $x^8$ term with coefficient -12. This means the degree of $p(p(x)) - p(x^3)$ is indeed 8. The value of $p(-1)$ is -15. The options provided are: A ( ) -5. B ( ) -1. C ( ) -3. D ( ) -15. E ( ) -2. The calculation gives -15, which corresponds to option D. My thought process confirms the degree of the polynomial is 3, and with the given conditions, I can uniquely determine the polynomial and evaluate $p(-1)$. **Confirming Polynomial Value** My analysis has now confirmed the polynomial's degree is 3. I've used the complex root and the point $p(1)=-1$ to derive the specific coefficients of $p(x)$. Crucially, the cancellation required for $p(p(x)) - p(x^3)$ to be degree 8 led to the leading coefficient being 1, allowing me to solve for the remaining unknown, ultimately yielding $p(-1) = -15$. **Deducing Polynomial Coefficients** I have successfully determined that the polynomial, $p(x)$, must be of degree 3. Leveraging the given complex root and the point $p(1)=-1$, I've established the form $p(x) = (x^2-2x+2)(ax+b)$. The critical condition that $p(p(x)) - p(x^3)$ has degree 8 has fixed the leading coefficient to 1, allowing me to solve for $a$ and $b$, thus uniquely defining $p(x)$. **Confirming Polynomial Value** My analysis has now confirmed the polynomial's degree is 3. I've used the complex root and the point $p(1)=-1$ to derive the specific coefficients of $p(x)$. Crucially, the cancellation required for $p(p(x)) - p(x^3)$ to be degree 8 led to the leading coefficient being 1, allowing me to solve for the remaining unknown, ultimately yielding $p(-1) = -15$. **Finding Third Root** My focus is now on pinpointing the third real root of $p(x)$. I've confirmed the polynomial's leading coefficient is 1 and that $(x^2-2x+2)$ is a factor. Using the condition $p(1)=-1$, I've derived that the third root, $r$, must be 2, which will fully define my polynomial.