CARA 1

Misal f (x) = x³ + ax² + bx + c

\(\begin{aligned} f(1)&=1\\ 1^3+a(1)^2+b(1)+c&=1\\ a+b+c&=0&\qquad{\color{red}(1)} \end{aligned}\)

\(\begin{aligned} f(2)&=4\\ 2^3+a(2)^2+b(2)+c&=4\\ 4a+2b+c&=-4&\qquad{\color{red}(2)} \end{aligned}\)

\(\begin{aligned} f(3)&=9\\ 3^3+a(3)^2+b(3)+c&=9\\ 9a+3b+c&=-18&\qquad{\color{red}(3)} \end{aligned}\)

(1) dan (2)
\(\begin{aligned} a+b+c&=0\\ 4a+2b+c&=-4&\ {\color{red}-}\\\hline -3a-b&=4&\qquad{\color{red}(4)} \end{aligned}\)

(2) dan (3)
\(\begin{aligned} 4a+2b+c&=-4\\ 9a+3b+c&=-18&\ {\color{red}-}\\\hline -5a-b&=14&\qquad{\color{red}(5)} \end{aligned}\)

(4) dan (5)
\(\begin{aligned} -3a-b&=4\\ -5a-b&=14&\ {\color{red}-}\\\hline 2a&=-10\\ a&=-5 \end{aligned}\)

Substitusikan nilai a ke persamaan (4)
\(\begin{aligned} -3a-b&=4\\ -3(-5)-b&=4\\ 15-b&=4\\ -b&=-11\\ b&=11 \end{aligned}\)

Substitusikan nilai a dan b ke persamaan (1)
\(\begin{aligned} a+b+c&=0\\ -5+11+c&=0\\ c&=-6 \end{aligned}\)

Substitusikan nilai a, b, c, dan d ke fungsi
f (x) = x³ − 5x² + 11x − 6

\(\begin{aligned} f(4)&=4^3-5(4)^2+11(4)-6\\ &=64-80+44-6\\ &=\boxed{\boxed{22}} \end{aligned}\)

CARA 2

Kita lihat bahwa f (k) = k² untuk k ∈ {1, 2, 3}, sehingga bisa ditulis,
f (x) = (x − 1)(x − 2)(x − 3) + x²

\(\begin{aligned} f(4)&=(4-1)(4-2)(4-3)+4^2\\ &=\boxed{\boxed{22}} \end{aligned}\)

\(\begin{aligned} x^4+ax^2+b&=0\\ \left(x^2+\dfrac{a}2\right)^2&=\left(\dfrac{a}2\right)^2-b \end{aligned}\)

\(\begin{aligned} \left(\dfrac{a}2\right)^2-b&=0\\ b&=\dfrac{a^2}4 \end{aligned}\)

\(\begin{aligned} a+2b&=a+\dfrac{a^2}2\\[3.8pt] &=\dfrac12(a+1)^2-\dfrac12 \end{aligned}\)

\(\begin{aligned} a-b&=a-\dfrac{a^2}4\\[3.8pt] &=-\dfrac14(a-2)^2+1 \end{aligned}\)

\begin{aligned} P(x)&=Q(x)(x-r_1)(x-r_2)+220\\ &=Q(x)\left(x^2+x-25\right)+220\\ P(1)&=Q(1)\left(1^2+1-25\right)+220\\ &=Q(1)(-23)+220\\ P(1)\pmod{23}&=\boxed{\boxed{220}} \end{aligned}

Misal akar-akarnya adalah x₁, x₂, x₃, x₄

x₁ + x₂ + x₃ + x₄ = 18

x₁x₂ = −32

\(\begin{aligned} x_1x_2x_3x_4&=-1984\\ -32x_3x_4&=-1984\\ x_3x_4&=62 \end{aligned}\)

\(\begin{aligned} x_1x_2x_3+x_1x_2x_4+x_1x_3x_4+x_2x_3x_4&=-200\\ -32x_3-32x_4+62x_1+62x_2&=-200\\ -16x_3-16x_4+31x_1+31x_2&=-100\\ 31\left(x_1+x_2+x_3+x_4\right)-47\left(x_3+x_4\right)&=-100\\ 31(18)-47\left(x_3+x_4\right)&=-100\\ 558-47\left(x_3+x_4\right)&=-100\\ x_3+x_4&=14 \end{aligned}\)

x₁ + x₂ = 18 − 14 = 4

\(\begin{aligned} k&=x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4\\ &=x_1x_2+(x_1+x_2)(x_3+x_4)+x_3x_4\\ &=-32+(4)(14)+62\\ &=\boxed{\boxed{86}} \end{aligned}\)

\(\begin{aligned} P(-2)&=-2023\\ (a+b)(-2)^2+b(-2)-a-b&=-2023\\ 4a+4b-2b-a-b&=-2023\\ 3a+b&=-2023\\ \end{aligned}\)

\(\begin{aligned} Q(-2)&=c(-2)^2+(c+a)(-2)-a-b-2c\\ &=4c-2c-2a-a-b-2c\\ &=-3a-b\\ &=-\left(3a+b\right)\\ &=-(-2023)\\ &=\boxed{\boxed{2023}} \end{aligned}\)

\(\begin{aligned} S&=(x-2)^4+8(x-2)^3+24(x-2)^2+32(x-2)+16\\ &=(x-2)^4+4\cdot2(x-2)^3+6\cdot2^2(x-2)^2+4\cdot2^3(x-2)+2^4\\ &=(x-2+2)^4\\ &=\color{blue}\boxed{\boxed{\color{black}x^4}} \end{aligned}\)

Perhatikan bahwa agar x² − 73x + k mempunyai akar bilangan real, maka diskriminannya harus setidaknya 0.
\(\begin{aligned} 73^2-4k&\geq0\\ k&\leq1332,... \end{aligned}\)
Dengan menggunakan vieta, didapat bahwa perkalian semua akar real dari soal adalah 1332!. Maka
n = 1332

Lemma. Jika P(x) adalah polinom dengan koefisien bulat, maka a − b | P(a) − P(b) untuk setiap bilangan bulat a, b.

BUKTI

Misalkan P(x) = a_nxⁿ + a_{n − 1}x^{n − 1} + ⋯ + a₀. Maka
P(a) − P(b) = a_n(aⁿ − bⁿ) + a_{n − 1}(a^{n − 1} − b^{n − 1}) + ⋯ + a₁(a − b).
Karena a − b | aⁿ − bⁿ untuk setiap n, maka jelas a − b | P(a) − P(b)

TUTUP

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Dengan menggunakan lemma tersebut, maka n − 7 | P(n) − P(7) = 2045 − 2021 = 24.
Sehingga, semua bilangan menarik yang memenuhi adalah 1, 3, 4, 5, 6, 8, 9, 10, 11, 13, 15, 19, 31.
Maka, bilangan prima menarik adalah 3, 5, 11, 13, 19, 31.

a + b + c = −3
ab + bc + ac = 4
abc = 11

\(\begin{aligned} (a+b)(b+c)(a+c)&=-t\\ (-3-c)(-3-a)(-3-b)&=-t\\ (3+c)(3+a)(3+b)&=t\\ 3^3+3^2(a+b+c)+3(ab+bc+ac)+abc&=t\\ 27+9(-3)+3(4)+11&=t\\ t&=\color{blue}\boxed{\boxed{\color{black}23}} \end{aligned}\)