Contoh 03

Jabarkan \((2x + 3y + z)^2\)

\begin{equation*} \begin{split} &(2a)^2 + (3b)^2 + c^2 + 2 \:.\: 2a \:.\: 3b + 2 \:.\: 2a \:.\: c + 2 \:.\: 3b \:.\: c \\\\ &4a^2 + 9b^2 + c^2 + 12ab + 4ac + 6bc \end{split} \end{equation*}

Contoh 04

Jabarkan \((m - 2n + 5r)^2\)

\begin{equation*} \begin{split} &m^2 + (-2n)^2 + (5r)^2 + 2 \:.\: m \:.\: -2n + 2 \:.\: m \:.\: 5r + 2 \:.\: -2n \:.\: 5r \\\\ & m^2 + 4n^2 + 25r^2 - 4mn + 10mr - 20nr \end{split} \end{equation*}

Ekspansi Binomial

Segitiga Pascal

Baris pada segitiga pascal menunjukkan koefisien penjabaran \((a + b)^n\). Koefisien untuk \((a + b)^n\) dapat dibaca pada baris ke\(-(n + 1)\)

Contoh,

• \(\color{purple}(a + b)^2 = a^2 + 2ab + b^2\)  dari baris 1, 2, 1
• \(\color{purple}(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\)  dari baris 1, 3, 3, 1
• \(\color{purple}(a + b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4\)  dari baris 1, 4, 6, 4, 1
• \(\color{purple}(a + b)^5 = a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5\)  dari baris 1, 5, 10, 10, 5, 1
• \(\color{purple}(a + b)^6 = a^6 + 6a^5b + 15a^4b^2 + 20a^3b^3 + 15a^2b^4 + 6ab^5+b^6\) dari baris 1, 6, 15, 20, 15, 6, 1

Contoh 01

Jabarkan \((2x + 3)^3\)

Misal \(a = 2x\) dan \(b = 3\)

maka,

\begin{equation*} \begin{split} (2x + 3)^3&= (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\\\\ (2x + 3)^3&= (2x)^3 + 3(2x)^2(3) + 3(2x)(3)^2 + 3^3\\\\ (2x + 3)^3& = 8x^3 + 3(4x^2)(3) + 3(2x)(9) + 27\\\\ (2x + 3)^3& = 8x^3 + 36x^2 + 54x + 27 \end{split} \end{equation*}

Contoh 02

Jabarkan \((3m - 5)^5\)

Misal \(a = 3m\) dan \(b = -5\)

maka,

\begin{equation*} \begin{split} (3m - 5)^5&= (a + b)^5 = a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5\\\\ (3m - 5)^5& = (3m)^5 + 5(3m)^4(-5) + 10(3m)^3(-5)^2 + 10(3m)^2(-5)^3 + 5(3m)(-5)^4 + (-5)^5\\\\ (3m - 5)^5& = 243m^5 + 5(81m^4)(-5) + 10(27m^3)(25) + 10(9m^2)(-125) + 5(3m)(625) - 3125\\\\ (3m - 5)^5& = 243m^5 -2025m^4 + 6750m^3 - 11250m^2 + 9375m - 3125\\\\ \end{split} \end{equation*}