# Aljabar Bentuk 5

## Konsep Dasar

#### 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 1

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 2

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*}