Menyelesaikan Soal Suku Banyak (Polinomial) dengan Cepat

CONTOH SOAL

Suku banyak berderajad 3, jika dibagi $\left( {{x^2} + 2x - 3} \right)$ bersisa $\left( {3x - 4} \right)$, jika dibagi $\left( {{x^2} - x - 2} \right)$ bersisa $\left( {2x + 3} \right)$. Suku banyak tersebut adalah ....

1. ${x^3} - {x^2} - 2x - 1$
2. ${x^3} + {x^2} - 2x - 1$
3. ${x^3} + {x^2} + 2x - 1$
4. ${x^3} + 2{x^2} - x - 1$
5. ${x^3} + 2{x^2} + x + 1$

Penyelesaian:

(CARA BIASA)

Misal:

• $P\left( x \right)$ adalah suku banyak berderajat $3$.
• $Q\left( x \right)$ adalah hasil pembagian.

Suku banyak berderajad 3, dibagi $\left( {{x^2} + 2x - 3} \right)$ bersisa $\left( {3x - 4} \right)$ artinya:

$\begin{array}{l} P\left( x \right) = \left( {{x^2} + 2x - 3} \right).Q\left( x \right) + \left( {3x - 4} \right)\\ \Leftrightarrow P\left( x \right) = \left( {x + 3} \right)\left( {x - 1} \right).Q\left( x \right) + \left( {3x - 4} \right)\\ \Leftrightarrow P\left( { - 3} \right) = \left( { - 3 + 3} \right)\left( { - 3 - 1} \right).Q\left( x \right) + \left( {3.\left( { - 3} \right) - 4} \right)\\ \Leftrightarrow P\left( { - 3} \right) = \left( 0 \right)\left( { - 4} \right).Q\left( x \right) + \left( { - 9 - 4} \right)\\ \Leftrightarrow P\left( { - 3} \right) = 0 + \left( { - 13} \right)\\ \therefore P\left( { - 3} \right) = - 13\\ P\left( x \right) = \left( {{x^2} + 2x - 3} \right).Q\left( x \right) + \left( {3x - 4} \right)\\ \Leftrightarrow P\left( x \right) = \left( {x + 3} \right)\left( {x - 1} \right).Q\left( x \right) + \left( {3x - 4} \right)\\ \Leftrightarrow P\left( 1 \right) = \left( {1 + 3} \right)\left( {1 - 1} \right).Q\left( x \right) + \left( {3.1 - 4} \right)\\ \Leftrightarrow P\left( 1 \right) = \left( 4 \right)\left( 0 \right).Q\left( x \right) + \left( {3 - 4} \right)\\ \Leftrightarrow P\left( 1 \right) = 0 + \left( { - 1} \right)\\ \therefore P\left( 1 \right) = - 1 \end{array}$

Suku banyak berderajad $3$, jika dibagi $\left( {{x^2} - x - 2} \right)$ bersisa $\left( {2x + 3} \right)$ artinya:

$\begin{array}{l} P\left( x \right) = \left( {{x^2} - x - 2} \right).Q\left( x \right) + \left( {2x + 3} \right)\\ \begin{array}{*{20}{c}} {Misal:}&{Q\left( x \right)} \end{array} = ax + b\\ \Leftrightarrow P\left( x \right) = \left( {{x^2} - x - 2} \right).\left( {ax + b} \right) + \left( {2x + 3} \right)\\ \Leftrightarrow P\left( x \right) = \left( {x - 2} \right)\left( {x + 1} \right).\left( {ax + b} \right) + \left( {2x + 3} \right)\\ P\left( { - 3} \right) = \left( { - 3 - 2} \right)\left( { - 3 + 1} \right).\left( {a\left( { - 3} \right) + b} \right) + \left( {2\left( { - 3} \right) + 3} \right)\\ \Leftrightarrow P\left( { - 3} \right) = \left( { - 5} \right)\left( { - 2} \right).\left( { - 3a + b} \right) + \left( { - 6 + 3} \right)\\ \Leftrightarrow P\left( { - 3} \right) = 10.\left( { - 3a + b} \right) + \left( { - 3} \right)\\ \Leftrightarrow P\left( { - 3} \right) = - 30a + 10b - 3\\ \Leftrightarrow - 13 + 3 = - 30a + 10b\\ \Leftrightarrow - 30a + 10b = - 10\left( {\begin{array}{*{20}{c}} {dibagi}&{ - 10} \end{array}} \right)\\ \Leftrightarrow 3a - b = 1 \cdots \left( {\begin{array}{*{20}{c}} {Persamaan}&1 \end{array}} \right)\\ P\left( 1 \right) = \left( {1 - 2} \right)\left( {1 + 1} \right).\left( {a\left( 1 \right) + b} \right) + \left( {2\left( 1 \right) + 3} \right)\\ \Leftrightarrow P\left( 1 \right) = \left( { - 1} \right)\left( 2 \right).\left( {a + b} \right) + \left( {2 + 3} \right)\\ \Leftrightarrow P\left( 1 \right) = - 2.\left( {a + b} \right) + 5\\ \Leftrightarrow P\left( 1 \right) = - 2a - 2b + 5\\ \Leftrightarrow - 1 - 5 = - 2a - 2b\\ \Leftrightarrow - 2a - 2b = - 6\left( {\begin{array}{*{20}{c}} {dibagi}&{ - 2} \end{array}} \right)\\ \Leftrightarrow a + b = 3 \cdots \left( {\begin{array}{*{20}{c}} {Persamaan}&2 \end{array}} \right) \end{array}$

Eliminasi Persamaan (1) dan (2) diperoleh:

$\begin{array}{l} \left( 1 \right)3a - b = 1\\ \left( 2 \right)\underline {a + b = 3} \left( + \right)\\ \Leftrightarrow 4a = 4\\ \Leftrightarrow a = \frac{4}{4} = 1 \end{array}$

Substitusi nilai $a=1$ ke persamaan (2) diperoleh:

$\begin{array}{l} a + b = 3 \cdots \left( 2 \right)\\ \Leftrightarrow 1 + b = 3\\ \Leftrightarrow b = 3 - 1 = 2 \end{array}$

Dengan substitusi $a=1$ dan $b=2$ ke persamaan awal diperoleh:

$\begin{array}{l} P\left( x \right) = \left( {{x^2} - x - 2} \right).\left( {ax + b} \right) + \left( {2x + 3} \right)\\ \Leftrightarrow P\left( x \right) = \left( {{x^2} - x - 2} \right).\left( {1.x + 2} \right) + \left( {2x + 3} \right)\\ \Leftrightarrow P\left( x \right) = \left( {{x^2} - x - 2} \right).\left( {x + 2} \right) + \left( {2x + 3} \right)\\ \Leftrightarrow P\left( x \right) = x\left( {{x^2} - x - 2} \right) + 2\left( {{x^2} - x - 2} \right) + \left( {2x + 3} \right)\\ \Leftrightarrow P\left( x \right) = {x^3} - {x^2} - 2x + 2{x^2} - 2x - 4 + 2x + 3\\ \Leftrightarrow P\left( x \right) = {x^3} + {x^2} - 2x - 1 \end{array}$

Jawaban yang benar adalah B

(CARA SUPER CEPAT)

Suku banyak berderajad 3, jika dibagi $\left( {{x^2} + 3x - 2} \right) = \left( {x + 3} \right)\left( {x - 1} \right)$ bersisa $\left( {3x - 4} \right)$ artinya:

$\begin{array}{l} P\left( { - 3} \right) = 3.\left( { - 3} \right) - 4 = - 13\\ P\left( 1 \right) = 3.1 - 4 = - 1 \end{array}$

Misalkan kita pilih satu persamaan yaitu $P\left( 1 \right) = - 1$. Jika kita substitusikan $x = 1$ hasilnya adalah $-1$. Dari pilihan jawaban yang memenuhi hanya jawaban B saja.