PAS MATEMATIKA BLOG: Contoh Soal Polinom (Bagian 6)

$\begin{array}{ll} 26. & Banyaknya akar rasional bulat \\ dari 4 x^{4} - 15 x^{2} + 5 x + 6 = 0 adalah . . . . \\ \begin{array}{lll} a . 0 & d . 3 \\ b . 1 & e . 4 \\ c . 2 \end{array} \\ Jawab : \\ \begin{matrix} \begin{aligned} Perhatikan uraian berikut \\ \frac{4 x^{4} - 15 x^{2} + 5 x + 6}{(x - 1)} \end{aligned} \\ \begin{array}{llll} pembagi & 4 x^{3} + 4 x^{2} - 11 x - 6 & hasil \\ x - 1 & 4 x^{4} - 15 x^{2} + 5 x + 6 \\ 4 x^{4} - 4 x^{3} & - \\ 4 x^{3} - 15 x^{2} + 5 x + 6 \\ 4 x^{3} - 4 x^{2} & - \\ - 11 x^{2} + 5 x + 6 \\ - 11 x^{2} + 11 x & - \\ - 6 x + 6 \\ - 6 x + 6 & - \\ Sisa & 0 & (habis) \end{array} \\ \begin{aligned} ∴ f (x) & = 4 x^{4} - 15 x^{2} + 5 x + 6 \\ = (x - 1) (4 x^{3} + 4 x^{2} - 11 x - 6) \end{aligned} \end{matrix} \\ \begin{matrix} \begin{aligned} Selanjutnya perhatikan pula \\ \frac{4 x^{3} + 4 x^{2} - 11 x - 6}{(x + 2)} \end{aligned} \\ \begin{array}{llll} pembagi & 4 x^{2} - 4 x - 3 & hasil \\ x + 2 & 4 x^{3} + 4 x^{2} - 11 x - 6 \\ 4 x^{3} + 8 x^{2} & - \\ - 4 x^{2} - 11 x - 6 \\ - 4 x^{2} - 8 x & - \\ - 3 x - 6 \\ - 3 x - 6 & - \\ Sisa & 0 & (habis) \end{array} \\ \begin{aligned} ∴ f (x) & = 4 x^{3} + 4 x^{2} - 11 x - 6 \\ = (x + 2) (4 x^{2} - 4 x - 3) \\ = (x + 2) (2 x - 3) (2 x + 1) \end{aligned} \end{matrix} \end{array}$

$\begin{array}{ll} 27. & Salah satu akar dari polinomial \\ 2 x^{3} + 5 x^{2} + x - 2 = 0 adalah \frac{1}{2} \\ Jumlah dua akar yang lain adalah . . . . \\ \begin{array}{lll} a . 6 & d . - 3 \\ b . 3 & e . - 6 \\ c . 2 \end{array} \\ Jawab : \\ \begin{matrix} \begin{aligned} Perhatikan uraian berikut \\ \frac{2 x^{3} + 5 x^{2} + x - 2}{(2 x - 1)} \end{aligned} \\ \begin{array}{llll} pembagi & x^{2} + 3 x + 2 & hasil \\ 2 x - 1 & 2 x^{3} + 5 x^{2} + x - 2 \\ 2 x^{3} - x^{2} & - \\ 6 x^{2} + x - 2 \\ 6 x^{2} - 3 x & - \\ 4 x - 2 \\ 4 x - 2 & - \\ Sisa & 0 & (habis) \end{array} \\ \begin{aligned} ∴ f (x) & = 2 x^{3} + 5 x^{2} + x - 2 \\ = (2 x - 1) (x^{2} + 3 x + 2) \\ = (2 x - 1) (x + 1) (x + 2) \end{aligned} \end{matrix} \\ Sehingga x_{2} = - 1, x_{3} = - 2, maka \\ x_{2} + x_{3} = - 1 + (- 2) = - 3 \\ atau dapat juga ditentukan dengan rumus \\ ABC pada persamaan kuadrat, yaitu \\ x^{2} + 3 x + 2 = a_{2} x^{2} + a_{1} x + a_{0} \\ \Leftrightarrow x^{2} + 3 x + 2 = - \frac{a_{1}}{a_{1}} = - \frac{3}{1} = - 3 \end{array}$ .

$\begin{array}{ll} 28. & Salah satu akar dari polinomial \\ x^{4} - 5 x^{3} + 5 x^{2} + 5 x - 6 = 0 \\ adalah 2 . Jumlah akar-akar yang \\ lain adalah . . . . \\ \begin{array}{lll} a . 6 & d . 3 \\ b . 5 & e . 2 \\ c . 4 \end{array} \\ Jawab : \\ \begin{matrix} \begin{aligned} Perhatikan uraian berikut \\ \frac{x^{4} - 5 x^{3} + 5 x^{2} + 5 x - 6}{(x - 2)} \end{aligned} \\ \begin{array}{llll} pembagi & x^{3} - 3 x^{2} - x + 3 & hasil \\ x - 2 & x^{4} - 5 x^{3} + 5 x^{2} + 5 x - 6 \\ x^{4} - 2 x^{3} & - \\ - 3 x^{3} + 5 x^{2} + 5 x - 6 \\ - 3 x^{3} + 6 x^{2} & - \\ - x^{2} + 5 x - 6 \\ - x^{2} + 2 x & - \\ 3 x - 6 \\ 3 x - 6 & - \\ Sisa & 0 & (habis) \end{array} \\ \begin{aligned} ∴ f (x) & = x^{4} - 5 x^{3} + 5 x^{2} + 5 x - 6 \\ = (x - 2) (x^{3} - 3 x^{2} - x + 3) \end{aligned} \end{matrix} \\ Sehingga \\ x^{3} - 3 x^{2} - x + 3 = a_{3} x^{3} + a_{2} x^{2} + a_{1} x + a_{0} \\ \Leftrightarrow x_{1} + x_{2} + x_{3} = - \frac{a_{2}}{a_{2}} = - \frac{- 3}{1} = 3 \end{array}$ .

$\begin{array}{ll} 29. & Akar-akar dari persamaan polinom \\ x^{3} - 3 x^{2} + 4 x + 5 = 0 adalah x_{1}, x_{2} \\ dan x_{3} . Nilai x_{1}^{2} + x_{2}^{2} + x_{3}^{2} = . . . . \\ \begin{array}{lll} a . 1 & d . 17 \\ b . 3 & e . 19 \\ c . 9 \end{array} \\ Jawab : \\ \begin{aligned} x^{3} - 3 x^{2} + 4 x + 5 = a x^{3} + b x^{2} + c x + d \\ Perhatikan bahwa : \\ x_{1}^{2} + x_{2}^{2} + x_{3}^{2} \\ = (x_{1} + x_{2} + x_{3})^{2} - 2 (x_{1} x_{2} + x_{2} x_{3} + x_{1} x_{3}) \\ = {(\frac{- b}{a})}^{2} - 2 (\frac{c}{a}) \\ = {(\frac{- (- 3)}{1})}^{2} - 2 (\frac{4}{1}) = 9 - 8 = 1 \end{aligned} \end{array}$ .

$\begin{array}{ll} 30. & Diketahui persamaan polinom \\ x^{3} - 4 x^{2} + 6 x - 12 = 0 mempunyai \\ akar-akar x_{1}, x_{2} dan x_{3} . \\ Nilai \frac{1}{x_{1}} + \frac{1}{x_{2}} + \frac{1}{x_{3}} = . . . . \\ \begin{array}{lll} a . - 3 & d . \frac{1}{2} \\ b . - 2 & e . 3 \\ c . \frac{1}{3} \end{array} \\ Jawab : \\ \begin{aligned} x^{3} - 4 x^{2} + 6 x - 12 = a x^{3} + b x^{2} + c x + d \\ Perhatikan bahwa : \\ \frac{1}{x_{1}} + \frac{1}{x_{2}} + \frac{1}{x_{3}} \\ = \frac{x_{2} x_{3} + x_{1} x_{3} + x_{1} x_{2}}{x_{1} x_{2} x_{3}} \\ = \frac{\frac{c}{a}}{- \frac{d}{a}} = - \frac{c}{d} = - \frac{6}{- 12} = \frac{1}{2} \end{aligned} \end{array}$ .

PAS MATEMATIKA BLOG

Contoh Soal Polinom (Bagian 6)

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