The cubic polynomial $ \displaystyle f(x)$ is such that the coefficient of $ \displaystyle x^3$ is $ \displaystyle -1$ and the roots of the equation $ \displaystyle f(x)=0$ are $ \displaystyle 1$, $ \displaystyle 2$ and $ \displaystyle k$. Given that $ \displaystyle f(x)$ has a remainder of $ \displaystyle 8$ when divided by $ \displaystyle x-3$, find the value of $ \displaystyle k$ and the remainder when $ \displaystyle f(x)$ is divided by $ \displaystyle x+3$ .
$ \displaystyle f(x)$ ဟာ အႀကီးဆံုးထပ္ကိန္း $ \displaystyle 3$ ထပ္ပါတဲ့ $ \displaystyle \text{polynomial}$ ကိန္းတန္း တစ္ခုျဖစ္တယ္။ $ \displaystyle f(x)=0$ ကို ေျဖရွင္းလို႔ရတဲ့ $ \displaystyle x$ တန္ဖိုးေတြက $ \displaystyle 1$, $ \displaystyle 2$ နဲ႔ $ \displaystyle k$ ျဖစ္ၾကတယ္။ $ \displaystyle x^3$ ရဲ့ ေျမႇာက္ေဖၚကိန္းက $ \displaystyle -1$ ျဖစ္တယ္။
$ \displaystyle f(x)$ ကို $ \displaystyle x-3$ နဲ႔ စားလို႔ရတဲ့ အႂကြင္း $ \displaystyle \text{(remainder)}$ က $ \displaystyle 8$ ျဖစ္တယ္ဆိုရင္ $ \displaystyle k$ ကို ရွာေပးပါ။
အဲဒီေနာက္ $ \displaystyle f(x)$ ကို $ \displaystyle x+3$ နဲ႔ စားလို႔ရတဲ့ အႂကြင္း $ \displaystyle \text{(remainder)}$ ကိုလည္း ရွာေပးပါ။
ဒီအခ်က္ေတြကိုေပါင္းလိုက္ရင္
$ \displaystyle \begin{array}{l}\text{By the problem,}\\\\f(x)=-1(x-1)(x-2)(x-k)\\\\f(3)=8\\\\\therefore -1(3-1)(3-2)(3-k)=8\\\\\therefore -1(2)(1)(3-k)=8\\\\\therefore 3-k=-4\Rightarrow k=7\\\\\therefore f(x)=-1(x-1)(x-2)(x-7)\\\\\text{When }f(x)\ \text{is divided by }x+3,\\\\\text{the remainder = }f(-3)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-1(-3-1)(-3-2)(-3-7)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-1(-4)(-5)(-10)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =200\end{array}$
- $ \displaystyle f(x)=-1(x-1)(x-2)(x-k)$ လို႔သိရပါမယ္။
- $ \displaystyle f(3)= 8$ ျဖစ္တယ္လို႔ သိရမယ္။
- $ \displaystyle f(3)$ မွာ ကို $ \displaystyle \text{function}$ မွာ အစားသြင္းၿပီး $ \displaystyle f(3)$ နဲ႔ ညီေပးလိုက္ရင္ $ \displaystyle k$ တန္ဖိုးရၿပီေပါ့။
- $ \displaystyle k$ ရတဲ့အခါ $ \displaystyle f(x)$ ကို $ \displaystyle x+3$ နဲ႔ စားလို႔ရတဲ့ အႂကြင္း $ \displaystyle \text{(remainder)}$ ဆိုတာ $ \displaystyle f(-3)$ ကို ရွာခိုင္းတာ ျဖစ္ပါတယ္။
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Solution$ \displaystyle \begin{array}{l}\text{By the problem,}\\\\f(x)=-1(x-1)(x-2)(x-k)\\\\f(3)=8\\\\\therefore -1(3-1)(3-2)(3-k)=8\\\\\therefore -1(2)(1)(3-k)=8\\\\\therefore 3-k=-4\Rightarrow k=7\\\\\therefore f(x)=-1(x-1)(x-2)(x-7)\\\\\text{When }f(x)\ \text{is divided by }x+3,\\\\\text{the remainder = }f(-3)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-1(-3-1)(-3-2)(-3-7)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-1(-4)(-5)(-10)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =200\end{array}$
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