Question Number 84430 by M±th+et£s last updated on 12/Mar/20
Commented by mr W last updated on 12/Mar/20
$${f}\left(\mathrm{2}\right)=\mathrm{2}^{\mathrm{2}} =\mathrm{4} \\ $$
Commented by M±th+et£s last updated on 12/Mar/20
$${thank}\:{you}\:{sir}\:{but}\:{can}\:{you}\:{tell}\:{me}\:{how}\: \\ $$$${did}\:{you}\:{get}\:{this}\:{answer} \\ $$
Commented by mr W last updated on 13/Mar/20
$${what}\:{is}\:{a}\:{polynomial}? \\ $$$${a}\:{polynomial}\:{is}\:{a}\:{expression}\:{consisting} \\ $$$${of}\:{variables}\:{and}\:{coefficients},\:{that} \\ $$$${involves}\:{only}\:{the}\:{operations}\:{of} \\ $$$${addition},\:{subtraction},\:{multiplication}, \\ $$$${and}\:{non}−{negative}\:{integer}\:{exponents} \\ $$$${of}\:{variables}. \\ $$$$ \\ $$$${f}\left({x}\right)=\frac{{a}+\mathrm{3}}{{x}^{\mathrm{4}} }+\left({b}−\mathrm{2}\right){x}^{−\mathrm{3}} +{x}^{{n}−{a}+{b}} \\ $$$${such}\:{that}\:{f}\left({x}\right)\:{is}\:{a}\:{polynomial},\:{the} \\ $$$${terms}\:{with}\:\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\:{and}\:{x}^{−\mathrm{3}} \:{must}\:{vanish}, \\ $$$${i}.{e}.\:{a}=−\mathrm{3}\:{and}\:{b}=\mathrm{2}. \\ $$$$\Rightarrow{f}\left({x}\right)={x}^{{n}−{a}+{b}} \\ $$$${since}\:{it}\:{is}\:{a}\:{quadratic}\:{polynomial},\:{the} \\ $$$${highest}\:{exponent}\:{of}\:{variables}\:{is}\:\mathrm{2}, \\ $$$${but}\:{it}\:{has}\:{only}\:{one}\:{term},\:{therefore} \\ $$$${this}\:{term}\:{must}\:{be}\:{x}^{\mathrm{2}} ,\:{i}.{e}.\:{n}−{a}+{b}=\mathrm{2}, \\ $$$${or}\:{n}=−\mathrm{3}. \\ $$$$\Rightarrow{f}\left({x}\right)={x}^{\mathrm{2}} \\ $$
Commented by M±th+et£s last updated on 13/Mar/20
$${thank}\:{you}\:{so}\:{much}\:{sir} \\ $$