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Question Number 120355 by TITA last updated on 30/Oct/20

Commented by TITA last updated on 30/Oct/20

please help

$${please}\:{help} \\ $$

Commented by TITA last updated on 30/Oct/20

please help

$${please}\:{help} \\ $$

Answered by john santu last updated on 30/Oct/20

(1) f(2)=4 (2)f(−3)=−1 (3) f(x): (x+3)(x−2) ⇒ r(x) = ((x+3)/(2+3)) f(2)+((x−2)/(−3−2))f(−3) ⇒r(x)=(4/5)(x+3)+(1/5)(x−2) ⇒r(x) = ((4x+12+x−2)/5)=((5x+10)/5) ⇒r(x)=x+2

$$\left(\mathrm{1}\right)\:{f}\left(\mathrm{2}\right)=\mathrm{4} \\ $$$$\left(\mathrm{2}\right){f}\left(−\mathrm{3}\right)=−\mathrm{1} \\ $$$$\left(\mathrm{3}\right)\:{f}\left({x}\right):\:\left({x}+\mathrm{3}\right)\left({x}−\mathrm{2}\right) \\ $$$$\Rightarrow\:{r}\left({x}\right)\:=\:\frac{{x}+\mathrm{3}}{\mathrm{2}+\mathrm{3}}\:{f}\left(\mathrm{2}\right)+\frac{{x}−\mathrm{2}}{−\mathrm{3}−\mathrm{2}}{f}\left(−\mathrm{3}\right) \\ $$$$\Rightarrow{r}\left({x}\right)=\frac{\mathrm{4}}{\mathrm{5}}\left({x}+\mathrm{3}\right)+\frac{\mathrm{1}}{\mathrm{5}}\left({x}−\mathrm{2}\right) \\ $$$$\Rightarrow{r}\left({x}\right)\:=\:\frac{\mathrm{4}{x}+\mathrm{12}+{x}−\mathrm{2}}{\mathrm{5}}=\frac{\mathrm{5}{x}+\mathrm{10}}{\mathrm{5}} \\ $$$$\Rightarrow{r}\left({x}\right)={x}+\mathrm{2} \\ $$

Commented by TITA last updated on 30/Oct/20

thank sir

$${thank}\:{sir} \\ $$$$ \\ $$

Answered by benjo_mathlover last updated on 30/Oct/20

in other way let f(x) = g(x)(x+3)(x−2)+px+q where px+q is remainder when f(x) divided by x^2 +x−6 for x=2⇒2p+q=4 for x=−3⇒−3p+q=−1 ((( 2 1)),((−3 1)) ) ((p),(q) ) = ((( 4)),((−1)) ) → { ((p=( determinant ((( 4 1)),((−1 1)))/5)= (5/5)=1)),((q=( determinant ((( 2 4)),((−3 −1)))/5)=((10)/5)=2)) :} thus r(x)=1.x+2=x+2

$${in}\:{other}\:{way}\: \\ $$$${let}\:{f}\left({x}\right)\:=\:{g}\left({x}\right)\left({x}+\mathrm{3}\right)\left({x}−\mathrm{2}\right)+{px}+{q} \\ $$$${where}\:{px}+{q}\:{is}\:{remainder}\:{when}\:{f}\left({x}\right)\:{divided} \\ $$$${by}\:{x}^{\mathrm{2}} +{x}−\mathrm{6} \\ $$$${for}\:{x}=\mathrm{2}\Rightarrow\mathrm{2}{p}+{q}=\mathrm{4} \\ $$$${for}\:{x}=−\mathrm{3}\Rightarrow−\mathrm{3}{p}+{q}=−\mathrm{1} \\ $$$$\begin{pmatrix}{\:\:\:\mathrm{2}\:\:\:\:\:\:\mathrm{1}}\\{−\mathrm{3}\:\:\:\:\mathrm{1}}\end{pmatrix}\:\begin{pmatrix}{{p}}\\{{q}}\end{pmatrix}\:=\:\begin{pmatrix}{\:\:\:\mathrm{4}}\\{−\mathrm{1}}\end{pmatrix} \\ $$$$\rightarrow\begin{cases}{{p}=\frac{\begin{vmatrix}{\:\:\:\mathrm{4}\:\:\:\:\:\:\:\mathrm{1}}\\{−\mathrm{1}\:\:\:\:\:\mathrm{1}}\end{vmatrix}}{\mathrm{5}}=\:\frac{\mathrm{5}}{\mathrm{5}}=\mathrm{1}}\\{{q}=\frac{\begin{vmatrix}{\:\:\:\:\mathrm{2}\:\:\:\:\:\:\mathrm{4}}\\{−\mathrm{3}\:\:\:\:−\mathrm{1}}\end{vmatrix}}{\mathrm{5}}=\frac{\mathrm{10}}{\mathrm{5}}=\mathrm{2}}\end{cases} \\ $$$${thus}\:{r}\left({x}\right)=\mathrm{1}.{x}+\mathrm{2}={x}+\mathrm{2} \\ $$

Answered by mathmax by abdo last updated on 30/Oct/20

f(x)=(x−2)q(x)+4 and f(x)=(x+3)v(x)−1 ⇒ f(2)=4 and f(−3)=−1 f(x)=h(x)(x^2 +x−6)+ax+b we have f(2) =2a+b=4 and f(−3)=−3a+b =−1 ⇒ { ((2a+b=4)),((3a−b=1 ⇒ { ((5a=5)),((b=4−2a ⇒ { ((a=1)),((b=2 ⇒)) :})) :})) :} f(x)=h(x^2 +x−6)+x+2 ⇒r(x)=x+2

$$\mathrm{f}\left(\mathrm{x}\right)=\left(\mathrm{x}−\mathrm{2}\right)\mathrm{q}\left(\mathrm{x}\right)+\mathrm{4}\:\mathrm{and}\:\mathrm{f}\left(\mathrm{x}\right)=\left(\mathrm{x}+\mathrm{3}\right)\mathrm{v}\left(\mathrm{x}\right)−\mathrm{1}\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{2}\right)=\mathrm{4}\:\mathrm{and}\:\mathrm{f}\left(−\mathrm{3}\right)=−\mathrm{1} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{h}\left(\mathrm{x}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}−\mathrm{6}\right)+\mathrm{ax}+\mathrm{b} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{f}\left(\mathrm{2}\right)\:=\mathrm{2a}+\mathrm{b}=\mathrm{4}\:\mathrm{and}\:\mathrm{f}\left(−\mathrm{3}\right)=−\mathrm{3a}+\mathrm{b}\:=−\mathrm{1}\:\Rightarrow \\ $$$$\begin{cases}{\mathrm{2a}+\mathrm{b}=\mathrm{4}}\\{\mathrm{3a}−\mathrm{b}=\mathrm{1}\:\Rightarrow\begin{cases}{\mathrm{5a}=\mathrm{5}}\\{\mathrm{b}=\mathrm{4}−\mathrm{2a}\:\Rightarrow\begin{cases}{\mathrm{a}=\mathrm{1}}\\{\mathrm{b}=\mathrm{2}\:\Rightarrow}\end{cases}}\end{cases}}\end{cases} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{h}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}−\mathrm{6}\right)+\mathrm{x}+\mathrm{2}\:\Rightarrow\mathrm{r}\left(\mathrm{x}\right)=\mathrm{x}+\mathrm{2} \\ $$

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