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If-x-2-2-is-a-factor-of-the-polynomial-f-x-mx-3-x-2-x-n-find-the-values-of-m-and-n-

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Question Number 57000 by pieroo last updated on 28/Mar/19
If (x+2)^2 is a factor of the polynomial f(x)=mx^3 +x^2 +x+n, find; the values of m and n.
$$\mathrm{If}\:\left(\mathrm{x}+\mathrm{2}\right)^{\mathrm{2}} \:\mathrm{is}\:\mathrm{a}\:\mathrm{factor}\:\mathrm{of}\:\mathrm{the}\:\mathrm{polynomial} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{mx}^{\mathrm{3}} +\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{n},\:\mathrm{find}; \\ $$$$\:\mathrm{the}\:\mathrm{values}\:\mathrm{of}\:\mathrm{m}\:\mathrm{and}\:\mathrm{n}. \\ $$
Commented by maxmathsup by imad last updated on 29/Mar/19
⇒f(−2)=f^′ (−2)=0 but f(−2)=−8m+4−2+n =−8m+n+2 f^′ (x)=3mx^2 +2x+1 ⇒f^′ (−2)=12m−4+1 =12m−3 ⇒12m−3=0 and −8m +n+2 =0 ⇒m=(1/4) and n=0 ⇒f(x)=(1/4) x^3 +x^2 +x .
$$\Rightarrow{f}\left(−\mathrm{2}\right)={f}^{'} \left(−\mathrm{2}\right)=\mathrm{0}\:\:\:{but}\:{f}\left(−\mathrm{2}\right)=−\mathrm{8}{m}+\mathrm{4}−\mathrm{2}+{n}\:=−\mathrm{8}{m}+{n}+\mathrm{2} \\ $$$${f}^{'} \left({x}\right)=\mathrm{3}{mx}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{1}\:\Rightarrow{f}^{'} \left(−\mathrm{2}\right)=\mathrm{12}{m}−\mathrm{4}+\mathrm{1}\:=\mathrm{12}{m}−\mathrm{3}\:\Rightarrow\mathrm{12}{m}−\mathrm{3}=\mathrm{0}\:{and} \\ $$$$−\mathrm{8}{m}\:+{n}+\mathrm{2}\:=\mathrm{0}\:\Rightarrow{m}=\frac{\mathrm{1}}{\mathrm{4}}\:{and}\:{n}=\mathrm{0}\:\Rightarrow{f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{4}}\:{x}^{\mathrm{3}} \:+{x}^{\mathrm{2}} \:+{x}\:. \\ $$
Answered by Smail last updated on 28/Mar/19
f(−2)=8m+4+2+n f′(x)=3mx^2 +2x+1 Since (x+2) is a double factor of f the (x+2) is a simple factor of ((df(x))/dx) Which means f′(−2)=0⇔12m+4+1=0 m=−(5/(12)) f(−2)=8×(−(5/(12)))+6+n n=(8/3) f(x)=((−5)/(12))x^3 +x^2 +x+(8/3)
$${f}\left(−\mathrm{2}\right)=\mathrm{8}{m}+\mathrm{4}+\mathrm{2}+{n} \\ $$$${f}'\left({x}\right)=\mathrm{3}{mx}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1} \\ $$$${Since}\:\left({x}+\mathrm{2}\right)\:\:{is}\:{a}\:{double}\:{factor}\:{of}\:{f} \\ $$$${the}\:\left({x}+\mathrm{2}\right)\:{is}\:\:{a}\:{simple}\:{factor}\:{of}\:\frac{{df}\left({x}\right)}{{dx}} \\ $$$${Which}\:{means}\:\:{f}'\left(−\mathrm{2}\right)=\mathrm{0}\Leftrightarrow\mathrm{12}{m}+\mathrm{4}+\mathrm{1}=\mathrm{0} \\ $$$${m}=−\frac{\mathrm{5}}{\mathrm{12}} \\ $$$${f}\left(−\mathrm{2}\right)=\mathrm{8}×\left(−\frac{\mathrm{5}}{\mathrm{12}}\right)+\mathrm{6}+{n} \\ $$$${n}=\frac{\mathrm{8}}{\mathrm{3}} \\ $$$${f}\left({x}\right)=\frac{−\mathrm{5}}{\mathrm{12}}{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\frac{\mathrm{8}}{\mathrm{3}} \\ $$
Commented by MJS last updated on 28/Mar/19
something went wrong
$$\mathrm{something}\:\mathrm{went}\:\mathrm{wrong} \\ $$
Commented by Smail last updated on 28/Mar/19
I made a mistake on the 5th line f′(−2)=3m(−2)^2 +2×(−2)+1=0 12m−4+1=0⇔m=(1/4) f(−2)=(1/4)×(−2)^3 +4−2+n=0 n=0
$${I}\:{made}\:{a}\:{mistake}\:{on}\:{the}\:\mathrm{5}{th}\:{line} \\ $$$${f}'\left(−\mathrm{2}\right)=\mathrm{3}{m}\left(−\mathrm{2}\right)^{\mathrm{2}} +\mathrm{2}×\left(−\mathrm{2}\right)+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{12}{m}−\mathrm{4}+\mathrm{1}=\mathrm{0}\Leftrightarrow{m}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${f}\left(−\mathrm{2}\right)=\frac{\mathrm{1}}{\mathrm{4}}×\left(−\mathrm{2}\right)^{\mathrm{3}} +\mathrm{4}−\mathrm{2}+{n}=\mathrm{0} \\ $$$${n}=\mathrm{0} \\ $$
Answered by mr W last updated on 28/Mar/19
mx^3 +x^2 +x+n=(x+2)^2 (ax+b) mx^3 +x^2 +x+n=(x^2 +4x+4)(ax+b) mx^3 +x^2 +x+n=ax^3 +(4a+b)x^2 +4(a+b)x+4b 4a+b=4(a+b)=1 ⇒b=0 ⇒a=(1/4) ⇒m=a=(1/4) ⇒n=4b=0 (1/4)x^3 +x^2 +x=(1/4)(x+2)^2 x
$${mx}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+{n}=\left({x}+\mathrm{2}\right)^{\mathrm{2}} \left({ax}+{b}\right) \\ $$$${mx}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+{n}=\left({x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{4}\right)\left({ax}+{b}\right) \\ $$$${mx}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+{n}={ax}^{\mathrm{3}} +\left(\mathrm{4}{a}+{b}\right){x}^{\mathrm{2}} +\mathrm{4}\left({a}+{b}\right){x}+\mathrm{4}{b} \\ $$$$\mathrm{4}{a}+{b}=\mathrm{4}\left({a}+{b}\right)=\mathrm{1} \\ $$$$\Rightarrow{b}=\mathrm{0} \\ $$$$\Rightarrow{a}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow{m}={a}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow{n}=\mathrm{4}{b}=\mathrm{0} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}=\frac{\mathrm{1}}{\mathrm{4}}\left({x}+\mathrm{2}\right)^{\mathrm{2}} {x} \\ $$
Commented by pete last updated on 29/Mar/19
thanks prof
$$\mathrm{thanks}\:\mathrm{prof} \\ $$
Answered by malwaan last updated on 29/Mar/19
by long division we have n−8m+2=0 (1) 12m−3=0 (2) ⇒m=(1/4) ; n=0
$${by}\:{long}\:{division}\:{we}\:{have} \\ $$$${n}−\mathrm{8}{m}+\mathrm{2}=\mathrm{0}\:\left(\mathrm{1}\right) \\ $$$$\mathrm{12}{m}−\mathrm{3}=\mathrm{0}\:\left(\mathrm{2}\right) \\ $$$$\Rightarrow{m}=\frac{\mathrm{1}}{\mathrm{4}}\:;\:\:{n}=\mathrm{0} \\ $$

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