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Question Number 57000 by pieroo last updated on 28/Mar/19

$$\mathrm{If}{(\mathrm{x}+2)}^2\mathrm{is}\mathrm{a}\mathrm{factor}\mathrm{of}\mathrm{the}\mathrm{polynomial}$$$$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{mx}}^3+{\mathrm{x}}^2+\mathrm{x}+\mathrm{n},\mathrm{find};$$$$\mathrm{the}\mathrm{values}\mathrm{of}\mathrm{m}\mathrm{and}\mathrm{n}.$$

Commented by maxmathsup by imad last updated on 29/Mar/19

$$\Rightarrow f(-2)=f^{{}^{\prime }}(-2)=0butf(-2)=-8m+4-2+n=-8m+n+2$$$$f^{{}^{\prime }}\left(x\right)=3{mx}^2+2x+1\Rightarrow f^{{}^{\prime }}(-2)=12m-4+1=12m-3\Rightarrow 12m-3=0and$$$$-8m+n+2=0\Rightarrow m=\frac{1}{4}andn=0\Rightarrow f\left(x\right)=\frac{1}{4}{x}^3+x^2+x.$$

Answered by Smail last updated on 28/Mar/19

$$f(-2)=8m+4+2+n$$$$f^{\prime }\left(x\right)=3{mx}^2+2x+1$$$$Since(x+2)isadoublefactoroff$$$$the(x+2)isasimplefactorof\frac{df\left(x\right)}{dx}$$$$Whichmeans{f}^{\prime }(-2)=0\iff 12m+4+1=0$$$$m=-\frac{5}{12}$$$$f(-2)=8\times (-\frac{5}{12})+6+n$$$$n=\frac{8}{3}$$$$f\left(x\right)=\frac{-5}{12}{x}^3+x^2+x+\frac{8}{3}$$

Commented by MJS last updated on 28/Mar/19

$$\mathrm{something}\mathrm{went}\mathrm{wrong}$$

Commented by Smail last updated on 28/Mar/19

$$Imadeamistakeonthe5thline$$$$f^{\prime }(-2)=3m{(-2)}^2+2\times (-2)+1=0$$$$12m-4+1=0\iff m=\frac{1}{4}$$$$f(-2)=\frac{1}{4}\times {(-2)}^3+4-2+n=0$$$$n=0$$

Answered by mr W last updated on 28/Mar/19

$${mx}^3+x^2+x+n={(x+2)}^2(ax+b)$$$${mx}^3+x^2+x+n=(x^2+4x+4)(ax+b)$$$${mx}^3+x^2+x+n={ax}^3+(4a+b)x^2+4(a+b)x+4b$$$$4a+b=4(a+b)=1$$$$\Rightarrow b=0$$$$\Rightarrow a=\frac{1}{4}$$$$\Rightarrow m=a=\frac{1}{4}$$$$\Rightarrow n=4b=0$$$$\frac{1}{4}{x}^3+x^2+x=\frac{1}{4}{(x+2)}^{2}x$$

Commented by pete last updated on 29/Mar/19

$$\mathrm{thanks}\mathrm{prof}$$

Answered by malwaan last updated on 29/Mar/19

$$bylongdivisionwehave$$$$n-8m+2=0\left(1\right)$$$$12m-3=0\left(2\right)$$$$\Rightarrow m=\frac{1}{4};n=0$$

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