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Question Number 14775 by tawa tawa last updated on 04/Jun/17
Using the remainder theorem to factorize completely the expression x^3 (y − z) + y^3 (z − x) + z^3 (x − y)
$$\mathrm{Using}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{theorem}\:\mathrm{to}\:\mathrm{factorize}\:\mathrm{completely}\:\mathrm{the}\:\mathrm{expression}\: \\ $$$$\mathrm{x}^{\mathrm{3}} \left(\mathrm{y}\:−\:\mathrm{z}\right)\:+\:\mathrm{y}^{\mathrm{3}} \left(\mathrm{z}\:−\:\mathrm{x}\right)\:+\:\mathrm{z}^{\mathrm{3}} \left(\mathrm{x}\:−\:\mathrm{y}\right)\: \\ $$
Answered by ajfour last updated on 04/Jun/17
we can see that, if y=z 0+z^3 (z−x)+z^3 (x−z)=0 so, (y−z) or (z−y) is a factor similarly for (x−y) or (y−x) and for (z−x) or (x−z) ; hence : ±(x−y)(y−z)(z−x) is a factor of the expression; what can come next is of degree one, first let us see if (x+y+z) happens to be a factor or not, for that let z=−x−y x^3 (2y+x)+y^3 (−2x−y)+ (x+y)^2 (y^2 −x^2 ) =2x^3 y+x^4 −2xy^3 −y^4 + +x^2 y^2 −x^4 +2xy^3 −2x^3 y+y^4 −x^2 y^2 =0 so the expression is ±(x−y)(y−z)(z−x)(x+y+z) to check for the sign let us compare coefficient of x^3 y we find that they agree when we choose the −ve sign , so factorised form of expression in question is (y−x)(z−y)(x−z)(x+y+z) .
$${we}\:{can}\:{see}\:{that},\:{if}\:\boldsymbol{{y}}=\boldsymbol{{z}} \\ $$$$\:\:\:\mathrm{0}+\boldsymbol{{z}}^{\mathrm{3}} \left(\boldsymbol{{z}}−\boldsymbol{{x}}\right)+\boldsymbol{{z}}^{\mathrm{3}} \left(\boldsymbol{{x}}−\boldsymbol{{z}}\right)=\mathrm{0} \\ $$$${so},\:\:\:\left({y}−{z}\right)\:{or}\:\left({z}−{y}\right)\:{is}\:{a}\:{factor} \\ $$$${similarly}\:{for}\:\:\:\left({x}−{y}\right)\:{or}\:\left({y}−{x}\right) \\ $$$${and}\:{for}\:\:\:\left({z}−{x}\right)\:{or}\:\left({x}−{z}\right)\:; \\ $$$${hence}\::\:\:\pm\left({x}−{y}\right)\left({y}−{z}\right)\left({z}−{x}\right)\:{is}\:{a} \\ $$$${factor}\:{of}\:{the}\:{expression}; \\ $$$${what}\:{can}\:{come}\:{next}\:{is} \\ $$$${of}\:{degree}\:\:{one},\:{first}\:{let}\:{us}\:{see}\:{if} \\ $$$$\left({x}+{y}+{z}\right)\:{happens}\:{to}\:{be}\:{a}\:{factor} \\ $$$${or}\:{not},\:{for}\:{that}\:{let}\:{z}=−{x}−{y} \\ $$$$\:{x}^{\mathrm{3}} \left(\mathrm{2}{y}+{x}\right)+{y}^{\mathrm{3}} \left(−\mathrm{2}{x}−{y}\right)+ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({x}+{y}\right)^{\mathrm{2}} \left({y}^{\mathrm{2}} −{x}^{\mathrm{2}} \right) \\ $$$$=\mathrm{2}{x}^{\mathrm{3}} {y}+{x}^{\mathrm{4}} −\mathrm{2}{xy}^{\mathrm{3}} −{y}^{\mathrm{4}} + \\ $$$$+{x}^{\mathrm{2}} {y}^{\mathrm{2}} −{x}^{\mathrm{4}} +\mathrm{2}{xy}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{3}} {y}+{y}^{\mathrm{4}} −{x}^{\mathrm{2}} {y}^{\mathrm{2}} \\ $$$$\:\:\:\:\:=\mathrm{0} \\ $$$${so}\:{the}\:{expression}\:{is}\: \\ $$$$\pm\left({x}−{y}\right)\left({y}−{z}\right)\left({z}−{x}\right)\left({x}+{y}+{z}\right) \\ $$$${to}\:{check}\:{for}\:{the}\:{sign}\:{let}\:{us}\: \\ $$$${compare}\:{coefficient}\:{of}\:\boldsymbol{{x}}^{\mathrm{3}} \boldsymbol{{y}} \\ $$$$\:{we}\:{find}\:{that}\:{they}\:{agree}\:{when} \\ $$$${we}\:{choose}\:{the}\:−{ve}\:{sign}\:,\:{so} \\ $$$${factorised}\:{form}\:{of}\:{expression}\:{in} \\ $$$${question}\:{is} \\ $$$$\:\:\left(\boldsymbol{{y}}−\boldsymbol{{x}}\right)\left(\boldsymbol{{z}}−\boldsymbol{{y}}\right)\left(\boldsymbol{{x}}−\boldsymbol{{z}}\right)\left(\boldsymbol{{x}}+\boldsymbol{{y}}+\boldsymbol{{z}}\right)\:. \\ $$$$\: \\ $$
Commented by tawa tawa last updated on 04/Jun/17
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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