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If-2x-2-mxy-3y-2-5y-2-have-two-rational-factor-then-find-m-

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Question Number 75602 by peter frank last updated on 13/Dec/19
If 2x^2 −mxy+3y^2 −5y−2 have two rational factor then find m
$${If}\:\mathrm{2}{x}^{\mathrm{2}} −{mxy}+\mathrm{3}{y}^{\mathrm{2}} −\mathrm{5}{y}−\mathrm{2} \\ $$$${have}\:{two}\:{rational}\:{factor}\:{then} \\ $$$${find}\:\:\:{m} \\ $$
Answered by mr W last updated on 14/Dec/19
2x^2 −mxy+3y^2 −5y−2=(ax+by+c)(ex+fy+g) a,b,c,.... are integer. expand right hand side: aex^2 +bexy+cex +afxy +bfy^2 +cfy +agx +bgy+cg compare coefficients of all terms: ae=2 be+af=−m ce+ag=0 (coefficient of x term) bf=3 cf+bg=−5 cg=−2 a=1,e=2 (or a=2,e=1) 2b+f=−m 2c+g=0 ⇒g=−2c bf=3 cf+bg=−5 cg=−2 ⇒c(−2c)=−2 ⇒c^2 =1 with c=1: g=−2 2b+f=−m ⇒m=−6−1=−7 f−2b=−5 ⇒f=2b−5 ⇒f=1 ∨ f=−6 bf=3 ⇒b(2b−5)−3=0 ⇒2b^2 −5b−3=0 ⇒(2b+1)(b−3)=0 ⇒b=3 ∨ b=−(1/2) with c=−1: g=2 2b+f=−m ⇒m=6+1=7 −f+2b=−5 ⇒f=2b+5 ⇒f=−1 ∨ f=6 bf=3 ⇒b(2b+5)−3=0 ⇒2b^2 +5b−3=0 ⇒(2b−1)(b+3)=0 ⇒b=−3 ∨ b=(1/2) ⇒m=−7 or 7. 2x^2 +7xy+3y^2 −5y−2=(x+3y+1)(2x+y−2) 2x^2 −7xy+3y^2 −5y−2=(x−3y−1)(2x−y+2)
$$\mathrm{2}{x}^{\mathrm{2}} −{mxy}+\mathrm{3}{y}^{\mathrm{2}} −\mathrm{5}{y}−\mathrm{2}=\left({ax}+{by}+{c}\right)\left({ex}+{fy}+{g}\right) \\ $$$${a},{b},{c},….\:{are}\:{integer}. \\ $$$${expand}\:{right}\:{hand}\:{side}: \\ $$$${aex}^{\mathrm{2}} +{bexy}+{cex} \\ $$$$\:\:\:\:\:\:\:\:\:\:+{afxy}\:\:\:\:\:\:\:\:\:\:\:+{bfy}^{\mathrm{2}} +{cfy} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+{agx}\:\:\:\:\:\:\:\:\:\:\:\:\:+{bgy}+{cg} \\ $$$${compare}\:{coefficients}\:{of}\:{all}\:{terms}: \\ $$$${ae}=\mathrm{2} \\ $$$${be}+{af}=−{m} \\ $$$${ce}+{ag}=\mathrm{0}\:\:\:\left({coefficient}\:{of}\:{x}\:{term}\right) \\ $$$${bf}=\mathrm{3} \\ $$$${cf}+{bg}=−\mathrm{5} \\ $$$${cg}=−\mathrm{2} \\ $$$$ \\ $$$${a}=\mathrm{1},{e}=\mathrm{2}\:\left({or}\:{a}=\mathrm{2},{e}=\mathrm{1}\right) \\ $$$$\mathrm{2}{b}+{f}=−{m} \\ $$$$\mathrm{2}{c}+{g}=\mathrm{0}\:\Rightarrow{g}=−\mathrm{2}{c} \\ $$$${bf}=\mathrm{3} \\ $$$${cf}+{bg}=−\mathrm{5} \\ $$$${cg}=−\mathrm{2}\:\Rightarrow{c}\left(−\mathrm{2}{c}\right)=−\mathrm{2}\:\Rightarrow{c}^{\mathrm{2}} =\mathrm{1} \\ $$$$ \\ $$$${with}\:{c}=\mathrm{1}: \\ $$$${g}=−\mathrm{2} \\ $$$$\mathrm{2}{b}+{f}=−{m}\:\Rightarrow{m}=−\mathrm{6}−\mathrm{1}=−\mathrm{7} \\ $$$${f}−\mathrm{2}{b}=−\mathrm{5}\:\Rightarrow{f}=\mathrm{2}{b}−\mathrm{5}\:\Rightarrow{f}=\mathrm{1}\:\vee\:{f}=−\mathrm{6} \\ $$$${bf}=\mathrm{3}\:\Rightarrow{b}\left(\mathrm{2}{b}−\mathrm{5}\right)−\mathrm{3}=\mathrm{0}\:\Rightarrow\mathrm{2}{b}^{\mathrm{2}} −\mathrm{5}{b}−\mathrm{3}=\mathrm{0}\:\Rightarrow\left(\mathrm{2}{b}+\mathrm{1}\right)\left({b}−\mathrm{3}\right)=\mathrm{0}\:\Rightarrow{b}=\mathrm{3}\:\vee\:{b}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$$${with}\:{c}=−\mathrm{1}: \\ $$$${g}=\mathrm{2} \\ $$$$\mathrm{2}{b}+{f}=−{m}\:\Rightarrow{m}=\mathrm{6}+\mathrm{1}=\mathrm{7} \\ $$$$−{f}+\mathrm{2}{b}=−\mathrm{5}\:\Rightarrow{f}=\mathrm{2}{b}+\mathrm{5}\:\Rightarrow{f}=−\mathrm{1}\:\vee\:{f}=\mathrm{6} \\ $$$${bf}=\mathrm{3}\:\Rightarrow{b}\left(\mathrm{2}{b}+\mathrm{5}\right)−\mathrm{3}=\mathrm{0}\:\Rightarrow\mathrm{2}{b}^{\mathrm{2}} +\mathrm{5}{b}−\mathrm{3}=\mathrm{0}\:\Rightarrow\left(\mathrm{2}{b}−\mathrm{1}\right)\left({b}+\mathrm{3}\right)=\mathrm{0}\:\Rightarrow{b}=−\mathrm{3}\:\vee\:{b}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$$$\Rightarrow{m}=−\mathrm{7}\:{or}\:\mathrm{7}. \\ $$$$\mathrm{2}{x}^{\mathrm{2}} +\mathrm{7}{xy}+\mathrm{3}{y}^{\mathrm{2}} −\mathrm{5}{y}−\mathrm{2}=\left({x}+\mathrm{3}{y}+\mathrm{1}\right)\left(\mathrm{2}{x}+{y}−\mathrm{2}\right) \\ $$$$\mathrm{2}{x}^{\mathrm{2}} −\mathrm{7}{xy}+\mathrm{3}{y}^{\mathrm{2}} −\mathrm{5}{y}−\mathrm{2}=\left({x}−\mathrm{3}{y}−\mathrm{1}\right)\left(\mathrm{2}{x}−{y}+\mathrm{2}\right) \\ $$
Commented by peter frank last updated on 14/Dec/19
GOD bless you.thank you very much
$${GOD}\:{bless}\:{you}.{thank}\:{you}\:{very}\:{much} \\ $$

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