Menu Close

Question-59070

Tinku Tara 0 Comments
Pin




Question Number 59070 by Tawa1 last updated on 04/May/19
Answered by tanmay last updated on 04/May/19
14)f(x)=q(x)(x−α) x−α is factor of f(x) if f(α)=0 abc−(a+b+c)(bc+ca+ab)←put a=−b here (−b)(b)c−(−b+b+c)(bc+c×−b+b×−b) −b^2 c−c(bc−bc−b^2 ) =−b^2 c−bc^2 +bc^2 +b^2 c =0 so[(a+b) is a facor abc−(a+b+c)(bc+ca+ab) =abc−abc−a(ca+ab)−b(bc+ca+ab)−c(bc+ca+ab) =−a^2 (b+c)−b^2 c−abc−ab^2 −bc^2 −c^2 a−abc =(−)[a^2 b+a^2 c+b^2 c+bc^2 +2abc+ab^2 +ac^2 ] =(−1)[a^2 (b+c)+bc(b+c)+a(b+c)^2 ] =(−b−c)[a^2 +bc+ab+ac] =(−1)(b+c)[a(a+b))+c(a+b)] =(−1)(b+c)(a+b)(a+c)
$$\left.\mathrm{14}\right){f}\left({x}\right)={q}\left({x}\right)\left({x}−\alpha\right) \\ $$$${x}−\alpha\:{is}\:{factor}\:{of}\:{f}\left({x}\right)\:{if}\:{f}\left(\alpha\right)=\mathrm{0} \\ $$$${abc}−\left({a}+{b}+{c}\right)\left({bc}+{ca}+{ab}\right)\leftarrow{put}\:{a}=−{b}\:{here} \\ $$$$\left(−{b}\right)\left({b}\right){c}−\left(−{b}+{b}+{c}\right)\left({bc}+{c}×−{b}+{b}×−{b}\right) \\ $$$$−{b}^{\mathrm{2}} {c}−{c}\left({bc}−{bc}−{b}^{\mathrm{2}} \right) \\ $$$$=−{b}^{\mathrm{2}} {c}−{bc}^{\mathrm{2}} +{bc}^{\mathrm{2}} +{b}^{\mathrm{2}} {c} \\ $$$$=\mathrm{0} \\ $$$${so}\left[\left({a}+{b}\right)\:{is}\:{a}\:{facor}\:\right. \\ $$$${abc}−\left({a}+{b}+{c}\right)\left({bc}+{ca}+{ab}\right) \\ $$$$={abc}−{abc}−{a}\left({ca}+{ab}\right)−{b}\left({bc}+{ca}+{ab}\right)−{c}\left({bc}+{ca}+{ab}\right) \\ $$$$=−{a}^{\mathrm{2}} \left({b}+{c}\right)−{b}^{\mathrm{2}} {c}−{abc}−{ab}^{\mathrm{2}} −{bc}^{\mathrm{2}} −{c}^{\mathrm{2}} {a}−{abc} \\ $$$$=\left(−\right)\left[{a}^{\mathrm{2}} {b}+{a}^{\mathrm{2}} {c}+{b}^{\mathrm{2}} {c}+{bc}^{\mathrm{2}} +\mathrm{2}{abc}+{ab}^{\mathrm{2}} +{ac}^{\mathrm{2}} \right] \\ $$$$=\left(−\mathrm{1}\right)\left[{a}^{\mathrm{2}} \left({b}+{c}\right)+{bc}\left({b}+{c}\right)+{a}\left({b}+{c}\right)^{\mathrm{2}} \right] \\ $$$$=\left(−{b}−{c}\right)\left[{a}^{\mathrm{2}} +{bc}+{ab}+{ac}\right] \\ $$$$\left.=\left(−\mathrm{1}\right)\left({b}+{c}\right)\left[{a}\left({a}+{b}\right)\right)+{c}\left({a}+{b}\right)\right] \\ $$$$=\left(−\mathrm{1}\right)\left({b}+{c}\right)\left({a}+{b}\right)\left({a}+{c}\right) \\ $$
Commented by Tawa1 last updated on 04/May/19
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by tanmay last updated on 04/May/19
f(x)=px^4 +qx^3 +13x^2 +4x−4=0 f(1)=p+q+13+4−4=0→p+q=−13 f((1/2))=(p/(16))+(q/8)+((13)/4)+(4/2)−4=0 p+2q+52−32=0 p+2q+20=0 p+q+13=0 q+7=0→q=−7 p−7+13=0 p=−6
$${f}\left({x}\right)={px}^{\mathrm{4}} +{qx}^{\mathrm{3}} +\mathrm{13}{x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{4}=\mathrm{0} \\ $$$${f}\left(\mathrm{1}\right)={p}+{q}+\mathrm{13}+\mathrm{4}−\mathrm{4}=\mathrm{0}\rightarrow{p}+{q}=−\mathrm{13} \\ $$$${f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{{p}}{\mathrm{16}}+\frac{{q}}{\mathrm{8}}+\frac{\mathrm{13}}{\mathrm{4}}+\frac{\mathrm{4}}{\mathrm{2}}−\mathrm{4}=\mathrm{0} \\ $$$${p}+\mathrm{2}{q}+\mathrm{52}−\mathrm{32}=\mathrm{0} \\ $$$${p}+\mathrm{2}{q}+\mathrm{20}=\mathrm{0} \\ $$$${p}+{q}+\mathrm{13}=\mathrm{0} \\ $$$${q}+\mathrm{7}=\mathrm{0}\rightarrow{q}=−\mathrm{7} \\ $$$${p}−\mathrm{7}+\mathrm{13}=\mathrm{0}\:\:{p}=−\mathrm{6} \\ $$
Commented by Tawa1 last updated on 04/May/19
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by tanmay last updated on 04/May/19
px^4 +qx^3 −8x^2 +6=q(x)(x^2 −1)+2x+1 so px^4 +qx^3 −8x^2 +6−2x−1 is divisible by (x^2 −1) f(x)=px^4 +qx^3 −8x^2 −2x+5 f(1)=p+q−8−2+5=0 p+q−5=0 f(1−)=p−q−8+2+5=0 p−q−1=0 p+q−5=0 p−q−1=0 2p−6=0→p=3 3+q−5=0 q=2 p=3 and q=2
$${px}^{\mathrm{4}} +{qx}^{\mathrm{3}} −\mathrm{8}{x}^{\mathrm{2}} +\mathrm{6}={q}\left({x}\right)\left({x}^{\mathrm{2}} −\mathrm{1}\right)+\mathrm{2}{x}+\mathrm{1} \\ $$$${so} \\ $$$${px}^{\mathrm{4}} +{qx}^{\mathrm{3}} −\mathrm{8}{x}^{\mathrm{2}} +\mathrm{6}−\mathrm{2}{x}−\mathrm{1}\:\:{is}\:{divisible}\:{by}\:\left({x}^{\mathrm{2}} −\mathrm{1}\right) \\ $$$${f}\left({x}\right)={px}^{\mathrm{4}} +{qx}^{\mathrm{3}} −\mathrm{8}{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{5} \\ $$$${f}\left(\mathrm{1}\right)={p}+{q}−\mathrm{8}−\mathrm{2}+\mathrm{5}=\mathrm{0} \\ $$$${p}+{q}−\mathrm{5}=\mathrm{0} \\ $$$${f}\left(\mathrm{1}−\right)={p}−{q}−\mathrm{8}+\mathrm{2}+\mathrm{5}=\mathrm{0} \\ $$$${p}−{q}−\mathrm{1}=\mathrm{0} \\ $$$${p}+{q}−\mathrm{5}=\mathrm{0} \\ $$$${p}−{q}−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{2}{p}−\mathrm{6}=\mathrm{0}\rightarrow{p}=\mathrm{3} \\ $$$$\mathrm{3}+{q}−\mathrm{5}=\mathrm{0} \\ $$$${q}=\mathrm{2} \\ $$$${p}=\mathrm{3}\:\:\:{and}\:\:{q}=\mathrm{2} \\ $$
Commented by Tawa1 last updated on 04/May/19
God bless you sir. I appreciate your time
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{time} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *