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Question Number 205018 by BaliramKumar last updated on 06/Mar/24
For what value of ′k′ can be expression x^3 + kx^2 −7x +6 be resolved into three linear factors? (a) 0 (b) 1 (c) 2 (d) 3
$$\mathrm{For}\:\mathrm{what}\:\mathrm{value}\:\mathrm{of}\:\:'\mathrm{k}'\:\mathrm{can}\:\mathrm{be}\:\mathrm{expression}\:{x}^{\mathrm{3}} \:+\:{kx}^{\mathrm{2}} \:−\mathrm{7}{x}\:+\mathrm{6}\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\mathrm{be}\:\mathrm{resolved}\:\mathrm{into}\:\mathrm{three}\:\mathrm{linear}\:\mathrm{factors}? \\ $$$$\left(\mathrm{a}\right)\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{b}\right)\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{c}\right)\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{d}\right)\:\mathrm{3} \\ $$
Commented by Rasheed.Sindhi last updated on 06/Mar/24
(a) k=0
$$\left(\mathrm{a}\right)\:\mathrm{k}=\mathrm{0} \\ $$
Commented by mr W last updated on 06/Mar/24
question is not clear. in complex numbers with any real value of k the expression can always be resolved into three linear factors. in real numbers the expression can be resolved into three linear factors if k_1 <k<k_2 .
$${question}\:{is}\:{not}\:{clear}.\:{in}\:{complex} \\ $$$${numbers}\:{with}\:{any}\:{real}\:{value}\:{of}\:{k} \\ $$$${the}\:{expression}\:{can}\:{always}\:{be}\:{resolved} \\ $$$${into}\:{three}\:{linear}\:{factors}.\:{in}\:{real} \\ $$$${numbers}\:{the}\:{expression}\:{can}\:{be} \\ $$$${resolved}\:{into}\:{three}\:{linear}\:{factors} \\ $$$${if}\:{k}_{\mathrm{1}} <{k}<{k}_{\mathrm{2}} . \\ $$
Answered by Rasheed.Sindhi last updated on 06/Mar/24
(a)k=0 x^3 + kx^2 −7x +6⇒x^3 −7x +6 x^3 −1 −7x +7 (x−1)(x^2 +x+1)−7(x−1) (x−1)(x^2 +x−6) (x−1)(x+3)(x−2) For k=1,2,3 the expression has no linear factor.
$$\left({a}\right){k}=\mathrm{0} \\ $$$${x}^{\mathrm{3}} \:+\:{kx}^{\mathrm{2}} \:−\mathrm{7}{x}\:+\mathrm{6}\Rightarrow{x}^{\mathrm{3}} \:\:−\mathrm{7}{x}\:+\mathrm{6} \\ $$$${x}^{\mathrm{3}} \:−\mathrm{1}\:−\mathrm{7}{x}\:+\mathrm{7} \\ $$$$\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)−\mathrm{7}\left({x}−\mathrm{1}\right) \\ $$$$\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}−\mathrm{6}\right) \\ $$$$\left({x}−\mathrm{1}\right)\left({x}+\mathrm{3}\right)\left({x}−\mathrm{2}\right) \\ $$$${For}\:{k}=\mathrm{1},\mathrm{2},\mathrm{3}\:{the}\:{expression}\:{has} \\ $$$${no}\:{linear}\:{factor}. \\ $$
Commented by mr W last updated on 06/Mar/24
linear factors (x+a)(x+b)(x+c) don′t request that a, b, c should be integer numbers. they may be real numbers or even complex numbers.
$${linear}\:{factors}\:\left({x}+{a}\right)\left({x}+{b}\right)\left({x}+{c}\right) \\ $$$${don}'{t}\:{request}\:{that}\:{a},\:{b},\:{c}\:{should}\:{be} \\ $$$${integer}\:{numbers}.\:{they}\:{may}\:{be}\:{real} \\ $$$${numbers}\:{or}\:{even}\:{complex}\:{numbers}. \\ $$
Commented by Rasheed.Sindhi last updated on 06/Mar/24
Sir, I considered only numbers given as options: (a) k=0 (b) k=1 (c) k=2 (d) k=3
$$\boldsymbol{\mathrm{Sir}},\:\mathrm{I}\:\mathrm{considered}\:\mathrm{only}\:\mathrm{numbers}\:\mathrm{given} \\ $$$$\mathrm{as}\:\mathrm{options}: \\ $$$$\left(\mathrm{a}\right)\:{k}=\mathrm{0}\:\:\:\:\left(\mathrm{b}\right)\:{k}=\mathrm{1}\:\:\left(\mathrm{c}\right)\:{k}=\mathrm{2}\:\:\:\left(\mathrm{d}\right)\:{k}=\mathrm{3} \\ $$
Commented by mr W last updated on 06/Mar/24
yes. from these 4 given options only k=0 is suitable.
$${yes}.\:{from}\:{these}\:\mathrm{4}\:{given}\:{options}\:{only} \\ $$$${k}=\mathrm{0}\:{is}\:{suitable}. \\ $$

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