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Which-of-the-following-is-not-a-factor-of-x-6-56x-55-A-x-1-B-x-2-x-5-C-x-3-2x-2-2x-11-D-x-4-x-3-4x-2-9x-11-E-x-5-x-4-x-3-x-2-x-55-Please-show-all-workings-clearly-Thanks-

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Question Number 110425 by Aina Samuel Temidayo last updated on 28/Aug/20
Which of the following is not a factor of x^6 −56x+55 A. x−1 B. x^2 −x+5 C. x^3 +2x^2 −2x−11 D. x^4 +x^3 +4x^2 −9x+11 E. x^5 +x^4 +x^3 +x^2 +x−55 Please show all workings clearly. Thanks.
$$ \\ $$$$\mathrm{Which}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following}\:\mathrm{is}\:\mathrm{not}\:\mathrm{a}\:\mathrm{factor} \\ $$$$\mathrm{of}\:\mathrm{x}^{\mathrm{6}} −\mathrm{56x}+\mathrm{55} \\ $$$$\mathrm{A}.\:\mathrm{x}−\mathrm{1}\:\mathrm{B}.\:\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{5}\:\mathrm{C}. \\ $$$$\mathrm{x}^{\mathrm{3}} +\mathrm{2x}^{\mathrm{2}} −\mathrm{2x}−\mathrm{11}\:\mathrm{D}. \\ $$$$\mathrm{x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{3}} +\mathrm{4x}^{\mathrm{2}} −\mathrm{9x}+\mathrm{11}\:\mathrm{E}. \\ $$$$\mathrm{x}^{\mathrm{5}} +\mathrm{x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{3}} +\mathrm{x}^{\mathrm{2}} +\mathrm{x}−\mathrm{55} \\ $$$$ \\ $$$$\mathrm{Please}\:\mathrm{show}\:\mathrm{all}\:\mathrm{workings}\:\mathrm{clearly}. \\ $$$$\mathrm{Thanks}. \\ $$
Commented by bobhans last updated on 29/Aug/20
you can use Remainder theorem
$${you}\:{can}\:{use}\:{Remainder}\:{theorem} \\ $$
Commented by Aina Samuel Temidayo last updated on 29/Aug/20
Can you please do that for me?
$$\mathrm{Can}\:\mathrm{you}\:\mathrm{please}\:\mathrm{do}\:\mathrm{that}\:\mathrm{for}\:\mathrm{me}? \\ $$
Commented by john santu last updated on 29/Aug/20
not a factor is D
$${not}\:{a}\:{factor}\:{is}\:{D} \\ $$
Answered by bobhans last updated on 29/Aug/20
Σcoefficient = 0, so x−1 is one of factor this polynome. 1 0 0 0 0 −56 55 1 ∗ 1 1 1 1 1 −55 _________________________ 1 1 1 1 1 −55 0 then p(x) = (x−1)(x^5 +x^4 +x^3 +x^2 +x−55)
$$\:\Sigma{coefficient}\:=\:\mathrm{0},\:{so}\:{x}−\mathrm{1}\:{is}\:{one}\:{of}\: \\ $$$${factor}\:{this}\:{polynome}. \\ $$$$\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\mathrm{0}\:\:\:\:\:−\mathrm{56}\:\:\:\:\:\mathrm{55} \\ $$$$\mathrm{1}\:\:\:\:\:\:\ast\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:−\mathrm{55} \\ $$$$\:\:\:\:\:\:\:\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:−\mathrm{55}\:\:\:\:\mathrm{0} \\ $$$${then}\:{p}\left({x}\right)\:=\:\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{5}} +{x}^{\mathrm{4}} +{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}−\mathrm{55}\right) \\ $$
Commented by Aina Samuel Temidayo last updated on 29/Aug/20
So, what′s next? How do you get the other factors?
$$\mathrm{So},\:\mathrm{what}'\mathrm{s}\:\mathrm{next}?\:\mathrm{How}\:\mathrm{do}\:\mathrm{you}\:\mathrm{get}\:\mathrm{the} \\ $$$$\mathrm{other}\:\mathrm{factors}? \\ $$
Commented by bemath last updated on 29/Aug/20
p(x)=(x−1)(x^2 −x+5)(x^3 +2x^2 −2x−11)
$$\mathrm{p}\left(\mathrm{x}\right)=\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{5}\right)\left(\mathrm{x}^{\mathrm{3}} +\mathrm{2x}^{\mathrm{2}} −\mathrm{2x}−\mathrm{11}\right) \\ $$
Commented by Aina Samuel Temidayo last updated on 29/Aug/20
How did you do this?
$$\mathrm{How}\:\mathrm{did}\:\mathrm{you}\:\mathrm{do}\:\mathrm{this}? \\ $$
Answered by 1549442205PVT last updated on 29/Aug/20
Using division algorithm for two polynomials we obtain: x^6 −56x+55=(x^2 −x+5)(x^4 +x^3 +4x^2 −9x+11) x^6 −56x+55=(x−1)(x^5 +x^4 +x^3 +x^2 +x−55) x^6 −56x+55=(x^3 +2x^2 −2x−11)(x^3 −2x^2 +6x−5) From above result we see that all given polynomials are factors of x^6 −56x+55
$$\mathrm{Using}\:\mathrm{division}\:\mathrm{algorithm}\:\mathrm{for}\:\mathrm{two} \\ $$$$\mathrm{polynomials}\:\mathrm{we}\:\mathrm{obtain}: \\ $$$$\mathrm{x}^{\mathrm{6}} −\mathrm{56x}+\mathrm{55}=\left(\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{5}\right)\left(\mathrm{x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{3}} +\mathrm{4x}^{\mathrm{2}} −\mathrm{9x}+\mathrm{11}\right) \\ $$$$\mathrm{x}^{\mathrm{6}} −\mathrm{56x}+\mathrm{55}=\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{5}} +\mathrm{x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{3}} +\mathrm{x}^{\mathrm{2}} +\mathrm{x}−\mathrm{55}\right) \\ $$$$\mathrm{x}^{\mathrm{6}} −\mathrm{56x}+\mathrm{55}=\left(\mathrm{x}^{\mathrm{3}} +\mathrm{2x}^{\mathrm{2}} −\mathrm{2x}−\mathrm{11}\right)\left(\mathrm{x}^{\mathrm{3}} −\mathrm{2x}^{\mathrm{2}} +\mathrm{6x}−\mathrm{5}\right) \\ $$$$\mathrm{From}\:\mathrm{above}\:\mathrm{result}\:\mathrm{we}\:\mathrm{see}\:\mathrm{that}\:\mathrm{all}\:\mathrm{given} \\ $$$$\mathrm{polynomials}\:\mathrm{are}\:\mathrm{factors}\:\mathrm{of}\:\:\mathrm{x}^{\mathrm{6}} −\mathrm{56x}+\mathrm{55} \\ $$
Commented by malwan last updated on 29/Aug/20
Not all ! x^6 −56x+55=(x^2 −x+5)(x^4 +x^3 −4x^2 −9x+11) so D is not a factor
$${Not}\:{all}\:! \\ $$$${x}^{\mathrm{6}} −\mathrm{56}{x}+\mathrm{55}=\left({x}^{\mathrm{2}} −{x}+\mathrm{5}\right)\left({x}^{\mathrm{4}} +{x}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{2}} −\mathrm{9}{x}+\mathrm{11}\right) \\ $$$${so}\:{D}\:{is}\:{not}\:{a}\:{factor}\: \\ $$

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