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

$$$$$$\mathrm{Which}\mathrm{of}\mathrm{the}\mathrm{following}\mathrm{is}\mathrm{not}\mathrm{a}\mathrm{factor}$$$$\mathrm{of}{\mathrm{x}}^6-56\mathrm{x}+55$$$$\mathrm{A}.\mathrm{x}-1\mathrm{B}.{\mathrm{x}}^2-\mathrm{x}+5\mathrm{C}.$$$${\mathrm{x}}^3+{2\mathrm{x}}^2-2\mathrm{x}-11\mathrm{D}.$$$${\mathrm{x}}^4+{\mathrm{x}}^3+{4\mathrm{x}}^2-9\mathrm{x}+11\mathrm{E}.$$$${\mathrm{x}}^5+{\mathrm{x}}^4+{\mathrm{x}}^3+{\mathrm{x}}^2+\mathrm{x}-55$$$$$$$$\mathrm{Please}\mathrm{show}\mathrm{all}\mathrm{workings}\mathrm{clearly}.$$$$\mathrm{Thanks}.$$

Commented by bobhans last updated on 29/Aug/20

$$youcanuseRemaindertheorem$$

Commented by Aina Samuel Temidayo last updated on 29/Aug/20

$$\mathrm{Can}\mathrm{you}\mathrm{please}\mathrm{do}\mathrm{that}\mathrm{for}\mathrm{me}?$$

Commented by john santu last updated on 29/Aug/20

$$notafactorisD$$

Answered by bobhans last updated on 29/Aug/20

$$\mathrm{\Sigma }coefficient=0,sox-1isoneof$$$$factorthispolynome.$$$$10000-5655$$$$1\ast 11111-55$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}$$$$11111-550$$$$thenp\left(x\right)=(x-1)(x^5+x^4+x^3+x^2+x-55)$$

Commented by Aina Samuel Temidayo last updated on 29/Aug/20

$$\mathrm{So},{\mathrm{what}}^{\prime }\mathrm{s}\mathrm{next}?\mathrm{How}\mathrm{do}\mathrm{you}\mathrm{get}\mathrm{the}$$$$\mathrm{other}\mathrm{factors}?$$

Commented by bemath last updated on 29/Aug/20

$$\mathrm{p}\left(\mathrm{x}\right)=(\mathrm{x}-1)({\mathrm{x}}^2-\mathrm{x}+5)({\mathrm{x}}^3+{2\mathrm{x}}^2-2\mathrm{x}-11)$$

Commented by Aina Samuel Temidayo last updated on 29/Aug/20

$$\mathrm{How}\mathrm{did}\mathrm{you}\mathrm{do}\mathrm{this}?$$

Answered by 1549442205PVT last updated on 29/Aug/20

$$\mathrm{Using}\mathrm{division}\mathrm{algorithm}\mathrm{for}\mathrm{two}$$$$\mathrm{polynomials}\mathrm{we}\mathrm{obtain}:$$$${\mathrm{x}}^6-56\mathrm{x}+55=({\mathrm{x}}^2-\mathrm{x}+5)({\mathrm{x}}^4+{\mathrm{x}}^3+{4\mathrm{x}}^2-9\mathrm{x}+11)$$$${\mathrm{x}}^6-56\mathrm{x}+55=(\mathrm{x}-1)({\mathrm{x}}^5+{\mathrm{x}}^4+{\mathrm{x}}^3+{\mathrm{x}}^2+\mathrm{x}-55)$$$${\mathrm{x}}^6-56\mathrm{x}+55=({\mathrm{x}}^3+{2\mathrm{x}}^2-2\mathrm{x}-11)({\mathrm{x}}^3-{2\mathrm{x}}^2+6\mathrm{x}-5)$$$$\mathrm{From}\mathrm{above}\mathrm{result}\mathrm{we}\mathrm{see}\mathrm{that}\mathrm{all}\mathrm{given}$$$$\mathrm{polynomials}\mathrm{are}\mathrm{factors}\mathrm{of}{\mathrm{x}}^6-56\mathrm{x}+55$$

Commented by malwan last updated on 29/Aug/20

$$Notall!$$$$x^6-56x+55=(x^2-x+5)(x^4+x^3-4x^2-9x+11)$$$$soDisnotafactor$$

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