Question

write the degree of this polynomial with solving.

QUESTION IS :-
(i) ( x-2)^3
(ii) 1/x^2 ( x^3 - x^4 + x^8 )

Is question me ye jo polynomials diye h inku degrees btani h... pr mujhse ye solve n horha.. or kaise solve krte h please ye solve krke bta dijiye..
please answer anyone ​


IN QUES (i) I THINK SOLVING IS WRONG:- so let me know how to solve it....

IN QUES (ii) I DON'T KNOW HOW TO SOLVE THE EQUATION .. I MEAN HOW TO DIVIDE THE EXPONENTS.. SO LET ME KNOW ...

PLEASE SOLVE THIS QUESTION..
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Answer 1

$\huge\bf\underline\red{Answer:-}$

1.

$= (x - 2)^{3}$

$= {x}^{3} - 3 {x}^{2} \times 2 + 3x \times {2}^{2} - {2}^{3}$

$= {x}^{3} - 6 {x}^{2} + 3x \times 4 - 8$

$= {x}^{3} - 6 {x}^{2} + 12x - 8$

2.

$= \frac{1}{x ^{2} } ( {x}^{3} - {x}^{4} + {x}^{8} )$

$= \frac{1}{ {x}^{2} } \times {x}^{2} \times (x - {x}^{2} + {x}^{6} )$

$= x - {x}^{2} + {x}^{6}$

$= {x}^{6} - {x}^{2} + x$

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Answer 2

(x - 2)³

(x)³ - (2)³ - 3*x*2(x - 2)

x³ - 8 - 6x² + 12x

Degree of Polynomial → 3

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1/x² ( x³ - x⁴ + x⁸ )

1/x² * x² ( x - x³ + x⁶ )

( x - x³ + x⁶ )

Degree of Polynomial → 6

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