Question

What must be added to X ki power 4 + 2 x power 3 minus 2 X square + X - 1 so that the result is exactly divisible by X ki power 2 + 2 x minus 3
please give the answer because it is very urgent​

Answer 1

Answer:

On dividing X ki power 4 + 2 x power 3 minus 2 X square + X - 1,we get the quotient x power 2 + 1 and remainder -x+2.

Since the remainder is -x+2,-x+2 must be added to X ki power 4 + 2 x power 3 minus 2 X square + X - 1 so that the result is exactly divisible by X ki power 2 + 2 x minus 3.

Hope this helps you.

Answer 2

$\sf Correct\:Question :$

What must be added to the polynomial x⁴ + 2x³ - 2x² + x - 1 so that the resulting polynomial is exactly divisible by x² + 2x - 3

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⠀⠀⠀⠀ Polynomial Division

$\boxed{\begin{array}{l | n | r}\sf x^2+2x-3 &\sf x^4+2x^3-2x^2+x-1&\sf x^2+1\\ &\sf x^4+2x^3-3x^2\\ &( - )\:\:(-)\:\:( + )\\&\rule{100}{0.8}\\&\sf\qquad\qquad\quad x^2+x-1\\ &\sf\qquad\qquad\quad x^2+2x-3\\ &\qquad\qquad\:\: (-)\:\:( - )\:\:( + )\\&\qquad\quad\rule{70}{0.8}\\ &\sf\qquad\qquad\qquad\sf-\:x\:+\:2\end{array}}$

$\rule{100}{0.8}$

☯ $\underline{\boldsymbol{According\: to \:the\: Question\:now :}}$

$:\implies\sf Dividend=Divisor \times Quotient+Remainder\\\\\\:\implies\sf x^4+2x^3-2x^2+x-1=\bigg\lgroup (x^2+2x-3) \times (x^2+1)\bigg\rgroup+(-\:x+2)\\\\\\:\implies\sf x^4+2x^3-2x^2+x-1-(-\:x+2) =\bigg\lgroup (x^2+2x-3) \times (x^2+1)\bigg\rgroup\\\\\\:\implies\sf x^4+2x^3-2x^2+x-1+x -2 =\bigg\lgroup (x^2+2x-3) \times (x^2+1)\bigg\rgroup\\\\\\:\implies\sf x^4+2x^3-2x^2+2x-3=\bigg\lgroup (x^2+2x-3) \times (x^2+1)\bigg\rgroup$

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$\therefore\:\underline{\textsf{Hence, \textbf{(x - 2)} must be added to divide completely}}.$