Question

Divide (16y⁵-8y²)÷4 please tell answer​

Answer 1

$\textbf{To find:}$

$\mathsf{Divide\;16y^5-8y^2\;by\;4}$

$\textbf{Solution:}$

$\mathsf{Consider,}$

$\mathsf{\dfrac{16\,y^5-8\,y^2}{4}}$

$\mathsf{Taking\;8y^2\;as\;common}$

$\mathsf{=\dfrac{8y^2(2y^3-1)}{4}}$

$\mathsf{=4y^2(2y^3-1)}$

$\mathsf{=8y^5-4y^2}$

$\textbf{Answer:}$

$\mathsf{when\;16\,y^5-8\,y^2\;is\;divided\;by\;4\;we\;get\;8y^5-4y^2}$

$\textbf{Find more:}$

If 16x4-8x3+x2-8x+4 is divided by 2x-1 then the remainder is​

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Find the remainder when x cube + 3 X square + 3 X + 1 is divided by X using long division method​

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Answer 2

SOLUTION

TO DETERMINE

$\sf{(16 {y}^{5} - 8 {y}^{2} ) \div 4}$

EVALUATION

$\sf{(16 {y}^{5} - 8 {y}^{2} ) \div 4}$

$\displaystyle \: \sf{ = \frac{16 {y}^{5} - 8 {y}^{2} }{4} }$

$\displaystyle \: \sf{ = \frac{16 {y}^{5}}{4} - \frac{ 8 {y}^{2} }{4}} \: \: \bigg( \because \: \frac{a - b}{c} = \frac{a}{c} - \frac{b}{c} \bigg)$

$\displaystyle \: \sf{ = \frac{4 \times 4 \times {y}^{5}}{4} - \frac{ 4 \times 2 \times {y}^{2} }{4}} \:$

$\displaystyle \: \sf{ = 4 {y}^{5} - 2 {y}^{2} } \:$

FINAL ANSWER

$\boxed{ \: \sf{(16 {y}^{5} - 8 {y}^{2} ) \div 4} = (4 {y}^{5} - 2 {y}^{2}) \: \: }$

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