Question

Q1.m²n²(2²- n²) - mn²(4mn - 2m²) + m³n (4 - 3n).

Answer 1

Step-by-step explanation:

Question:

Simplify m²n²(2²- n²) - mn²(4mn - 2m²) + m³n (4 - 3n).

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Solution:

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$\sf \dashrightarrow m^2n^2(\textsf{\textbf{2}}^2- n^2) - mn^2(4mn - 2m^2) + m^3n (4 - 3n)$

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$\sf \dashrightarrow m^2n^2(\textsf{\textbf{4}}- n^2) - mn^2(4mn - 2m^2) + m^3n (4 - 3n) </p><p>$

m is common in 4mn and - 2m², therefore it can be re-written as m(4n - 2m).

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$\sf \dashrightarrow m^2n^2(4- n^2) - mn^2(\textsf{\textbf{m(4n - 2m)}}) + m^3n (4 - 3n)$

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$\sf \dashrightarrow m^2n^2(4- n^2) - \textsf{\textbf{(mn}}^2 \times \textsf{\textbf{m)}}(4n - 2m) + m^3n (4 - 3n)$

$\sf \dashrightarrow m^2n^2(4- n^2) - \textsf{\textbf{m}}^2 \textsf{\textbf{n}}^2(4n - 2m) + m^3n (4 - 3n)$

m²n² is common in both (4 - n²) and -(4n - 2m), therefore it can be written as m²n²[4 - n² - (4n - 2m)].

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$\sf \dashrightarrow \textsf{\textbf{m}}^2\textsf{\textbf{n}}^2\Big[\textsf{\textbf{(4 - n}}^2\textsf{\textbf{) - (4n - 2m)}}\Big] + m^3n (4 - 3n)$

On opening the brackets for m³n(4 - 3n) we get;

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$\sf \dashrightarrow m^2n^2\Big[(4 - n^2) - (4n - 2m)\Big] + \textsf{\textbf{4(m}}^3\textsf{\textbf{n) - 3n(m}}^3\textsf{\textbf{n)}}$

$\sf \dashrightarrow m^2n^2\Big[(4- n^2) - (4n - 2m)\Big] + \textsf{\textbf{4m}}^3\textsf{\textbf{n}} - \textsf{\textbf{3m}}^3\textsf{\textbf{n}}^2$

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$</p><p>\sf \dashrightarrow m^2n^2\Big[(4- n^2) - (4n - 2m)\Big] \textsf{\textbf{- 3m}}^3\textsf{\textbf{n}}^2 \textsf{\textbf{+ 4m}}^3\textsf{\textbf{n}}$

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m²n² is common in -3m³n², therefore it can be written as m²n²(-3m)

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$\sf \dashrightarrow m^2n^2\Big[(4- n^2) - (4n - 2m)\Big] + \textsf{\textbf{m}}^2\textsf{\textbf{n}}^2(\textsf{\textbf{-3m}}) + 4m^3n$

m²n² is common in both [4 - n² - (4n - 2m)] and -3m, therefore it can be taken out as common from both the expressions.

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$\sf \dashrightarrow \textsf{\textbf{m}}^2\textsf{\textbf{n}}^2\Big[\textsf{\textbf{(4 - n}}^2\textsf{\textbf{) - (4n - 2m) - 3m}}\Big] + 4m^3n$

$\sf \dashrightarrow m^2n^2\Big[\textsf{\textbf{4 - n}}^2 \textsf{\textbf{ - 4n + 2m - 3m}}\Big] + 4m^3n$

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$</p><p>\sf \dashrightarrow m^2n^2\Big[\textsf{\textbf{4 - n}}^2 \textsf{\textbf{ - 4n - m}}\Big] + 4m^3n$

On opening the brackets of m²n²[4 - n² - 4n - m] we get;

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$\sf \dashrightarrow \textsf{\textbf{4(m}}^{2}\textsf{\textbf{n}}^2\textsf{\textbf{)}} - \textsf{\textbf{n}}^2\textsf{\textbf{(m}}^2\textsf{\textbf{n}}^2\textsf{\textbf{) - 4n(m}}^2\textsf{\textbf{n}}^2\textsf{\textbf{) - m(m}}^2\textsf{\textbf{n}}^2\textsf{\textbf{)}} + 4m^3n$

$\sf \dashrightarrow 4m^2n^2 - m^2n^4 - 4m^2n^3 - m^3n^2 + 4m^3n$

The expression cannot be simplified further, therefore the answer is:

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$\sf \dashrightarrow \underline{\underline{4m^2n^2 - m^2n^4 - 4m^2n^3 - m^3n^2 + 4m^3n}} </p><p>$

Answer 2

Answer:-

(2m^3 n^2-m^2 n^4) - (4m^2 n^3+2m^3 n^2)+ (4m^3 n-3m^3 n^2)

so,(1m^3 n^2)-(m^2 n^4)+(4m^3n) m^2

n(mn-n^3+4m) [by taking common]

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