Question

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



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Answer 1

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Answer 2

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

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Solution:

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$⇢m2n2(22−n2)−mn2(4mn−2m2)+m3n(4−3n)\sf \dashrightarrow m^2n^2(\textsf{\textbf{2}}^2- n^2) - mn^2(4mn - 2m^2) + m^3n (4 - 3n)⇢m$

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$⇢m2n2(4−n2)−mn2(4mn−2m2)+m3n(4−3n)\sf \dashrightarrow m^2n^2(\textsf{\textbf{4}}- n^2) - mn^2(4mn - 2m^2) + m^3n (4 - 3n)⇢m$

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

‎‎

$⇢m2n2(4−n2)−mn2(m(4n - 2m))+m3n(4−3n)\sf \dashrightarrow m^2n^2(4- n^2) - mn^2(\textsf{\textbf{m(4n - 2m)}}) + m^3n (4 - 3n)⇢m✓$

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$⇢m2n2(4−n2)−(mn2×m)(4n−2m)+m3n(4−3n)\sf \dashrightarrow m^2n^2(4- n^2) - \textsf{\textbf{(mn}}^2 \times \textsf{\textbf{m)}}(4n - 2m) + m^3n (4 - 3n)⇢m$

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$⇢m2n2(4−n2)−m2n2(4n−2m)+m3n(4−3n)\sf \dashrightarrow m^2n^2(4- n^2) - \textsf{\textbf{m}}^2 \textsf{\textbf{n}}^2(4n - 2m) + m^3n (4 - 3n)⇢m$

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

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$⇢m2n2[(4 - n2) - (4n - 2m)]+m3n(4−3n)\sf \dashrightarrow \textsf{\textbf{m}}^2\textsf{\textbf{n}}^2\Big[\textsf{\textbf{(4 - n}}^2\textsf{\textbf{) - (4n - 2m)}}\Big] + m^3n (4 - 3n)⇢m$

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

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$⇢m2n2[(4−n2)−(4n−2m)]+4(m3n) - 3n(m3n)\sf \dashrightarrow m^2n^2\Big[(4 - n^2) - (4n - 2m)\Big] + \textsf{\textbf{4(m}}^3\textsf{\textbf{n) - 3n(m}}^3\textsf{\textbf{n)}}⇢m$

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$⇢m2n2[(4−n2)−(4n−2m)]+4m3n−3m3n2\sf \dashrightarrow m^2n^2\Big[(4- n^2) - (4n - 2m)\Big] + \textsf{\textbf{4m}}^3\textsf{\textbf{n}} - \textsf{\textbf{3m}}^3\textsf{\textbf{n}}^2⇢m$

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$⇢m2n2[(4−n2)−(4n−2m)]- 3m3n2+ 4m3n\sf \dashrightarrow m^2n^2\Big[(4- n^2) - (4n - 2m)\Big] \textsf{\textbf{- 3m}}^3\textsf{\textbf{n}}^2 \textsf{\textbf{+ 4m}}^3\textsf{\textbf{n}}⇢m$

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

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$⇢m2n2[(4−n2)−(4n−2m)]+m2n2(-3m)+4m3n\sf \dashrightarrow m^2n^2\Big[(4- n^2) - (4n - 2m)\Big] + \textsf{\textbf{m}}^2\textsf{\textbf{n}}^2(\textsf{\textbf{-3m}}) + 4m^3n⇢m$

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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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$⇢m2n2[(4 - n2) - (4n - 2m) - 3m]+4m3n\sf \dashrightarrow \textsf{\textbf{m}}^2\textsf{\textbf{n}}^2\Big[\textsf{\textbf{(4 - n}}^2\textsf{\textbf{) - (4n - 2m) - 3m}}\Big] + 4m^3n⇢m$

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$⇢m2n2[4 - n2 - 4n + 2m - 3m]+4m3n\sf \dashrightarrow m^2n^2\Big[\textsf{\textbf{4 - n}}^2 \textsf{\textbf{ - 4n + 2m - 3m}}\Big] + 4m^3n⇢m$

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$⇢m2n2[4 - n2 - 4n - m]+4m3n\sf \dashrightarrow m^2n^2\Big[\textsf{\textbf{4 - n}}^2 \textsf{\textbf{ - 4n - m}}\Big] + 4m^3n⇢m$

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

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$⇢4(m2n2)−n2(m2n2) - 4n(m2n2) - m(m2n2)+4m3n\sf \dashrightarrow \textsf{\textbf{4(m}}^{2}\textsf{\textbf{n}}^2\textsf{\textbf{)}} - \textsf{\textbf{n}}^2\textsf{\textbf{(m}}^2\textsf{\textbf{n}}^2\textsf{\textbf{) - 4n(m}}^2\textsf{\textbf{n}}^2\textsf{\textbf{) - m(m}}^2\textsf{\textbf{n}}^2\textsf{\textbf{)}} + 4m^3n⇢4$

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$⇢4m2n2−m2n4−4m2n3−m3n2+4m3n\sf \dashrightarrow 4m^2n^2 - m^2n^4 - 4m^2n^3 - m^3n^2 + 4m^3n⇢4m$

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

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$⇢4m2n2−m2n4−4m2n3−m3n2+4m3n‾‾\sf \dashrightarrow \underline{\underline{4m^2n^2 - m^2n^4 - 4m^2n^3 - m^3n^2 + 4m^3n}}⇢ </p><p>4m$