Answer
453.3k+ views
Hint: Use the properly, ${a^n} \times {a^m} = {a^{n + m}}$
We have given the equation $(3{m^2} - 7)(4n + 3 - 5{m^2})$ . Observe that, we can directly start to solve by opening the brackets as follows:
\[
(3{m^2} - 7)(4n + 3 - 5{m^2}) \\
\Rightarrow 3{m^2} \times 4n + 3{m^2} \times 3 - 3{m^2} \times 5{m^2} - 7 \times 4n - 7 \times 3 + 7 \times 5{m^2} \\
\Rightarrow 12{m^2}n + 9{m^2} - 15{m^4} - 28n - 21 + 35{m^2} \\
\Rightarrow 12{m^2}n + 44{m^2} - 15{m^4} - 28n - 21 \\
\Rightarrow {m^2}(12n + 44 - 15{m^2}) - 28n - 21 \\
\]
Hence the required multiplication is \[{m^2}(12n + 44 - 15{m^2}) - 28n - 21\] .
Note: Here the underlying assumption was the commutative properties. Not all algebraic structures preserve competitiveness but here we assume because questions have not mentioned about m, n are coming from where.
We have given the equation $(3{m^2} - 7)(4n + 3 - 5{m^2})$ . Observe that, we can directly start to solve by opening the brackets as follows:
\[
(3{m^2} - 7)(4n + 3 - 5{m^2}) \\
\Rightarrow 3{m^2} \times 4n + 3{m^2} \times 3 - 3{m^2} \times 5{m^2} - 7 \times 4n - 7 \times 3 + 7 \times 5{m^2} \\
\Rightarrow 12{m^2}n + 9{m^2} - 15{m^4} - 28n - 21 + 35{m^2} \\
\Rightarrow 12{m^2}n + 44{m^2} - 15{m^4} - 28n - 21 \\
\Rightarrow {m^2}(12n + 44 - 15{m^2}) - 28n - 21 \\
\]
Hence the required multiplication is \[{m^2}(12n + 44 - 15{m^2}) - 28n - 21\] .
Note: Here the underlying assumption was the commutative properties. Not all algebraic structures preserve competitiveness but here we assume because questions have not mentioned about m, n are coming from where.
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