Question

Simplify the following algebraic expression: (only for brilliant)

1. (2m + 1) (m-2) - (3n - 1) (n+3) - [4 (m - 2n) + {1 - n² + 2 (m+n) (m-n)}]

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Answer 1

$\large\bf\underline\red{Answer \: :-}$

Given :

(2m + 1) (m-2) - (3n - 1) (n+3) - [4 (m - 2n) + {1 - n² + 2 (m+n) (m-n)}]

Solving (2m + 1) (m - 2)

$\sf\longrightarrow{(2m + 1)(m - 2)} \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf\longrightarrow{2m(m - 2) + 1(m - 2)} \\ \\ \sf\longrightarrow{2m {}^{2} - 4m + 1m - 2 } \: \: \: \: \\ \\ \bf\longrightarrow \red{2m {}^{2} - 3m - 2 } \: \: \: \: \: \: \: \: \: \: \: \:$

Solving (3n - 1) (n + 3)

$\sf\longrightarrow{(3n - 1)(n + 3)} \: \: \: \: \: \: \: \: \: \: \\ \\ \sf\longrightarrow{3n(n - 3) - 1(n + 3)} \\ \\ \sf\longrightarrow{3n {}^{2} - 6n - 1n - 3 } \: \: \: \: \\ \\ \bf\longrightarrow \red{3n {}^{2} - 7n - 3} \: \: \: \: \: \: \: \: \: \: \:$

Solving [4 (m - 2n) + {1 - n² + 2 (m+n) (m-n)}]

→ 4m - 8n + [ 3 - n² (m + n) (m - n) ]

We know (a + b) (a - b) = a² - b²

→ 4m - 8n + [ 3 - n² (m² - n²) ]

→ 4m - 8n + [ 3 (m² - n)² - n² (m² - n²)

→ 4m - 8n + [ 3m² - n² - n²m² - 2n²]

→ 4m - 8n + 3m² - 3n² - n²m²

Now,

We have,

(2m + 1) (m-2) = 2m² - 3m - 2

(3n - 1) (n+3) = 3n² - 7n - 3

[4 (m - 2n) + {1 - n² + 2 (m+n) (m-n)}] = 4m - 8n + 3m² - 3n² - n²m²

Putting all values

Solving (2m + 1) (m-2) - (3n - 1) (n+3)

$\small \sf \longrightarrow{2m {}^{2} - 3m - 2 - (3n {}^{2} - 7n - 3)} \\ \\ \small \sf \longrightarrow{2m {}^{2} - 3m - 2 - 3n {}^{2} - 7n - 3 } \: \: \\ \\ \small \bf \longrightarrow \red{2m {}^{2} - 3m - 3n {}^{2} - 7n - 5 } \: \: \: \: \: \:$

Solving

2m² - 3m - 3n² - 7n - 5 -( 4m - 8n + 3m² - 3n² - n²m² )

$\small\sf\longrightarrow{2m {}^{2} - 3m - 3n {}^{2} - 7n - 5 - 4m + 8n - 3m {}^{2} + 3n {}^{2} - n {}^{2} m {}^{2} } \\ \\ \small\sf\longrightarrow{ 2m {}^{2} - 3m \: \cancel{ - 3n {}^{2} } - 7n - 5 - 4m + 8n - 3m {}^{2} \: \cancel{ + 3n {}^{2} } - n {}^{2}m {}^{2} } \: \\ \\ \small\sf\longrightarrow{2m {}^{2} - 3m {}^{2} - 4m - 3m - 7n + 8n -5 - n {}^{2} m {}^{2} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \small\sf\longrightarrow{ - m {}^{2} - 7m + n - n {}^{2}m {}^{2} - 5 } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \small\bf\longrightarrow \red{ - m {}^{2} - 7m + n - n {}^{2} m { }^{2} - 5 } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Final answer : - m² - 7m + n - n²m² - 5

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Answer 2

Question :

• To Simplify : (2m + 1) (m-2) - (3n - 1) (n+3) - [4 (m - 2n) + {1 - n² + 2 (m+n) (m-n)}]

Solution :

➠ (2m + 1) (m-2) - (3n - 1) (n+3) - [4 (m - 2n) + {1 - n² + 2 (m+n) (m-n)}]

✠ Open the brackets by multiplying

➠ 2m(m-2) +1(m-2) -3n(n+3) -1(n+3) -[ 4m -8n + 1-n²+2(m+n)(m-n)]

➠ 2m²-4m + m - 2 -3n²+9n -n -3n -[4m -8n + 1-n²+2(m²-n²)]

$\because\sf \boxed{\sf (a-b)(a+b) = a^2-b^2}$

➠ 2m²-4m + m - 2 -3n²+9n -n -3n -[4m -8n + 1-n²+ 2m²-2n²]

➠ 2m²-4m + m - 2 -3n²+9n -n -3n -4m +8n -1 +n² -2m² +2n²

✠ Adding like terms :

➠ 2m² −4m+m−2−3n² −8n+3+−4m+8n −1+n²+−2m²+2n²

➠ 2m² −2m² +(−3n² +n²+2n²)+(−4m+m+−4m)+(−8n+8n)+(−2+3−1)

$\sf\implies\cancel{2m^2 }\cancel{-2m^2 } + ( \cancel{-3n^2} + \cancel{3n^2} ) + (-7m) + (\cancel{-8n}+\cancel{8n} )+(0)$

➠ -7m

Therefore ,

(2m + 1) (m-2) - (3n - 1) (n+3) - [4 (m - 2n) + {1 - n² + 2 (m+n) (m-n)}] = -7m