Question

Factorise : mn(3m2 – 4n2) – np(4n2 – 3m2) + pm(15m2 – 20n2)

Answer 1

Given :  mn(3m² – 4n²) – np(4n² – 3m²) + pm(15m² – 20n²)

To Find : Factorize

Solution:

mn(3m² – 4n²) – np(4n² – 3m²) + pm(15m² – 20n²)

= mn(3m² – 4n²) + np(3m² – 4n²) + pm(15m² – 20n²)

= mn(3m² – 4n²) + np(3m² – 4n²) + pm*5(3m² – 4n²)

= mn(3m² – 4n²) + np(3m² – 4n²) + 5pm(3m² – 4n²)

Taking (3m² – 4n²)  common

= (3m² – 4n²) (mn + np  +  5pm)

mn(3m² – 4n²) – np(4n² – 3m²) + pm(15m² – 20n²)  = (3m² – 4n²) (mn + np  +  5pm)

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Answer 2

Step-by-step explanation:

Given:

$mn(3 {m}^{2} - 4 {n}^{2} ) - np(4 {n}^{2} - 3 {m}^{2} ) + pm(15{m}^{2} - 20 {n}^{2} )$

To find: Factorise

Solution:

Step 1: Take (-) common from 2nd term and 5 common from 3rd term

$mn(3 {m}^{2} - 4 {n}^{2} ) + np(3 {m}^{2} - 4 {n}^{2} ) +5 pm(3 {m}^{2} - 4 {n}^{2} )$

Step 2: Take 3m²-4n² common from all terms

$(3 {m}^{2} - 4 {n}^{2} )(mn + np + 5pm)$

Final answer:

Factorisation of $mn(3 {m}^{2} - 4 {n}^{2} ) - np(4 {n}^{2} - 3 {m}^{2} ) + pm(15{m}^{2} - 20 {n}^{2} )$ is $\bold{(3 {m}^{2} - 4 {n}^{2} )(mn + np + 5pm)}$

Hope it helps you.

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