Question

mn(3m^2 – 4n^2) – np(4n^2 – 3m^2) + pm(15m^2 – 20n^2)

Answer 1

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

Hope it helps you.

To learn more:

1) x³ + 3x² - 4x solve this

https://brainly.in/question/43438473

2) if (X + 2) is a factor of X⁵ - 4a²x + 2 X +2a+3 find a.

https://brainly.in/question/12783153