Question

4. Factorise
8-4L-4m - 2Lm - 4n + 2Ln + 2mn - Lmn
(1) (2-L) (2 + m) (2 + n) ​

Answer 1

Answer:

you can find the answer from the picture attached

Answer 2

The factors are (2-L)(2-m)(2-n)

Step-by-step explanation:

The question can be solved by first grouping the terms( two terms) and then taking out whatever is common in the terms outside the bracket.

$8-4L-4m + 2Lm - 4n + 2Ln + 2mn - Lmn$

8-4L -4m + 2Lm - 4n + 2Ln + 2mn - Lmn

$4(2-L)-2m(2-L)-2n(2-L)+mn(2-L)$

(2-L) is common in all, so it can be taken out

$(2-L)(4-2m-2n+mn)$

$(2-L)$($2(2-m)-n(2-n)$)

$(2-L)(2-n)(2-m)$