NCERT Solutions Class 8 Mathematics Solutions for Exercise 14.2 in Chapter 14 - Factorisation

Question 6 Exercise 14.2

Factorise the expression:

$(1+m)^{2}-4 ! m\left(\text { Hint: Expand }(1+m)^{2} \text { first }\right)$

Answer:

Solution:

$\begin{aligned} &\text { Expand }(1+\mathrm{m})^{2} \text { using identity: }(x+y)^{2}=x^{2}+2 x y+y^{2}\\ &(1+m)^{2}-4 l m=l^{2}+m^{2}+2 m l-4 l m l\\ &=l^{2}+m^{2}-2 m l\\ &=(\mid-m)^{2}\\ & Usin g \text { identity: }(x-y)^{2}=x^{2}-2 x y+y^{2} \end{aligned}$

Video transcript

"hello students i am rita your math leader tutor and the today's question is factorize the expression l plus m whole square minus 4 l m first of all we use the identity that is x plus y whole square which is equal to x square plus 2 x y plus y square first of all we use this identity in the first one now when we open this identity then we got l square plus 2 l m plus m square and minus 4 l m now we solve these two both are same now we got the equation l square plus m square minus 2 l m now again we use a identity that is x minus y whole square now the identity is x square minus 2 x y plus y square now we got the answer that is l minus m whole square so this is the answer of this question i hope you like this video so please subscribe video for more updates and do comment your questions thank you "

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