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

Question 10 Exercise 14.2

Factorise:

$(1+m)^{2}-(1-m)^{2}$

Answer:

Solution:

$\begin{array}{l} =\{(1+\mathrm{m})-(1-\mathrm{m})\}\{(1+\mathrm{m})+(1-\mathrm{m})\} \\ \text { Using Identity: } x^{2}-y^{2}=(x+y)(x-y) \\ =(1+\mathrm{m}-1+\mathrm{m})(1+\mathrm{m}+1-\mathrm{m}) \\ =(2 \mathrm{m})(21) \\ =4 \mathrm{ml} \end{array}$

Video transcript

"hello students i am rita your math leader tutor and the today's question is factorize 1 plus m whole raised to power 2 minus bracket 1 minus m whole raised to power 2. we have to solve this question first of all we use the identity first of all x plus y whole square is equal to x square plus 2xy plus y square and the next identity is x minus y whole square that is equal to x square minus 2 x y plus y square and the last identity that is a square minus b square we got a plus b and a minus b we can use these three identity in this question there are two methods to solve first of all a plus b whole square we use first one identity and a minus b whole square we use second identity and then we solve the question and the last second method is we directly use the third identity that is x square minus y square is equal to x plus y and x minus y so we go through third identity is x square minus y square we take it this one is x and this one is y and we use this third one identity then we got a x square minus y square is x plus y and the second one is x minus y so m become plus because minus minus plus so plus and minus m cancel out 1 plus 1 2 and plus 1 minus 1 cancel out m plus m 2 n now the answer is 2 two dozer four amps so this is the answer of this question i hope you like this video so please subscribe d2 for more updates and do comment your questions thank you"

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