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NCERT Solutions Class 8 Mathematics Solutions for Exercise 14.3 in Chapter 14 - Factorisation
Prev
Factorize the expressions and divide them as directed:
5 p q\left(p^{2}-q^{2}\right) \div 2 p(p+q)
Answer:
Solution:
The factorization is made much simpler by employing several identities.
A number of expressions that can be factored take the form of or can be put into the following: a2 - b2 =(a+b)(a-b)
\mathrm{p}^{2}-\mathrm{q}^{2} \text { can be written as }(\mathrm{p}-\mathrm{q})(\mathrm{p}+\mathrm{q}) \text { using identity. }5 p q\left(p^{2}-q^{2}\right) \div 2 p(p+q)=5 p q(p-q)(p+q) / 2 p(p+q)=5 / 2 q(p-q)
Video transcript
"hello students i am rita your math
leader tutor and the today's question is
factorize the expressions and divide
them as directed
and the question is 39 y raised to power
3
bracket 50 y square minus
98 divided by 26
y raised to the power 2 bracket
5 y plus 7
first of all in this question we take
out common 2
from the 50 y square and 98 we take out
common
2 from them then we got 25 y
square minus 49
divided by 26 y
square five y plus seven
now 25 is the perfect
square of five and
25 is the perfect square of
5 and 49 is a perfect square of
7 so we can write it as 7 square
divided by 26 y raised to power 2
5 5 plus 7 now as we know
the formula of e square minus b square
is equal to a plus b bracket
a minus b now we use this identity
and solve the equation so 39
into 2 y raised to power 3
into a plus b 5 y plus
7 into 5 5 minus
7 divided by 26
y raised to power 2 bracket 5
y plus 7 now we cancel out the terms
5 y plus seven and
thirteen twos are thirteen threes are
and two ones are two ones are
so we got the answer and the answer is
3 and y is to power 3 and y is to power
2 so 3 minus 2 we got 1 so single y
y is 2 power 1 bracket 5 y
minus 7 so this is our answer i hope you
like this video so please subscribe leo
for more updates and do comment your
questions thank you"
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