Learning Standard Identities and their Applications Video Lecture - Class 8

hello friends this video on algebraic
expressions part 18 is brought to you by
example.com normal your gum exam so here
we are going to talk about the standard
identity so there are three standard
identities which are the basic
identities and are very very helpful in
to make our calculations easier
especially to make the calculations
associated with product of expressions
easier and each of these identities is
obtained by multiplying two binomials so
let us try to prove each of these
identities so the first identity is a
plus B whole square is equal to a
squared plus 2 a B plus B square so
normally these identities are people
generally memorize these identities but
what I ever see is just captures the
questions more understand from there you
are derived them and you will see it for
yourselves when you really don't need to
memorize it you will remember it on your
own let's try to prove this a plus B
whole square in the left hand side we
see that it is a plus B whole square
which means a plus B into a plus B that
is nothing but product of two binomials
so how do we find product of two
binomials each term of this expression
will get multiplied to each term of the
other expression similarly the second
term will also get multiplied to all the
terms so this would be a 2 8 plus a into
B plus B into 8 plus B into B so a into
a into a square a B plus a B plus B
squared so this is equal to a squared
plus 2 a B plus B Square this is equal
to what we have on the left hand side so
basically this is just a simplified
version of product of two binomials
but this identity helps us in a lot of
calculations so we will see that also
how do we use how do we use these
identities let's look at the next one a
minus B whole square is equal to a
square minus 2 a B plus B Square let's
do too tall for a minus B whole square
would mean a my
into a minus B so that would actually
need a into a minus a into B minus B
into 8 plus B to B so this is a square
minus a B minus a B plus B square that
is equal to a square minus 2 a B plus B
Square which is equal to R it and
therefore croute third one is a plus B
into a minus B is equal to a square
minus B Square so let's prove this so
again here is on the left hand side we
are actually multiplying two binomials
one binomial is a plus B the other
binomial is a minus B so here if you
multiply away with both these terms so
it becomes a into 8 minus a into B again
plus B into a minus B into B so this is
equal to e square minus 18 plus 18 minus
B squared so minus 80 plus 80 will get
cancelled so you get a square minus B
Square which is on the right hand side
so therefore these are the three
standard identities and we will next see
that how with the help of these
identities we made multiplications
easier so let us suppose that you have
been given a problem like 6x minus 7
into 6x plus 7 so you have to do this
you have to find out the product of
these two binomials now you have two
options one option is you solve this by
actual multiplication without making use
of the identity so let's see that is our
option one so in option man what we will
do fix it into 6x plus 7 minus 7 into
speak 6 plus 7 so this is equal to 6x
into 6x plus 6h into 7 minus 7 into 6x
minus forty-nine
which is 6 into 6 36 X into X X square
16 2014 2 X minus 18 42 X minus
forty-nine plus 42 X minus 42 X will get
cancelled
would be 36 extra square - 49 so this
would be our answer this is by option
one where you have not made use of
identities now in option two we made use
of the identity so in that case what do
we do so if you look at the expression
closely it is in the form of a minus B
where this stands for a and it stands
for B and this is a plus B so this is in
the form of a minus B into a plus B and
we know that a minus B into a plus B is
equal to a square minus B Square so a is
6x the 6 s square minus B Square D is 7
so 6 H whole square would be 36 X square
7 square would be 49 so you see just in
two simple steps you could cause this by
making use of the identity so this was
easier moreover in this case the chances
of making calculation mistake is also
less
so this is more than we need to discern
how the standard identities make our
calculations a lot easier thank you
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