L10 : Remainder theorem proof - Polynomials, Maths, Class 9 Video Lecture

hello friends this video on polynomials
part 10 is brought to you by exam viacom
no more fear from exam let's take some
more example you find the remainder by
dividing X cube plus 3 by X plus 1 so we
have two ways one is actually you divide
two that multiply with X square this is
equal X cube plus X square gets
cancelled you get minus X square plus X
minus X minus X square minus X this is
cancelled you get 2x check once again X
cube plus X square correct this becomes
1 this is X 1 actually 1 minus X square
and what you get is 1 plus X there is a
plus 1 1 plus X there is no remedy that
is one way or I know that if px is equal
to X cube plus 1 px by X plus 1 is
nothing but and then remainder 4 this is
nothing but P or minus 1
let's find P of minus M minus 1 cube
plus 1 that is it so using any of these
method you can find the remainder using
a metal theorem or you can actually do a
physical division ok so hope you
understood the concept of elemental
theorem and if I have two polynomials to
divide px and it suppose by X plus 1 or
X plus 3 kind of number so this X plus K
need to find P of minus K so P of minus
K actually or what is the number you
have it P
DX by GX right you say GX is equal to
zero you find the zeroes of GX and that
zero of Z X you put here that will
actually give you the remainder okay
when you say px by GX the degree of P X
has to be more than reading Rio TX so
let's do proof or remainder theorem okay
so let's take let's px be any polynomial
polynomial with degree greater than
equal to one okay so I am saying if px
is divided by X minus C I get quotient
is let's suppose QK and remainder s RX
excited
thus I can write that P X is nothing but
X minus C whatever you have into Q of X
plus remainder write dividend this is
divisor this is okay so dividend is
division into quotient plus iment one
more thing to notice as well general
rule we have seen that the degree of
remainder will always be less than the
degree of DBZ degree of divisor is
degree of D here is one maximum X to the
power is one so degree of remainder will
be zero so degree of remainder is zero I
can write our X as nothing but some are
okay because it has to be some constant
now I have to make this a zero to get
the remainder to make this a zero
QX is something which you don't know
this is unknown so X minus C is
something we know divisor so I can say X
minus C is zero so if you say X minus C
is X is zero there is X is equal to C
right so if I find P of C that will
become nothing but if you see find P of
C X minus C is zero zero into Q of X
plus R it is zero plus ah that is R so
as per this logic P of C will give me
the remainder okay so as we just proved
this is the remainder theorem so px be
any polynomial greater than or equal to
1 and ay be any real number if px is
divided by a linear polynomial X minus a
then the remainder is always PE
P X divided by X minus a for this you
want to find the remainder what this is
nothing but P okay let's take some
example find the remainder when this
whole expression is divided by X minus
one so one big ways you actually
physically do the division or I can see
that remainder when X minus one divides
px as you px is this is nothing but P of
1 as perimeter theorem
so let's find P of 1 P of 1 will be 1 to
the power 4 plus 1 to the power 3 minus
2 into 1 to the power 2 plus 1 plus 1 so
this becomes 2 so remainder is 2 pretty
easy shortcut to find remainder check
whether polynomial this is a multiple of
2 T plus 1 as I told you this polynomial
will be multiple of this if remainder is
equal to 0 if you divide Qt with this
number so if on QT you divide with 2d
plus okay so let's find the remainder so
as per remainder theorem Q of C 2 T plus
1 is equal to 0 gives T is equal to
minus 1 by 2 so Q of minus 1 by 2 is
nothing but the vendor when Qt divides 2
T plus 1 correct as parameter theorem so
let's find the remainder first so this
is 4 T cubed 4 into minus 1 by 2 Q plus
4 T Square t is in
- one by T Square minus T T again - 1 by
2 and n minus 1 solve it for in 2 minus
1 by 8 plus this becomes 4 into 4 1 by 4
this becomes plus 1 by 2 minus 1 this
gets canceled 1 minus 1 get canceled and
this is 4 by 8 is minus 1 by 2 this is
minus 1 by 2 plus 1 by 2 and this is 0
so the remainder is 0 I can say that
this whole QT is multiple of 2 T plus 1
same thing 9 by 3 is 0 no matter is 0
that means 3 is multiple of 9 8 by 3
Dementor is not 0 3 is not multiple of 8
same thing so Q T by 2 T plus 1 remember
is 0 that means 2t plus 1 is a factor of
Q T or Q T is a multiple of 2 T plus 1
both are same
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