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NCERT Exemplar Solutions Class 6 Mathematics Solutions for Exercise in Chapter 7 - Algebra
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0 is a solution of the equation x + 1 = 0
Answer:
False.
Consider the equation, x + 1 = 0
Then, x = -1
Video transcript
"hello students welcome to the lead or
doubt solving session
and in today's session we are going to
check
whether polynomial p of x is a multiple
of
okay polynomial j of x or not all right
okay so um okay to solve this we need
not to do the
division here right we need now to do
the actual division
by using the remainder theorem okay we
can check
all right so let's solve this
okay so for the first
few of x is given
that is x cube minus
phi u x square plus 4 x
minus 3 okay and g of x
is equals to x minus 2 right
so to apply the remainder theorem i will
quickly find out the
0 of the 0 0 of x so 0
of g of x
okay will be
x minus 2 i need to equate it to the 0
and that's why x will be the
2 right so x is equals to 2 that is 0
for
x minus j of x is equals to x minus 2
all right now as we are okay we are
dividing
p of x by g of x right so here
i will apply the remainder theorem so
when i divide p of x by g of x
remainder will be
remainder will be
p of 2 right where 2 is a
0 of j of x all right now we'll
okay just put the value instead of x we
put 2
so which is equals to
okay x cube that means 2 cube
minus 5 u into 2 square
plus 4 into 2
minus 3 all right ok now
let's do this calculations quickly so 2
cube is
8 minus
okay 5 ux square so 2 square is 4
and 4 multiplied by 5 will be 20
plus 4 into 2 8
minus 3 all right now
if i do this calculation 8
8 plus 8 this is this 8 and this 8
16 plus 16 minus 20 that is minus 4 and
minus 4
minus 3 is equal to minus 7 right
so here remainder
remainder is equals to
minus 7 and which is not equal to
zero right that means
okay p of x is
not multiple of not a multiple of
multiple of g of x all right
or g of x is a not a factor of
p of x all right
okay now we'll discuss the second
question
so for the second question
p of x is given p of x is equals to
2 x cube minus
11 x square
minus 4 x plus
5 all right and
g of x is equals to
2 x plus 1 so first
we'll find out the zero of g of x
okay and okay zero of j of x
so it is 0
of g of x
is equals to that is 2 x
plus 1 is equals to 0
and that's why x is equals to
minus 1 upon 2 all right
ok now what i will do i will apply the
remainder theorem so according to the
remainder theorem when i
divide p of x by j of x okay
the remainder will be
remainder will be
p of minus 1 by 2 right
so here minus 1 by 2 is a 0 of
j of x all right so
is equals to now what i will do instead
of x i will put
the value that is minus 1 by 2 okay so
it is
2 multiplied by
okay x cube that means minus 1 upon 2
cube
minus 11
okay and multiplied by minus
1 upon 2 square
minus 4 into
minus 1 upon 2
plus plus
5 all right
now i will i will quickly
do the calculation
all right so
okay the remainder equals to
okay minus 1 upon 2 cube that means
okay minus minus 1 upon 2 cube is
minus 1 upon 8 right so it will be minus
2 by 8 minus
okay minus 1 1 by 2 square means 1 upon
4 right so it will be
11 by 4
okay then minus 4 multiplied by minus 1
by 2 so minus minus will be
plus okay so it will be 4 upon 2
okay and phi u can be written as 5 by
1 right now what i will do i will find
out the lcm of
8 4 2 and 1 okay so here
lcm okay after finding the lcm lcm will
be
8 all right so now what i will do
i will make the denominator same right
so that's why this is minus
2 okay
minus 11 by 4 will become um
22 by 8 so minus 22
okay plus 4 by 2 will become
okay 16 by 2 so it is 16
okay plus 4t
okay and they will have the common
denominator that is
eight all right okay now if this
okay do this calculation
okay so it will gives
34 by 8 right so
minus 2 uh minus 22 that is minus 24
minus 24 plus 16 will be minus 6 and
minus 6 plus 40 will be
34 upon
8 okay that means the remainder
is not equal to 0 all right
so for this case as well
okay p of x is not a
multiple of g of x right so that's why
p of x okay is
not multiple of
it will be
okay so for the second case as well p of
x is
not a multiple of g of x all right
so that's all about today's session if
you have any doubt please leave your
leave your comment below and please do
subscribe to this channel
thanks for watching the video bye
"
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