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0 is a solution of the equation x + 1 = 0
Solution
Transcript
False.
Consider the equation, x + 1 = 0
Then, x = -1
"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 "
0 is a solution of the equation x + 1 = 0
Solution
Transcript
False.
Consider the equation, x + 1 = 0
Then, x = -1
"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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