Question

19. The polynomial bx³+3x²-3 and 2x³-5x+b , when divided by x-4 , leave the remainders R1 and R2 respectively. Find the value of b if 2R1 - R2 = 0.


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Answer 1

Step-by-step explanation:

Given :-

The polynomial bx³+3x²-3 and 2x³-5x+b , when divided by x-4 , leave the remainders R1 and R2 respectively.

To find :-

Find the value of b if 2R1 - R2 = 0 ?

Solution :-

Given polynomials are bx³+3x²-3 and 2x³-5x+b

Given divisor = (x-4)

Let p(x) = bx³+3x²-3

Let g(x) = 2x³-5x+b

We know that

Remainder Theorem:-

Let P(x) be a polynomial of the degree greater than or equal to 1 and x-a is another linear polynomial if P(x) is divided by x-a then the remainder is P(a).

If p(x) is divided by x-4 then the remainder = p(4)

=> p(4) = b(4)³+3(4)²-3

=> p(4) = b(64)+3(16)-3

=> p(4) = 64b +48-3

=> p(4) = 64b+45

According to the given problem

If bx³+3x²-3 is divided by x-4 then the remainder = R1

=> R1 = 64b+45 -------------(1)

If g(x) is divided by x-4 then the remainder = g(4)

=> g(4) = 2(4)³-5(4)+b

=> g(4) = 2(64)-20+b

=> g(4) = 128-20+b

=> g(4) = 108+b

According to the given problem

If 2x³-5x+n is divided by x-4 then the remainder = R2

=> R2 = 108+b -------------(2)

Given that

2R1-R2 = 0

From (1)&(2)

=> 2(64b+45) -(108+b) = 0

=> 128b+90-108-b = 0

=> (128b-b)+(90-108) = 0

=> (128-1)b +(-18) = 0

=> 127b -18 = 0

=> 127b = 18

=> b = 18/127

Answer:-

The value of b for the given problem is 18/127

Used formulae:-

Remainder Theorem:-

Let P(x) be a polynomial of the degree greater than or equal to 1 and x-a is another linear polynomial if P(x) is divided by x-a then the remainder is P(a).

Answer 2

$\large\underline{\sf{Solution-}}$

Given polynomials are

$\rm :\longmapsto\: {bx}^{3} + 3 {x}^{2} - 3$

and

$\rm :\longmapsto\: {2x}^{3} - 5x + b$

Let assume that

$\rm :\longmapsto\: p(x) = {bx}^{3} + 3 {x}^{2} - 3$

and

$\rm :\longmapsto\:q(x) = {2x}^{3} - 5x + b$

Now, it is given that when

$\red{\rm :\longmapsto\:p(x) \: is \: divided \: by \: (x - 4), \: leaves \: remainder \:\: R_1}$

We know,

Remainder Theorem states that when a polynomial p(x) is divided by linear polynomial x - a, then remainder is p(a).

So, using this Remainder Theorem, we have

$\rm :\longmapsto\:R_1 = p(4)$

$\rm :\longmapsto\: R_1 = {b(4)}^{3} + 3 {(4)}^{2} - 3$

$\rm :\longmapsto\: R_1 = 64b + 48 - 3$

$\rm :\longmapsto\: R_1 = 64b + 45 - - - (1)$

Also, given that when

$\red{\rm :\longmapsto\:q(x) \: is \: divided \: by \: (x - 4), \: leaves \: remainder \: R_2}$

So, by using Remainder Theorem, we have

$\rm :\longmapsto\:R_2 = q(4)$

$\rm :\longmapsto\:R_2 = {2(4)}^{3} - 5(4) + b$

$\rm :\longmapsto\:R_2 = 128 - 20 + b$

$\rm :\longmapsto\:R_2 = 108 + b - - - - (2)$

According to statement,

$\red{\rm :\longmapsto\:2R_1 - R_2 = 0}$

On substituting the values from equation (1) and (2), we get

$\rm :\longmapsto\:2(64b + 45) - (108 + b) = 0$

$\rm :\longmapsto\:128b + 90 - 108 - b = 0$

$\rm :\longmapsto\:127b - 18 = 0$

$\rm :\longmapsto\:127b = 18$

$\bf :\longmapsto\:b = \dfrac{18}{127}$

Additional Information :-

Factor Theorem states that when a polynomial p(x) is divided by linear polynomial x - a, then remainder is 0.

More Identities to know :-

$\boxed{ \rm{ {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy}}$

$\boxed{ \rm{ {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy}}$

$\boxed{ \rm{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y)}}$

$\boxed{ \rm{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}}$

$\boxed{ \rm{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2})}}$

$\boxed{ \rm{ {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + xy + {y}^{2})}}$

$\boxed{ \rm{ {x}^{2} - {y}^{2} = (x + y)(x - y)}}$