Question

p(x) is a polynomial of degree 3 such that p(1)=3 p(2) = 4 and p(3) =5 . if p(x) is divided by (x^2-4x+3) then the remainder is?



pls tell fast​

Answer 1

Let P(x) = $(x^2-4x+3)(ax+b)+(cx+d)$

Where, (ax + b) is quotient and  (cx + d) is remainder

We have to find, the value of remainder.

Solution:

∴ P(1) = $(1^2-4(1)+3)(a(1)+b)+(c(1)+d)$

⇒ c + d = 3                    ..... (i)

P(2) = $(2^2-4(2)+3)(a(2)+b)+(c(2)+d)$

⇒ - 2a - b + 2c + d = 4   ..... (ii)

P(3) = $(3^2-4(3)+3)(a(3)+b)+(c(3)+d)$

⇒ 3c + d = 5                    ..... (iii)

Subtracting (i) from (iii), we get

c + d - 3c - d = 3 - 5    

⇒ - 2c = - 2

⇒ c = 1

Put c = 1 in equation (i), we get

1 + d = 3  

⇒ d =  2  

∴ Remainder = cx + d = x + 2

Thus, the required "remainder is (x + 2)".

Answer 2

Given : p(x) is a polynomial of degree 3 such that p(1)=3 p(2) = 4 and p(3) =5  

To Find :  p(x) is divided by (x²-4x+3) then the remainder is

Solution:

Let say polynomial is

P(x) = ax³  + bx²  + cx + d

p(1)=3  

=> a + b + c + d  = 3            Eq1

p(2) = 4

=> 8a + 4b + 2c + d  = 4       Eq2

p(3) = 7

=> 27a + 9b + 3c + d  = 5    Eq3

7a + 3b + c = 1         Eq2 - Eq1

19a + 5b + c = 1       Eq3 - Eq2

=> 12a  + 2b  =  0  

=> b = - 6a

7a + 3b + c = 1 => 7a - 18a + c = 1  => c - 11a = 1

=> c = 11a + 1

a + b + c + d  = 3

=> a -6a + 11a + 1 + d = 3  => 6a + d  = 2

=> d = 2 - 6a

P(x) = ax³  + bx²  + cx + d

=>P(x)  =  ax³  -6ax²  + 11ax + x  + 2 - 6a

                       ax   -2a  

x² -4x + 3  _|   ax³  -6ax²  + 11ax + x  + 2 - 6a  |_

                      ax³   -4ax² + 3ax

                    ________________

                              -2ax² + 8ax  + x + 2 - 6a

                              -2ax²  + 8ax              -6a

                             _____________________

                                                   x + 2

remainder is x + 2

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