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A polynomial of degree more than 3 when divided by (x-1)^2 gives remainder 2x+1 , when divided by x-3 leaves remainder 15. When divided by (x-1)^2(x-3) gives remainder ax^2+bx+c then find a,b,c

A polynomial of degree more than 3 when divided by (x-1)^2 gives remainder 2x+1 , when divided by x-3 leaves remainder 15. When divided by (x-1)^2(x-3) gives remainder ax^2+bx+c then find a,b,c

Grade:10

2 Answers

mycroft holmes
272 Points
6 years ago
Let the polynomial be P(x).
 
Then P(x) = Q(x)(x-1)2+2x+1 from the 1st statement. Also, from this we get P(1)=3
 
Taking derivative and setting x =1, we get P’(1) = 2.
 
From the second statement, we have P(3)=15.
 
Now let P(x) = H(x)(x-1)2(x-3)+a(x-1)2+b(x-1)+c(x-3)
 
Then P(1) = 3 and hence c = -3/2
 
P’(1) = 2 and hence b+c = 2 gives b = 7/2
 
P(3) = 15, means 4a+2b = 15 and so a = 2.
 
Hence the required remainder is 2(x-1)2+7/2(x-1) – 3/2(x-3) which simplified gives you
 
2x2 -2x+3
Artemis
32 Points
one year ago
ATQ, When p(x) is divided by (x - 1), remainder is (2x - 1).
----->p(x) = q(x)(x - 1)+ (2x + 1)                         (i)                                           (using division algorithm for polynomials)
Again ATQ, when p(x) is divided by (x - 3), remainder is 15
----->p(3) = 15                                                           (ii)                                                                       (using remainder theorem)
 
----->p(3) = q(3)(3 - 1)2 + (2 × 3 +1)                                                                                                                                (using (ii))    
----->15 = 4 q(3) + 7
----->q(3) = 2                                                             (iii)
 
----->when q(x)is divided by (x - 3), remainder is 2                                            (using converse of remainder theorem)
----->q(x) = g(x)(x - 3) + 2                                      (iv)                                         (using division algorithm for polynomials)
 
----->p(x) = {g(x)(x - 3) + 2}(x - 1)+ (2x + 1)                                                                                         (substituting (iv) in (ì))
----->p(x) = g(x)(x - 3)(x - 1)+ {2(x - 1)+ (2x + 1)}
----->p(x) = g(x)(x - 3)(x - 1)+ (2x2 - 2x + 3)
 
----->when p(x) is divided by (x - 1)2(x - 3), remainder is 2x- 2x + 3              (using converse of division algorithm)
 
On comparing 2x- 2x + 3 with ax2 + bx + c,
a = 2, b = -2, c = 3
 
 
 

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