Answers

2014-10-17T16:18:10+05:30

This Is a Certified Answer

×
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
P(1) = a + b + c + d + 1 = p(2) = 16 + 8a + 4b + 2 c + d 

          7 a + 3 b +  c + 15  = 0     --  eq 1
             c  = - 7 a - 3b - 15    -- eq 4

p(3) = 81 + 27 a + 9 b + 3c +d = 0
          27 a + 9 b + 3 c + d = - 81       -- eq 2

4 * eq 1 - equ 2 gives    a + 3b + c - 21 = d        -- eq 3   substitute value of c.
              d = a + 3b -7a - 3b - 15 - 21  = - 6 a - 36   -- eq 5  

p(4)+p(0) = (256 + 64 a + 16 b + 4 c + d) + ( d)
               = 64 a + 16 b + 4 c + 2 d + 256

substituting values of c and d,
             = 24 a + 4b + 124

1 1 1
but sir it is given that p(1)=p(2) then how u have taken p(1)=0 and p(2)=0.Please explain
oh! i am sorry - the letters are so close, i mistook , for =
done
2014-10-17T18:38:10+05:30
P(1)= 1^4+a(1)^3+b(1)^2+c(1)+d=a+b+c+d+1
substitute the values of x=2 and 3
equate them. you will get P(4 )+ p(0) = 64a+16b+4c+2d+256
since, you have got the values of c and d, substitute them. you will get the final equation as 24a+4b+124
0