Question

let p nd q are the remainders when polynomials x³+2x²-5ax-7 and x³+ax²-12x+6are divided by x+1 and x-2 respectively. if 2p+q=6....find the value of a.

Answer 1

using remainder theorem

x = - 1

x^3 + 2x^2 - 5ax - 7 = p

( - 1) ^3 + 2 ( - 1 )^2 - 5 ( - 1 )a - 7 = p

- 1 + 2 - 7 + 5a = p

-6 + 5a = p

when x^2 + ax^2 - 12x + 6 is divided by x - 2

using remainder theorem

x = 2

x^3 + ax^2 - 12x + 6 = q

( 2 )^3 + a ( 2 )^2 - 12 ( 2 ) + 6 = q

8 + 4a - 24 + 6 = q

4a - 10 = q

2p + q = 6

2 ( 5a - 6 ) + (4a - 10 ) = 6

10a - 12 + 4a - 10 = 6

14a - 22 = 6

14a = 28

a = 2

Answer 2

Given p(x) = x^3 + 2x^2 - 5a - 7.

Given that p(x) is divisible by x + 1, the remainder is p(-1).

p(-1) = -1 + 2 + 5a - 7

        = 5a - 6


Given q(x) is divisible by x-2.

Given that q(x) is divisible by x - 2, the remainder is q(2).

q(2) = (2)^3 + a(2)^2 - 12(2) + 6

        = 8 + 4a - 24 + 6

        = 4a - 10.



Given Equation is 2p + q = 6

                              2(5a - 6) + (4a - 10) = 6

                              10a - 12 + 4a - 10 = 6

                              14a - 22 = 6

                              14a = 22 + 6

                              14a = 28

                                 a = 2.



Therefore the value of a = 2.


Hope this helps!