# `p(x)=9x^3+12x^2+hx+k` .When divided by (3x-1) there is a remainder of 2, when divided by (x+2) there is remainder is -5, find h and k

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### 2 Answers

The remainder of dividing p(x) by (3x-1) is 2

==> p(x)= (3x-1)h(x) +2

==> p(1/3)= (3*1/3 -1)h(1/3)+ 2 = 0*h(x)+2 = 2

==> p(1/3) = 2

==> p(1/3)= 9(1/3)^3 + 12(1/3)^2 + (1/3)h + k = 2

==> 1/3 + 4/3 + h/3 + k = 2

==> h/3 + k = 2- 5/3

==> h/3 + k = 1/3

==> h + 3k = 1.............(1)

Now given the remainder when dividing by (x+2) is -5

==> p(x)= (x+2)q(x) -5

==> p(-2) = -5

==> 9(-2)^3 +12(-2)^2 -2h + k = -5

==> -72 + 48 - 2h + k = -5

==> -2h + k = 19 ............(2)

Now we will solve the system (1) and (2).

Multiply (1) by 2 and add to (2).

==> 7k = 21

==> k= 3

==> -2h + k = 19 ==> -2h +3= 19 ==> -2h = 16==> h= -8

**==> k= 3 and h= -8**

Thanks a lot! I now see where I made my silly mistake!