Hello. I'm having a hard time with this problem. Please help. Thank you.
When a certain polynomial p(x) is divided by (x-1), remainder is 12. When the same polynomial is divided by (x-4), the remainder is 3. Find the remainder when the polynomial is divided by (x-1)(x-4).
When p(x) is divided by the quadratic polynomial (x-1)(x-4) ,
you get a quotient Q(x) and a remainder R(x).
Since the divisor, (x-1)(x-4) , has degree 2,
the degree of r(x) has to be less than 2.
So, P(x)=Q(x)(x-1)(x-4)+R(x) , and R(x)=ax+b .
Putting it all together,
P(x)=Q(x)(x-1)(x-4)+ax+b .
Dividing by (x-1) leaves a remainder of 12, so
P(1)=12 , meaning that
Q(1)(1-1)(1-4)+a+b=12 , and a+b=12 .
Dividing by (x-1) leaves a remainder of 12, so
P(4)=3 , meaning that
Q(4)(4-1)(4-4)+4a+b=3 , and 4a+b=3 .
The system of equations formed by a+b=12 and 4a+b=3 has as a solution
a=-3 and b=15 .
So, the remainder is R(x)=-3x+15 .
Thank you very much! :)