An algebra problem by Pankaj Joshi
A polynomial f ( x ) with rational coefficients leaves remainder 1 5 when divided by x − 3 and remainder 2 x + 1 when divided by ( x − 1 ) 2 . If f ( x ) is divided by ( x − 3 ) ( x − 1 ) 2 , then the remainder is r ( x ) . What is the value of r ( 5 ) .
The answer is 43.
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1 solution
Thanks for the solution.
An easier approach is to let
f
(
x
)
=
A
(
x
)
(
x
−
3
)
(
x
−
1
)
2
+
a
x
2
+
b
x
+
c
. Then, we know that
1. remainder when divided by (
(
x
−
3
)
:
9
a
+
3
b
+
c
=
1
5
2. Coefficient of
x
when divided by
(
x
−
1
)
2
:
b
+
2
a
=
2
3. Coefficient of
1
when divided by
(
x
−
1
)
2
:
c
−
a
=
1
Solving this gives us a = 2 , b = − 2 , c = 3 .
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I didn't understand steps 2 and 3. Could you please explain then.
Ah... I knew there had to be an easier way to solve this. Thanks.
Sir, but won't this be special case of f ( x ) , since the degree of f ( x ) is not mentioned.
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I had a typo and updated the form of f ( x ) , adding in a factor of A ( x ) .
thanks.............
We can write,
f ( x ) = g ( x ) ( x − 1 ) 2 + 2 x + 1
f ( x ) = g ( x ) ( x − 1 ) 2 + 2 ( x − 1 ) + 3
f ( x ) − 3 = ( x − 1 ) ( g ( x ) ( x − 1 ) + 2 )
x − 1 f ( x ) − 3 = g ( x ) ( x − 1 ) + 2
Now, let x − 1 f ( x ) − 3 = F ( x )
Then F ( x ) = g ( x ) ( x − 1 ) + 2
and F ( 1 ) = 2
Now, by remainder-factor theorem,
f ( 3 ) = 1 5
⇒ F ( 3 ) = 3 − 1 f ( 3 ) − 3 = 6
Now, let the remainder when F ( x ) is divided by ( x − 1 ) ( x − 3 ) be R ( x ) = a x + b
Then, R ( 1 ) = a + b = 2
And R ( 3 ) = 3 a + b = 6
Whose solution is a = 2 , b = 0
Therefore R ( x ) = 2 x
And F ( x ) = G ( x ) ( x − 1 ) ( x − 3 ) + 2 x
x − 1 f ( x ) − 3 = G ( x ) ( x − 1 ) ( x − 3 ) + 2 x
f ( x ) − 3 = G ( x ) ( x − 1 ) 2 ( x − 3 ) + 2 x 2 − 2 x
f ( x ) = G ( x ) ( x − 1 ) 2 ( x − 3 ) + 2 x 2 − 2 x + 3
Therefore r ( x ) = 2 x 2 − 2 x + 3
And r ( 5 ) = 2 ∗ 5 2 − 2 ∗ 5 + 3 = 4 3