Remainder theorem - Need Help

[Guest]

Need help on the below home work problem -
When P(x) is divided by x-1, the remainder is 1.
When divided by (x-2)(x-3), the remainder is 5.
What is the remainder when P(x) is divided by (x-1)(x-2)(x-5) ?

[Baltuilhe]

Good morning!

[tex]P(x)=q_1(x)(x-1)+1[/tex]
[tex]P(x)=q_2(x)(x-2)(x-3)+5[/tex]

[tex]P(x)=q_3(x)(x-1)(x-2)(x-3)+R(x)[/tex]
[tex]R(x)=ax^2+bx+c[/tex]

[tex]P(1)=1\\
P(2)=5\\
P(3)=5[/tex]

[tex]\begin{cases}a+b+c=1\\4a+2b+c=5\\9a+3b+c=5\\\end{cases}[/tex]

Solution:
[tex]\begin{cases}a=-2\\b=10\\c=-7\end{cases}[/tex]

So:
[tex]R(x)=-2x^2+10x-7[/tex]

[Angus53]

To find the remainder when \( P(x) \) is divided by \( (x-1)(x-2)(x-5) \), we'll use the Remainder Theorem.

Given that \( P(x) \) gives a remainder of 1 when divided by \( x-1 \) and a remainder of 5 when divided by \( (x-2)(x-3) \), we can write \( P(x) \) as:

\[ P(x) = (x - 1)Q_1(x) + 1 \]
\[ P(x) = (x - 2)(x - 3)Q_2(x) + 5 \]

Now, we want to express \( P(x) \) in terms of \( (x-1)(x-2)(x-5) \).

\[ P(x) = (x - 1)(x - 2)(x - 5)Q(x) + R(x) \]

We are looking for the remainder \( R(x) \).

Since \( P(x) \) and \( R(x) \) have the same remainders when divided by \( (x-1) \) and \( (x-2)(x-3) \), respectively, we can equate them:

\[ (x - 1)Q_1(x) + 1 = (x - 1)(x - 2)(x - 5)Q(x) + R(x) \]
\[ (x - 1)(Q_1(x) - (x - 2)(x - 5)Q(x)) = R(x) - 1 \]

Now, since \( (x-1) \) divides the left-hand side and \( (x-1)(x-2)(x-5) \) divides the right-hand side, the remainder \( R(x) - 1 \) must be a constant.

Let's denote \( R(x) - 1 = c \), where \( c \) is a constant.

Thus, the remainder when \( P(x) \) is divided by \( (x-1)(x-2)(x-5) \) is \( \boxed{c + 1} \), where \( c \) is the constant determined from the previous step.

[Guest]

To solve this problem, we can use the Remainder Theorem and Polynomial Remainder Theorem.

Let P(x) be the polynomial we're interested in.

When P(x) is divided by x−1, the remainder is 1. This means that P(1)=1.

When P(x) is divided by (x−2)(x−3), the remainder is 5. This means that P(2)=P(3)=5.

Now, let's construct a system of equations using these conditions.

From

P(1)=1,

We have:

P(1)=a(1−1)(1−2)(1−5)=1
a(1)(−1)(−4)=1
−4a=1
a=−1/4

From P(2)=P(3)=5,

we have:

P(2)=a(2−1)(2−2)(2−5)=5
a(1)(0)(−3)=5
−3a=5
a=−5/3

Since a should be consistent in both equations, we notice a contradiction. Therefore, the polynomial P(x) that satisfies the given conditions doesn't exist, or there's a mistake in the problem setup.

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