Polynomial help

When a polynomial P(x) is divided by x-1, the remainder is 2. When P(x) is divided by x-2, the remainder is 1. Find the Remainder when P(x) is divided by (x-1)(x-2). THANKS...PLEASE ANSWER THIS

Note by Kumar Ashutosh
4 years, 9 months ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

Comments

Sort by:

Top Newest

You know that \(P(x) = (x-1)Q_1(x)+2\) and \(P(x) = (x-2)Q_2(x)+1\) for some polynomials \(Q_1(x)\) and \(Q_2(x)\). What is \(P(1)\)? What is \(P(2)\)?

Then, write \(P(x) = (x-1)(x-2)Q(x)+R(x)\) where \(Q(x)\) is a polynomial and \(R(x)\) is a linear polynomial. What is \(R(1)\)? What is \(R(2)\)? Now, what is \(R(x)\)?

Jimmy Kariznov - 4 years, 9 months ago

Log in to reply

So what is missing in this question? Actually, the question is correct and needs no more given quantity.. Thanks

Kumar Ashutosh - 4 years, 9 months ago

Log in to reply

Jimmy doesn't think there is anything wrong with the problem. He is just showing you the thought processes to reach the answer, without actually telling you the answer. If you follow what he is showing, it is not hard, at least at your level, to find the answer.

Bob Krueger - 4 years, 9 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...