Dividing one polynomial by another gives a quotient and a remainder, as with integers. This page builds the process up from matching coefficients and then sets it out as long division.
Problem 10.4.1. Suppose are such that .
(a) Use the coefficient of to find .
(b) Use the coefficient of to find .
(c) Find .
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(a) We know , so .
(b) . Since , this is , so .
(c) We now have , so , therefore .
Let’s check this by expanding:
Problem 10.4.2. Let , so .
(a) What must the highest term of be? Why?
(b) Let , so that we have . Find by multiplying this equation by . Find the highest degree term of .
(c) Let , where is the highest degree term of from part (b), so we have . Find .
(d) Combine the previous parts to write , with (a polynomial) and .
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(a) must be the highest term, since when and ,. Therefore, as has degree , must have degree .
(b) Multiplying by :
so
Therefore , with highest term .
(c) Multiply out and compare with . The terms give , and with this is , so the highest term of is . But then the constant terms don’t match: . Hmmm.
(d) Let . Then
so ? (Let’s double check!)
So
i.e., and .
Note (Polynomial division in general). In general, for polynomial division we have
So , where either or , which means .
Long division of polynomials
Using the previous example:
At each step:
. We use since is the highest term: .
.
.
Therefore
Problem 10.4.3. Divide by .
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Problem 10.4.4.(The statement isn’t copied in the notes: it shows someone’s long division of by that contains a mistake.)
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He could tell he made a mistake since .
Corrected: , so the first term of the quotient is .
So the answer is . Check:
Problem 10.4.5.(The statement isn’t copied in the notes: it shows a mistaken long division of by .)
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He could tell because . Corrected:
So
Problem 10.4.6. Suppose . We divide into :, where either or . Now also let be such that , where or . We show that and .
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Now assume . Then the left side is of degree , but then must be such that , which cannot be, since
and both are such that . This contradicts . Therefore cannot be nonzero, so
so , and .
Exercises
Exercise 10.4.7. Divide:
(a) by ;
(b) by ;
(c) by ;
(d) by .
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(a)
(b)
(c)
(d)
Exercise 10.4.8. Find such that for .
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After the first step the coefficient is ; taking , the next quotient term is , and what is left, , divides exactly, so . So
Exercise 10.4.9. Divide by , and find .
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Since has degree less than , the division stops here: the remainder is . For no term to be left in the remainder, we need , and then the remainder is the constant .