Definition 6.1 (Polynomials and polynomial functions). is a polynomial function, and the right-hand side is the polynomial. Each summand () is called a term. When the highest power of is , the polynomial is a constant (e.g. ); the zero polynomial is the constant . All others are nonzero. is the degree of the polynomial function : the highest power of in it.
Degree
Name
0
constant
1
linear
2
quadratic
3
cubic
4
quartic
Number of terms
Name
1
monomial
2
binomial
3
trinomial
Problem 6.3. Suppose . Find .
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So , since . Additionally, .
Expanding the first few powers:
The sums of the coefficients of go
so
Oh wait: the sum of the coefficients is the value at , which is :/
Example 6.2. for all . Find .
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so ,,,. Therefore
Exercise 6.1.4. Find the degree of , given that and are polynomials such that and , with .
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Let and , with . Then the leading term is given by
WLOG let all with be , and , so that
Ah, so now we see we have a leading term of , therefore of degree . Thus
Problem 6.4. 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 6.5. 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 6.6. Divide by .
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Problem 6.7.(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 6.8.(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 6.9. 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 6.2.1. Divide:
(a) by ;
(b) by ;
(c) by ;
(d) by .
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(a)
(b)
(c)
(d)
Exercise 6.2.2. 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 6.2.3. 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 .
Synthetic Division on
(Also known as Ruffini’s rule.)
Consider , where and . Then synthetic division is given by
Note this is , not “”.
Consider divided by :
Each shape denotes a row.
* This only works when the divisor is , i.e. its leading coefficient is .
Filling it in: bring down the ; multiply by and write the product under the next coefficient; add down the column; repeat.
the long division from before: quotient , remainder .
Problem 6.10. Divide by .
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By long division:
By synthetic division (note ):
Problem 6.11. Divide into .
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So
(dividing by ). So we have ,.
Let’s test!
multiplying by , we get
So : dividing by gives quotient and remainder .
Problem 6.12. Use synthetic division:
(a) ;
(b) .
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(a)
(b) , so divide by and then by :
This gives quotient and remainder . Check:
Exercises
Exercise 6.3.1. Divide by .
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Exercise 6.3.2. Divide by .
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Exercise 6.3.4. Divide by .
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:
Dividing the quotient row by :
Exercise 6.3.5. Divide by .
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(, so .) Dividing the quotient row by :
Exercise 6.3.6. Divide by .
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, so divide by and negate the quotient:
Exercise 6.3.7.(The statement isn’t copied in the notes: it shows a synthetic division of by that went wrong.)
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Write :
He forgot to turn into .
Remainder Theorem
Problem 6.13. Let .
(a) Find the remainder of .
(b) Find .
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(a)
(b) .
So it looks like the value of at the divisor’s zero/root is equal to the remainder.
Problem 6.14. Prove the remainder theorem.
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Suppose and are the quotient and remainder, respectively, when the polynomial is divided by ,.
Note that , therefore (or ). Note too that
Ah! So now it is clear to see that when (the zero of ) the term is . Thus . Since , is a constant, so is that constant. Thus the remainder when the polynomial is divided by is .
Problem 6.15. Use synthetic division to find for .
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, therefore . Directly:
Problem 6.16. Find the remainder when is divided by .
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Synthetic division with , on the coefficients (laid out vertically; each product is times the previous result):
Power
Coefficient
previous
Result
So , and
Check with the remainder theorem:
Problem 6.17. Let .
(a) Divide by .
(b) Find .
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(a)
so the remainder is .
(b) Plugging in the zero of :
Problem 6.18. Let be a polynomial such that when is divided by the remainder is , and when is divided by the remainder is . We find the remainder when is divided by .
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and .
so .
We have
so and . Also
At and only survives:
Subtracting, , so ,,. Then :, so . The remainder is
Problem 6.19. Find the remainder when is divided by .
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The exponents are odd squares: .
, with zeros . Call the dividend ; at these zeros
The remainder has degree less than , so , and
So and . From the first, ; then ,,, and . Hence
Exercise 6.4.1.. Find when is divided by for each of the following ’s:
(a) ; (b) ; (c) ; (d) ; (e) ; (f) .
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(a) .
(b) .
(c) .
(d) , the same ( has only even powers).
(e) . With :
so ,,, and :
(f) . With :
so ,,, and :
Exercise 6.4.2. and , with . Find .
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, so , and .
Exercise 6.4.3. Find for divided by .
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With : and .
Therefore
Exercise 6.4.4.. Show that if is a root of , then .
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So if is a root of , setting in gives
Exercise 6.4.5. Find the remainder when is divided by .
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Factor the divisor:
and . So the zeros are .
, so and
Since :, so , i.e. . With , the third equation gives
so , and gives :
Problem 6.37. Find such that , is divisible by , and is itself divisible by .
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Take 1. and . The quotient of by has ; actually
so it is some . Write .
Since , try (the factor makes automatic). Then , so : for instance , or . For : ✓, but its quotient ✗; and for , synthetic division of by gives . The condition alone only makes divisible by , not by .
Take 2. Since ,. Since and , the quotient has :
So, comparing with :,, so , and . Then
Problem 6.36. Find the remainder when is divided by .
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, so
and the first term is at the zeros of the divisor. , so . The zeros: , and .
Since (the powers of cycle ), every term of is and every term of is :
From ,. Then gives , so and . So , and
Problem 6.35. Let . For how many polynomials does there exist with such that ?
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, so .
Let . So
where the remainder must be ; here has degree , has degree , and has degree .