The remainder on dividing by is the value . This page finds that pattern, proves it, and uses it to find remainders on division by divisors of higher degree and to answer some questions about divisibility.
Problem 10.6.1. 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 10.6.2. 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 10.6.3. Use synthetic division to find for .
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, therefore . Directly:
Problem 10.6.4. 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 10.6.5. Let .
(a) Divide by .
(b) Find .
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(a)
so the remainder is .
(b) Plugging in the zero of :
Problem 10.6.6. 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 10.6.7. 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
Problem 10.6.8. 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 10.6.9. 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 10.6.10. 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 .
Recall, for and :
must be at most a quadratic: .
Expanding :, and
so we need
That is, , and
Writing , the coefficients are
power
coefficient of
and on the other side, with ,
Exercises
Exercise 10.6.11.. 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 10.6.12. and , with . Find .
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, so , and .
Exercise 10.6.13. Find for divided by .
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With : and .
Therefore
Exercise 10.6.14.. Show that if is a root of , then .
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So if is a root of , setting in gives
Exercise 10.6.15. 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 :