When the remainder on dividing by is , the divisor is a factor. This page proves that, uses it to test and factor polynomials, and looks at how the roots of a factor relate to the roots of the polynomial it divides.
Problem 10.7.1. Let , where is a constant such that is divisible by . Find .
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If , then is a root of , therefore . So
Problem 10.7.2. Let be a polynomial.
(a) Show that if is a factor of , then .
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If is a factor of , then such that . By the remainder theorem, note
Thus .
(b) Show that if , then is a factor of .
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Divide by : , and since is linear, is of order , i.e. a constant. If , then by the remainder theorem
so . Thus is a factor of .
Theorem 10.7.3 (Factor Theorem). For , iff .
Proof. If , then , so . If is a root of , then ; since by the remainder theorem the remainder of is , but , the remainder is . Thus is a factor of . ∎
Example 10.7.4. ; , i.e. . Is the -value a root?
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Thus is a root. That means that is the remainder when you divide by the factor that creates, .
Long division. Divide the first terms, then multiply the result by the and subtract:
The quotient is . If you do not end with , then you did something wrong! So
Synthetic division (only when dividing by , that is, when is a root you are testing). Bring down the first coefficient, multiply it by , add to the next coefficient, and repeat:
The last number is the remainder. The bottom row gives , so .
Example 10.7.5. ; .
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Synthetic division (fill in zeros for the missing powers):
which gives , and .
Long division:
So the quotient is , and
Problem 10.7.6. has roots . Find .
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so , , , .
Unneeded, but the other way: , so
Using :
So , and then , so
and with : , so .
Problem 10.7.7. Let with .
(a) If is a factor of , then must every root of also be a root of ?
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Let such that . Now for the roots of we have
So we substitute the factored form for :
and by the remainder theorem we know that at a root of . Note that if is the for any , then
Ah! But for to divide , must be , thus for all root solutions to . Thus every is, by definition, a root of .
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If , then , which by our previous proof means either or . If , then trivial. If , then , where is any root of , always is . Now if for any , then for all such :
since by the remainder theorem .
(b) If every root of is a root of , then must evenly?
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Let and . Then every root of is a root of , but
is not evenly divided.
Theorem 10.7.8. For all , with roots , , we have
Back to part (a) of the problem above on the roots of a factor: if , then . Therefore if is a root of , , so , which by the factor theorem means . Thus all roots of divide .
Observe: , :
Problem 10.7.9. Let and note that .
(a) Find two linear factors of . Divide by one of these factors. Why are the roots of the quotient also roots of ?
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Since , and are factors of . The roots of the quotient are the remaining roots of , since , and like we showed, if , the roots of divide . So!
NOTE: .
Dividing by :
Dividing by :
Then dividing that by :
and .
So
Now to verify:
(the in comes from the in ). The constant term: , , . And
since the coefficient is and the coefficient is .
Exercises
Exercise 10.7.10. For what values of is a factor of ?
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Exercise 10.7.11. How can we quickly tell that is a factor of without using long division?
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Plug in : . Since the result is , is a factor of by the factor theorem.
Exercise 10.7.12. Let . One root is . Find all roots.
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Try in the cubic:
(On the side: .)
So
and the roots are .
Check: , and times this is . Also . Then
Exercise 10.7.13. Suppose and are polynomials such that . Must it be true that ?
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If , then
Exercise 10.7.14. is divisible by for some constant . Factor .
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So .
whose roots are
Exercise 10.7.15. Let . For which is a root?
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Synthetic division by :
For the remainder to be we need , so , so
Directly, , so