Remainder Theorem

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11.Remainder and Factor Theorem (A)

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The Remainder Theorem If is any polynomial and is divided by then the remainder is . If = 0, then is a factor of . We apply the Remainder Theorem to obtain the remainder when %( ’) = 2 4 + 7’-+2’ 9 was divided by (2’ + 3). By the Remainder Theorem, the remainder is %A−4-B. %F− 3 2 G = 2(− 3 2)4 +7(− 3 2)-+2(− 3 2)+ 9 %F− 3 2 ...

Theorem, Remainder, Remainder theorem

3.2 The Factor Theorem and The Remainder Theorem

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Theorem 3.5.The Remainder Theorem: Suppose pis a polynomial of degree at least 1 and cis a real number. When p(x) is divided by x cthe remainder is p(c). The proof of Theorem3.5is a direct consequence of Theorem3.4. When a polynomial is divided by x c, the remainder is either 0 or has degree less than the degree of x c. Since x cis degree

Theorem, Remainder, Remainder theorem

The Chinese Remainder Theorem - luc.edu

gauss.math.luc.edu

By solving this by the Chinese remainder theorem, we also solve the original system. (The solution is x 20 (mod 56).) Of course, the formula in the proof of the Chinese remainder theorem is not the only way to solve such problems; the technique presented at the beginning of this lecture is actually more general, and it requires no mem-orization.

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The Chinese Remainder Theorem - homepages.math.uic.edu

homepages.math.uic.edu

Chinese Remainder Theorem tells us that there is a unique solution modulo m, where m = 11 ⋅ 16 ⋅ 21 ⋅ 25 = 92400. We apply the technique of the Chinese Remainder Theorem with k = 4, m 1 = 11, m 2 = 16, m 3 = 21, m 4 = 25, a 1 = 6, a 2 = 13, a 3 = 9, a 4 = 19, to obtain the solution. We compute z 1 = m / m 1 = m 2 m 3 m 4 = 16 ⋅ 21 ⋅ ...

Chinese, Theorem, Remainder, Chinese remainder theorem

Math 127: Chinese Remainder Theorem

www.math.cmu.edu

Example 5. Use the Chinese Remainder Theorem to nd an x such that x 2 (mod5) x 3 (mod7) x 10 (mod11) Solution. Set N = 5 7 11 = 385. Following the notation of the theorem, we have m 1 = N=5 = 77, m 2 = N=7 = 55, and m 3 = N=11 = 35. We now seek a multiplicative inverse for each m i modulo n i. First: m 1 77 2 (mod5), and hence an inverse to m 1 ...

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The Remainder Theorem - cdn.kutasoftware.com

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The Remainder Theorem Date_____ Period____ Evaluate each function at the given value. 1) f (x) = −x3 + 6x − 7 at x = 2 2) f (x) = x3 + x2 − 5x − 6 at x = 2 3) f (a) = a3 + 3a2 + 2a + 8 at a = −3 4) f (a) = a3 + 5a2 + 10 a + 12 at a = −2 5) f (a) = a4 + 3a3 − 17 a2 + 2a − 7 at a = 3 6) f (x) = x5 − 47 x3 − 16 x2 + 8x + 52 at ...

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THE CHINESE REMAINDER THEOREM

kconrad.math.uconn.edu

The Chinese remainder theorem can be extended from two congruences to an arbitrary nite number of congruences, but we have to be careful about the way in which the moduli are relatively prime. Consider the three congruences x 1 mod 6; x 4 mod 10; x 7 mod 15:

Chinese, Theorem, Remainder, Chinese remainder theorem

Section 4.3 - The Chinese Remainder Theorem

zimmer.csufresno.edu

Thus the remainder is 10 when 7×8×9×15×16×17×23×24×25×43 is divided by 11. Exercise 12: Use Fermat’s Little Theorem to ﬁnd the least positive residue of 2 10 6 modulo 7. Solution: Note that 10 6 = 6(166,666)+4.

Theorem, Remainder, Remainder theorem

The Remainder Theorem and Bounds Date Period

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Find the remainder when f (x) is divided by x k. 5) f ( x ) x x x x x x k

Theorem, Remainder, Remainder theorem