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Published by krujum16kb, 2021-09-12 09:46:50

ร10-64

ร10-64

p(x) an xn + an−1xn−1 + ... + a1x + a0 n

an , an−1,.., a1, , a0 an  0

p(x) p(x)x − c x − c p(c) =p0(c) = 0

x−2 x2 + x −6

p(x) = x2 + x − 6

p(2) = (2)2 + 2 − 6
= 4+2−6

=0 x2 + x −6

x−2

2 x3 + 2x2 − 5x − 6

p(x) = x3 + 2x2 − 5x − 6

− 6 1,2,3,6
p(1)

p(1) = (1)3 + 2(1)2 − 5(1) − 6

= 1+ 2(1) − 5(1) − 6

=1+ 2 − 5 − 6
= −8

p(1)  0

x −1 p(x)

2

p(−1)
p(−1) = (−1)3 + 2(−1)2 − 5(−1) − 6

= −1+ 2(1) − 5(−1) − 6

= −1+ 2 + 5 − 6 x3 + 2x2 − 5x − 6
=0
x +1

2

x +1 x3 + 2x2 − 5x − 6

−−11 1 −1 2 6−5 − 6 6
−−−111 −−−−111 −6166 0

111 1111 −−16−66 −0060

x2 + x −6
xx33 ++22xx22 −−55xx−−66 = (x +1)(x2 + x − 6)

= (x +1)(x + 3)(x − 2)

3 x3 − x2 −14x + 24

p(x) = x3 − x2 −14x + 24

24 1,2,3,4,6,8,12,24

p(1), p(−1)

p(2)

p(2) = 23 − 22 −14(2) + 24

= 8 − 4 − 28 + 24

=0

x−2 x3 − x2 −14x + 24

3 x3 − x2 −14x + 24

x−2

22 12 −−124 −14 24
222 222 −−−222444 − 24

11 111 −1−1212 −1200 0

x2 + x −12

xx33 −− xx22 −14xx++2244==((xx−−22)()x(x2 2++x −x 1−21)2)

= (x − 2)(x − 3)(x + 4)

1. x −1 x3 − 2x2 − 5x + 6

2. x +1 x3 + x2 + x +1

3. x3 − x2 − 4x + 4

4. x4 − 5x2 + 4


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