
Polinomial II
Presentation
•
Mathematics
•
11th Grade
•
Medium
Christopher Delgado
Used 5+ times
FREE Resource
8 Slides • 19 Questions
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Multiple Choice
Nos dan dos funciones polinomiales P y D. Usa cualquier división sintética o larga para dividir P(x) entre D(x)
P(x)=3x2+5x−4
D(x)=x+3
3x−4+x+38
3x+14+x+338
3x−4−x+38
3x−4+x+338
6
Multiple Choice
Nos dan dos funciones polinomiales P y D. Usa cualquier división sintética o larga para dividir P(x) entre D(x)
P(x)=x3+4x2−6x+1
D(x)=x−1
x2+5x−1
x2+3x−9+x−110
x2+5x−1+x−110
x2+3x−9−x−110
7
Multiple Choice
Nos dan dos funciones polinomiales P y D. Usa cualquier división sintética o larga para dividir P(x) entre D(x)
P(x)=4x3+7x+9
D(x)=2x+1
2x2−x+4
2x2−x+4+2x+15
2x2+x+4+2x+113
2x2+x+4+2x+15
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9
Fill in the Blanks
Type answer...
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Fill in the Blanks
Type answer...
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Fill in the Blanks
Type answer...
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13
Multiple Choice
Determina si f(x)=x3+4x2+x−6 tiene por factor x+3
No es factor
Si es factor
14
15
Multiple Choice
Determina si f(x)=x3+3x2−4 tiene por factor x−1
Si es factor
No es factor
16
Multiple Choice
Determina si f(x)=x3+4x2+x−9 tiene por factor x+2
Si es factor
No es factor
17
Multiple Choice
Determina si f(x)=x3−5x2−x−1 tiene por factor x−3
Si es factor
No es factor
18
Multiple Choice
Determina si f(x)=x3+x2−16x−16 tiene por factor x−4
SÍ
19
Multiple Choice
Determina si f(x)=x2−3x2+2x+2 tiene por factor x−3
Sí
20
Multiple Choice
Determina si f(x)=x3−8x2+14x−4 tiene por factor x−2
Sí
No
21
Multiple Choice
Determina si (x+3) de un factor de f(x)=x3−6x+9
Sí
No
Hoy no, no lo creo
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23
Multiple Choice
Si x−4 es un factor de f(x)=x3−6x2+5x+12 . Factoriza completamente la función
(x-4)(x+1)(x-3)
(x-4)(x+1)(x+3)
(x-1)(x-4)(x-3)
(x+1)(x+4)(x+3)
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Multiple Choice
Si x−5 es un factor de P(x)=x3−2x2−13x−10 Factoriza completamente la función
(x−5)(x−1)(x−2)
(x−5)(x+1)(x+2)
(x−5)(x−1)(x+2)
(x−5)(x+1)(x−1)
25
Multiple Choice
Si (x−1) es un factor de f(x)=x3−7x2−x+7
Factoriza completamente la función
f(x)=
(x−1)(x−7)(x+1)
(x−1)(x+7)(x−1)
(x−1)(x−6)(x−1)
(x−1)(x−6)(x+1)
26
Multiple Choice
Factoriza f(x)=x3−3x2−4x+12 considerando que x=−2 es un cero.
f(x)=
(x−2)(x+2)(x+3)
(x−2)(x−2)(x+3)
(x+2)(x+2)(x+3)
(x−2)(x+2)(x−3)
27
Multiple Select
Si x=−3 es un cero de f(x) = x3−13x−12
Encuentra todos los ceros de f(x)
x=1
x=4
x=−1
x=−4
x=3
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