resol 2015 fun 8oano ativ supl alg l2

Álgebra

Resoluções das atividades
Ca

pít ulo

6

Polinômios

01 a) 2x + 4y

b) 0,25a + 0,05b + 0,10c

02 a) 16 · 6 + 11 · 1,8 + 10 · 4 = 96 + 19,8 + 40 = 155,8

O gasto da turma foi de R$ 155,80.

b) 6x + 1,8y + 4z

03 a) 3x3 – 16y – 7x3 + 8y + 9x3 + 4y = 3x3 – 7x3 + 9x3 – 16y + 8y + 4y = 5x – 4y
3

1 2 7 2 8 2 7 2 10 x 2 − 35 x 2 − 48 xy 2 + 35 xy 2 x − x − xy + xy =
=
b)
3
6
5
6
30
−25 x 2 − 13 xy 2
25
13 2
5
13 2
= − x2 − xy = − x 2 − xy 30
30
30
6
30

1
2
c) ⋅ ( − 4 x 3 + 6 x − 5) + x 3 − 3 x + 4 =
4
5
4 x 3 6 x 5 2x 3
−
+
− +
− 3x + 4 =
4
4 4
5
3
−20 x + 30 x − 25 + 8 x 3 − 60 x + 80
=
20
−12x 3 − 30 x + 55
=
20
12
30
55
− x3 −
=
x+
20
20
20
3
11
3
− x3 − x +
5
2
4
07 a) Somente o polinômio A. b) Somente o polinômio C. c) B é de grau 3. D é de grau 2 + 1 = 3. E é de grau 1 + 3 + 1 = 5.
08 a) ( y 3 + 3y 2 + 6y + 8) + ( y 3 − 3y 2 − 6y − 2) =
2

y 3 +3y + 6y + 8 + y 3 − 3y 2 − 6y − 2 =
2y 3 + 6

7
1
abc − abc − 2ab + ab
20
4
7abc − 20abc 8ab + ab
P=
−
20
4
13
9
P = − abc − ab
20
4

04 a) P =

b)
( 5 x 4 − x 2 + 9) − (2x 4 − x 3 + 2x 2 − x ) =
5 x 4 − x 2 + 9 − 2x 4 + x 3 − 2x 2 + x =
3x4 + x3 − 3x2 + x + 9 c)
( x 2 + x + 2) − (2 + x + x 2 ) = x 2 + x +2 −2 − x − x 2 = 0
2
( x + x + 2) − (2 + x
+ x ) = x 2 + x +2 −2 − x − x 2 = 0 0
2

b) P é um binômio.

Desafio x 9 → x = 9q1 + 5
5 qq11

x 3 → x = 3q2 + 2
2 q22

Logo: 9q1 + 5 = 3q2 + 2 → 9q1 – 3q2 = –3 → 3q1 – q2 = –1

3q 1 −–qq2==−1–1
3q
Como q1 + q2 = 9, resolvendo o sistema 
,
q1 ++qq2==9 9
q
1

1

2

2
 2 1
 1 d) x + x + 2 −  x 2 − x +  =

 3
4
5
1
1
2
x2 + x + 2 − x2 + x − =
4
3
5
8
2 2 5 x + x+
3
4
5
09 a) (A · B) + H – C = (–2 · x2) + 6 – (x2 – 1) = –2x2 + 6 – x2 + 1 = –3x2 + 7

b) (F – D) – G + D · B + A = [x – (–1)] – (1 – x2) + (–1) · x2 + (–2) =

2

encontram-se q1 = 2 e q2 = 7m, portanto x = 23.

x +1 −1 − 2 = x−2 2

7
7
14 · 3
05 a − 14 = 0 → a = 14 → 7a = 14 · 3 → a =
→a= 6
3
3
7

06 a) 2