ACTIVITY 34

Review (Sections 3.6+3.7+4.1+4.2+4.3)

Problems 3 and 5:

Sketch the graph of the function by transforming an appropriate function of the form y = xn. Indicate all x- and y-intercepts on each graph.

3)2()( xxP3xy 32 xy 32 xyintercept-y

0x

3)20(0 P 8

8,0

intercept-x

0y3)2(0 x

3)2(0 x

)2(0 x 2x

0,2

82)( 4 xxP

y = 2x4

y = x4

y = 2x4 + 8

No x – interceptsy – intercepts is (0,8)

Problems 11, 13, and 15:

Match the polynomial function with the below graphs

)4()( 2 xxxP xxxxR 45)( 35 34 2)( xxxT

xxxxR 45)( 35 )4()( 2 xxxP34 2)( xxxT

Problems 21:

Sketch the graph of the polynomial function. Make sure your graph shows all intercepts and exhibits the proper end behavior.

2323)( xxxxP

szero' / intercepts-x

3x

2x

3

2x

intercept-y

0x 2032030)0( P

223 12

The leading term is 3x3

Problem 31:

Factor the polynomial P(x) = −x3 + x2 + 12x and use the factored form to find the zeros. Then sketch the graph.

xxxxP 12)( 23 122 xxx

34 xxx

0 x 04 x 03 x0x 4x 3x

Zeros

Problem 3:

Let P(x) = x4 − x3 + 4x + 2 and Q(x) = x2 + 3. (a)Divide P(x) by Q(x). (b)Express P(x) in the form P(x) = D(x) · Q(x) + R(x).

x4 − x3 + 4x + 2x2 + 3

x2

x4 + 3x2 –x4 – 3x2

− x3 – 3x2 + 4x + 2

– x

– x3 – 3x + x3 + 3x

– 3x2 + 7x + 2

)(xQ

)(xR

)()()()( xRxDxQxP 1173324 2234 xxxxxxx

– 3

– 3x2 – 9 +3x2 + 9

7x + 11

Problem 9:

Find the quotient and remainder using long division for the expression

22

362

3

xx

xx

x3 + 6x + 3x2 – 2x+2

x

x3 – 2x2 + 2x –x3 + 2x2 – 2x

2x2 + 4x + 3

+ 2

2x2 – 4x + 4– 2x2 + 4x – 4

8x – 1

)(xQ

)(xR

)()()()( xRxDxQxP

)(

)()()(

)(

)(

xD

xRxDxQ

xD

xP

)(

)(

)(

)()(

xD

xR

xD

xDxQ

)(

)()(

xD

xRxQ

22

182

22

3622

3

xx

xx

xx

xx

Problem 21:

Find the quotient and remainder using synthetic division for the expression

3

283

x

xx 13 0 8

1

3

3

9

1

)(xQ )(xR13)( 2 xxxQ

2

3

1

1)( xR

Problem 35:

Use synthetic division and the Remainder Theorem to evaluate P(c) if P(x) = 5x4 + 30x3 − 40x2 + 36x + 14 and c = −7.

57 30 40

5

35

5

35

5)(xQ )(xR

)()( cRcP

36

35

71

)7(P

14

497

483

483

Problem 43:

Use the Factor Theorem to show that x − 1 is a factor of P(x) = x3 − 3x2 + 3x − 1.

11 3 3

1

1

2

2

1

1

1

0

Showing that x = 1 is a zero.

Therefore, (x – 1) is a factor

Problem 53:

Find a polynomial of degree 3 that has zeros 1, −2, and 3, and in which the coefficient of x2 is 3.

1x 2x 3xa

3222 xxxxa

322 xxxa 6233 223 xxxxxa

652 23 xxxa

aaxaxax 652 23

32 a

2

3

a

1x 2x 3x2

3

Problem 3:

List all possible rational zeros given by the Rational Zeros Theorem (but don’t check to see which actually are zeros).

8432)( 235 xxxxR

q

p ,1

,1 2

,1possible rational zeros:

2 of divisors

8 of divisors

,2

1 ,2 ,4

,2 ,4 8

8

Problems 13 and 23:

Find all rational zeros of the polynomial.

23)( 3 xxxP

q

p ,11

,1possible rational zeros:

1 of divisors

2 of divisors

2

2

11 0 3

1

1

1

1

2

2

2

4

11 0 3

1

11

1

2

2

2

0

233 xx 22 xx )1(x

22 xx 1x

233 xx 22 xx 1x

22 xx 12 xx

112233 xxxxx

Consequently, the zero’s are x = 2 and x = – 1

8676)( 234 xxxxxP

q

p

1 of divisors

8 of divisors

,1

1

,2 ,4 8

possible rational zeros: ,1 ,2 ,4 8

8676)( 234 xxxxxP possible rational zeros: ,1 ,2 ,4 8

11 6 7

1

1

7

7

14

6

14

8

8

8

0

8676 234 xxxx 8147 23 xxx 1x

So we need only factor 8147 23 xxx

possible rational zeros: ,1 ,2 ,4 8possible rational zeros: ,1 ,2 ,4 8

8147 23 xxx

11 7 14

1

16

6

8

8

8

0

8676 234 xxxx 8147 23 xxx 1x

1)1(862 xxxx

11862 xxxxSo we need only factor 862 xx 14 xx

possible rational zeros: ,1 ,2 ,4 8

11248676 234 xxxxxxxx

Consequently, the roots are x = -4, x = -2, x = -1, and x = 1