Polynomial Functions, algebra homework help

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Unit 3 Project Simplify each expression. 1. 2. 3. 4. (3b)5(8c)3 (7x ­ 2)(x + 4) (6x2 ­ 10x + 4) ­ (x2 + 12x ­ 7) (2x3 + 9x2 ­ 2x + 7) ÷ (x + 2) Factor completely. If the polynomial is not factorable, write prime. 5. 6. 3x3y + x2y2 + x2y 8r3 ­ 64s6 For each function, complete each of the following. a. Graph each function by making a table of values. b. Determine consecutive integer values of x between which each real zero is located. c. Estimate the x­coordinate at which the relative maxima and relative minima occur. 7. 8. g(x) = x3 + 6x2 + 6x ­ 4 h(x) = x4 ­ 2x3 ­ 6x2 + 8x + 5 Solve each equation. 9. 10. p3 + 8p2 = 18p 16x4 ­ x2 = 0 Use synthetic substitution to find f(­2) and f(3). 11. f(x) = 3x4 ­ 12x3 ­ 12x2 + 30x 12. Write the polynomial equation of degree 4 with leading coefficient 1 that has roots at ­2, ­1, 3, and 4. State the possible number of positive real zeros, negative real zeros, and imaginary zeros for each function. 13. 14. f(x) = ­x3 ­ x2 + 14x ­ 24 f(x) = 2x3 ­ x2 + 16x ­ 5 Find all rational zeros of each function. 15. 16. h(x) = x4 + 2x3 ­ 23x2 + 2x ­ 24 f(x) = 5x3 ­ 29x2 + 55x ­ 28 Determine whether each pair of functions are inverse functions. 17. 18. f(x) = 4x ­ 9, g(x) = x 4− 9 f(x) = x 1+ 2 , g(x) = x1 ­ 2 If f(x) = 2x ­ 4 and g(x) = x2 + 3, find each value. 19. 20. (f ­ g)(x) (f • g)(x)
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1. (3𝑏)5 (8𝑐)3 = 35 × (𝑏)5 × 83 ×(𝑐)3 = 243 × 𝑏 5 × 512 × 𝑐 3 = 124416 𝒃𝟓 𝒄𝟑
2. (7𝑥 − 2)(𝑥 + 4) = 7𝑥 2 + 28𝑥 − 2𝑥 − 8 = 𝟕𝒙𝟐 + 𝟐𝟔𝒙 − 𝟖
3. (6𝑥 2 − 10𝑥 + 4) − ( 𝑥 2 + 12𝑥 − 7) = 6𝑥 2 − 10𝑥 + 4 − 𝑥 2 − 12𝑥 + 7 =
𝟓𝒙𝟐 − 𝟐𝟐𝒙 + 𝟏𝟏
4. (2𝑥 3 + 9𝑥 2 − 2𝑥 + 7) ÷ (x+2) =𝟐𝒙𝟐 + 𝟓𝒙 − 𝟏𝟐 +

𝟑𝟏
𝒙+𝟐

5. 3𝑥 3 𝑦 + 𝑥 2 𝑦 2 + 𝑥 2 𝑦 = 𝒙𝟐 𝒚( 𝟑𝒙 + 𝒚 + 𝟏)
6. 8𝑟 3 -64 𝑠 6 = 8(𝑟 3 − 8𝑠 6 )= 8 (r-2𝒔𝟐 ) (𝒓𝟐 + 𝟐𝒓𝒔𝟐 + 𝟒𝒔𝟒 )

7.
ax
g(x)

-5
-9

-4
4

-3
5

-2
0

-1
-5

0
-4

1
9

2
40

3
95

70
60
50
40
30
20
10
0
-4

-3

-2

-1

0

1

2

3

4

-10
-20

b- consecutive integer values of x between which each real zero is located:
The integers are -5 and -4, also 0 and 1 also -2
c- The x- coordinate at which the relative maxima and relative minima occur.
We differentiate the function: 𝑔′ = 3𝑥 2 + 12𝑥 + 6, we then find the zeros of
the derivative. We find the relative maxima occur at x= -3.414 and the relative
minima occur when x= -0.586

8. ax
g(x)

-3
62

-2
-3

-1
-6

0
5

1
6

2
-3

3
2

70
60
50
40
30
20
10
0
-4

-3

-2

-1

0

1

2

3

4

-10
-20

b- consecutive integer values of x between which each real zero is located:
The integers are -3 and -2, also -1 and 0 also 1 and 2 also 2 and 3.
c- The x- coordinate at which the relative maxima and relative minima occur.
We differentiate the function: ℎ′ = 4𝑥 3 − 6𝑥 2 − 12𝑥 + 8, we then find the
zeros of the derivative. We find the relative minima occur at x= -1.469 and
x=2.4 and the relative maxima occur when x= 0.567

9. 𝑝3 + 8𝑝2 = 18𝑝
𝑝3 + 8𝑝2 − 18𝑝 = 0
𝑝(𝑝2 + 8𝑝 − 18) = 0
Either p=0 or from quadratic general law𝑏 =
Then the solution set is: {𝟎, 𝟏. 𝟖𝟑𝟏, −𝟗. 𝟖𝟑}
10.16𝑥 4 -𝑥 2 =0
𝑥 2 (16𝑥 2 -1) =0
𝑥 2 (4𝑥-1) (4𝑥+1) = 0

−𝑏±√𝑏2 −4𝑎𝑐
2𝑎

so p= 1.831 and p= -9.83

Then 𝑥 2 = 0

𝑠𝑜 𝑥 = 0

Or (4𝑥-1) = 0

so x=

Or (4𝑥+1) = 0

so x= −

1
4

𝟏

𝟏

𝟒

𝟒

1
4

The solution set is: {0, ,− }

11.
𝑓(−2) = 3 × (−2)4 − 12 × (−2)3 − 12 × (−2)2 + 30 × (−2) = 𝟑𝟔
𝑓(3) = 3 × (3)4 − 12 × (3)3 − 12 × (3)2 + 30 × (3) = −𝟗𝟗
12. 𝑓(𝑥) = 1 × (𝑥 + 2)(𝑥 + 1)(𝑥 − 3)(𝑥 − 4) = 𝒙𝟒 − 𝟒𝒙𝟑 − 𝟕𝒙𝟐 + 𝟐𝟐𝒙 + 𝟐𝟒

13.𝑓(𝑥) = −𝑥 3 − 𝑥 2 + 14𝑥 − 24 ...


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