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Unit 3 Project

Simplify each expression.

1.

(3b)5(8c)3

(3b)5(8c)3 = 243b5512c3

= 124416b5c3

2.

(7x - 2)(x + 4)

(7x - 2)(x + 4) = 7x2 – 2x + 28x – 8

= 7x2 + 26x - 8

3.

(6x2 - 10x + 4) - (x2 + 12x - 7)

(6x2 - 10x + 4) - (x2 + 12x - 7) = 5x2 + 2x + 11

4.

(2x3 + 9x2 - 2x + 7) ÷ (x + 2)

(2x3 + 9x2 - 2x + 7) ÷ (x + 2) =

-2

2

↓

2

9 -2 7

-4 -10 24

5 -12 31

Then,

𝟐𝒙𝟐 + 𝟓𝒙 − 𝟏𝟐 +

𝟑𝟏

𝒙+𝟐

Factor completely. If the polynomial is not factorable, write prime.

3x3y + x2y2 + x2y

5.

3x3y + x2y2 + x2y = xy( 3x2+ xy + x)

8r3 - 64s6

6.

8r3 - 64s6 = 8(r3 - 8s6)

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 x-coordinate at which the relative maxima

and relative minima occur.

g(x) = x3 + 6x2 + 6x - 4

7.

a.

x

g(x)

0

-4

1

9

2

40

-1

-5

-2

0

-3

5

-4

4

-5

-9

b. Consecutive integer values of x are: From -6 to -3. From -3 to

0. From 0 to 2.

c. Relative Maxima: x≈-3

Relative Minima: Near to 0

h(x) = x4 - 2x3 - 6x2 + 8x + 5

8.

a.

x

g(x)

-3

62

-2

-3

-1

-6

0

5

1

6

2

-3

3

2

4

69

b. Consecutive integer values of x are: From -3 to -2. From -1 to

0. From 1 to 2. From 2 to 3

c. Relative Maxima: x≈0.5

Relative Minima: x≈-1.5 and x≈2.5

Solve each equation.

9.

p3 + 8p2 = 18p

𝑝3 + 8𝑝2 − 18𝑝 = 0

𝒑(𝒑𝟐 + 𝟖𝒑 − 𝟏𝟖) = 𝟎

Using the zero factor principle:

𝑥=

−𝑏 ± √𝑏 2 − 4𝑎𝑐

2𝑎

For a = 1

b=8

c = -18

−8 + √82 − 4 ∗ 1 ∗ −18

𝑥=

2∗1

𝒙 = √𝟑𝟒 − 𝟒

−8 − √82 − 4 ∗ 1 ∗ −18

𝑥=

2∗1

𝒙 = −√𝟑𝟒 − 𝟒

10.

16x4 - x2 = 0

16x4 - x2 = 0

16x2(x2 - 1) = 0

16x2(x - 1)(x + 1) = 0

So, X1 = X2 = 0

X3 = 1

X4 = -1

Use synthetic substitution to find f(-2) and f(3).

11.

f(x) = 3x4 - 12x3 - 12x2 + 30x

For f(-2)

3(−2)4 − 12(−2)3 − 12(−2)2 + 30(−2) = 36

For f(3)

3(3)4 − 12(3)3 − 12(3)2 + 30(3) = 369

12.

Write the polynomial equation of degree 4...