Question\#1. For each area model below, fill in any missing boxes, then write the area as both a product and sum.
Product and sum: $\qquad$ Product and sum: $\qquad$

Studdy Solution

STEP 1

1. The area models are used to represent polynomial multiplication.
2. Each box in the area model represents a term in the expanded form of the product of two binomials.
3. The values given in the area models can be used to find the missing values through multiplication and addition.

STEP 2

1. Fill in the missing boxes for the first area model.
2. Write the product and sum for the first area model.
3. Fill in the missing boxes for the second area model.
4. Write the product and sum for the second area model.

STEP 3

Find the missing value in the top right box of the first area model.
Since the product of the left values and the right values should be equal, we have: $6y = -7 \cdot \text{(missing value)}$ Solving for the missing value: $\text{Missing value} = \frac{6y}{-7} = -\frac{6y}{7}$

STEP 4

Find the missing value in the bottom right box of the first area model.
Similar to the previous step: $-1 = 4y \cdot \text{(missing value)}$ Solving for the missing value: $\text{Missing value} = \frac{-1}{4y} = -\frac{1}{4y}$

STEP 5

Write the product for the first area model.
The original expressions being multiplied are: $(-7 + \text{top right box}) \cdot (4y + \text{bottom right box})$ Replacing the missing values: $(-7 - \frac{6y}{7}) \cdot (4y - \frac{1}{4y})$

STEP 6

Write the sum for the first area model.
The sum of the terms in the area model is: $-7 \cdot 4y + (-7 \cdot -\frac{1}{4y}) + (-\frac{6y}{7} \cdot 4y) + (-\frac{6y}{7} \cdot -\frac{1}{4y})$ Simplifying further: $-28y + \frac{7}{4y} - \frac{24y^2}{7} + \frac{6}{28y}$

STEP 7

Find the missing value in the top right box of the second area model.
Since the product of the left values and the right values should be equal, we have: $2x = 5 \cdot \text{(missing value)}$ Solving for the missing value: $\text{Missing value} = \frac{2x}{5} = \frac{2x}{5}$

STEP 8

Find the missing value in the bottom right box of the second area model.
Similar to the previous step: $-3 = x \cdot \text{(missing value)}$ Solving for the missing value: $\text{Missing value} = \frac{-3}{x} = -\frac{3}{x}$

STEP 9

Write the product for the second area model.
The original expressions being multiplied are: $(5 + \text{top right box}) \cdot (x + \text{bottom right box})$ Replacing the missing values: $(5 + \frac{2x}{5}) \cdot (x - \frac{3}{x})$

STEP 10

Write the sum for the second area model.
The sum of the terms in the area model is: $5 \cdot x + 5 \cdot -\frac{3}{x} + \frac{2x}{5} \cdot x + \frac{2x}{5} \cdot -\frac{3}{x}$ Simplifying further: $5x - \frac{15}{x} + \frac{2x^2}{5} - \frac{6}{5}$
Solution for the First Area Model: Product: $(-7 - \frac{6y}{7}) \cdot (4y - \frac{1}{4y})$ Sum: $-28y + \frac{7}{4y} - \frac{24y^2}{7} + \frac{6}{28y}$
Solution for the Second Area Model: Product: $(5 + \frac{2x}{5}) \cdot (x - \frac{3}{x})$ Sum: $5x - \frac{15}{x} + \frac{2x^2}{5} - \frac{6}{5}$

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