Exercise 59 Page 272

Practice makes perfect

a A rectangle's area is the product of its two sides. By breaking down the known areas into factors, where one factor matches a known side of the smaller rectangles, we can determine the second side.

- 6x &=- 3(2x) - 20xy &=- 4y(5x) - 15 &=- 3(5) We now have expressions for the unknown sides of the smaller rectangles.

Now we can find expressions for the unknown rectangle's areas.

We will begin by writing the area as a sum. This can be done by adding the smaller rectangle's areas. A=10x^2-20xy+25x-6x+12y-15 ⇕ A=10x^2-20xy+19x+12y-15 To write the area as a product, we add the lengths along the larger rectangle's sides, and multiply these sums. A=(5x+(-3))(2x+(- 4y)+5) ⇕ A=(5x-3)(2x-4y+5)

b Like in Part A, we know that the smaller rectangle's areas are a product of their sides. However, for the bottom right rectangle, we know its area, 12, but not its sides. Therefore we must choose its sides arbitrarily, as long as their product equals 12. We will choose 3 and 4 units.

By multiplying the rectangle's sides, we can find expressions for the unknown rectangle's areas.

Now we can write the area as a sum by adding the smaller rectangle's areas. A=x^2+4x+3x+12 ⇔ A=x^2+7x+12 To write the area as a product we add the lengths along the larger rectangle's sides, and multiply these sums. A=(x+3)(x+4)