8-5 Factoring Special Products 1. A rectangular fountain has an area of 16 x 2 8x 1 f t 2. The dimensions of the rectangle have the form ax b, where a and b are whole numbers. Write an expression for the perimeter of the fountain. Then find the perimeter when x 2 feet. 16x 4; 36 feet 3. The floor plan of a daycare center is shown. • Factor polynomials of the form x2 + bx + c by grouping. Topic 3: Factoring Trinomials by Grouping 2 Learning Objectives • Factor polynomials of the form ax2 + bx + c by grouping. Lesson 2: Factoring Special Products of Polynomials Topic 1: Factoring Special Products Learning Objectives • Identify and factor special products of binomials. Special products of polynomials In the previous section we showed you how to multiply binominals. There are a couple of special instances where there are easier ways to find the product of two binominals than multiplying each term in the first binomial with all terms in the second binomial.
1.The product of two consecutive integersis 272. Find the value of each integer.
The first thing you need to do is to define the integers.
Let n = the first integer Letn+1 = the 2nd integer
The product means to multiply so we need to multiply the twointegers together.
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(First integer)*(Second integer) = 272
(n)*(n + 1) = 272
Now multiply everything out and set it equal to zero.
n2 + n = 272
n2 + n -272 = 272 – 272
n2 + n -272 = 0
Now you need to factor and solve.
(n + 17)(n – 16) = 0
n + 17 = 0 or n – 16 = 0
n = -17 or n = 16
Now you need to go back and answer the questions using each answer.
If n = -17 then the 2nd integer is n + 1 or -17 + 1 = -16 sothe two integers are -17 and -16
If n = 16 then the 2nd integer is n + 1 or 16 + 1 = 17 so thetwo integers are 16 and 17.
Remember thatnegative answers are not always bad.
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2.The product of two consecutive evenintegers is 528. Find the value of each integer.
We are going to follow the same steps as in #1.
Let n = the first integer Letn+2 = the 2nd integer
(First integer)*(Second integer) = 528
(n)*(n + 2) = 528
n2 + 2n = 528
n2 + 2n -528 = 528 – 528
n2 + 2n -528 = 0
(n + 24)(n – 22) = 0
n + 24 = 0 or n – 22 = 0
n = -24 or n = 22
If n = -24 then the 2nd integer is n + 2 or -24 + 2 = -22 sothe two integers are -24 and -22
If n = 22 then the 2nd integer is n + 2 or 22 + 2 = 24 so thetwo integers are 22 and 24.
3. A rectangular swimming pool is twice as long as it iswide. A small concrete walkway surrounds the pool. The walkway is aconstant 2 feet wide and has an area of 196 square feet. Find thedimensions of the pool.
The first thing you need to do is to define yourvariables. In this example, the length is described in terms of the width sodefine the width.
Let w = the width of the pool
The problem says that the length is twice as long asthe width and remember that twice means to multiply by 2.
Let 2w = the length of the pool
This means that the area of the pool is A = (w)(2w)
The pool has a concrete walkway of 2 feet surroundingit so this means that there is 2 feet on EACH SIDE of the width and the lengthso we need to increase BOTH the length and width by 4 in order to get the lengthand width of the pool and the walkway.
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This means that the width is w + 4 and the length is 2w+ 4. This means that the walkway and pool area is: A = (w + 4)(2w + 4)
We are also given that the area of the walkway and thepool is 196 square feet.
In this case we have two areas. We have the area of thepool and the area of the walkway and pool (Seepicture).
In order to find the dimensions we need to use the twodifferent areas.
Area Total = Area Walkwayand Pool – Area Pool
196 = (w + 4)(2w + 4) -(w)(2w)
196 = (2w2 + 4w + 8w + 16) – 2w2
196 = 2w2 + 4w +8w + 16 – 2w2
196 = 12w+ 16
196 -16 = 12w+ 16- 16
180 = 12w
12 12
15 = w
So now we know that the width of the pool is 15 ft. Butwe aren’t done yet. The question asked for the dimensions so we still need tofind the length. The length is twice the width so 2 * 15 = 30ft.
The dimensions of the pool are 15 ft by 30 ft.
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