Station 1 – Recognizing patterns in multiplying binomials and

Station 1 – Recognizing patterns in multiplying binomials and factoring trinomials
We have learned two special patterns for multiplying binomials. FOIL is not one of these
patterns.
1. What are the two special patterns for multiplying binomials? Name them and give
examples of each
These special patterns help us with their counterpart factoring patterns.
2. Name the two factoring patterns that partner with the multiplying patterns and give
examples of each.
Station 1 key
1. Squaring a binomial
Sum and Difference
(πŸπ’™ βˆ’ πŸ“)𝟐 = πŸ’π’™πŸ βˆ’ πŸπŸŽπ’™ + πŸπŸ“
(πŸ’π’™ + πŸ”)(πŸ’π’™ βˆ’ πŸ”) = πŸπŸ”π’™πŸ βˆ’ πŸ‘πŸ”
2. Perfect square trinomial
Difference of two squares
πŸπŸ“π’™πŸ βˆ’ πŸ‘πŸŽπ’™ + πŸ— = (πŸ“π’™ βˆ’ πŸ‘)𝟐
πŸ’πŸ—π’™πŸ βˆ’ 𝟏𝟐𝟏 = (πŸ•π’™ βˆ’ 𝟏𝟏)(πŸ•π’™ + 𝟏𝟏)
Station 2 – Simplifying and naming polynomials
Simplify and write in standard form, then name each based on degree and number of terms.
1. (πŸπŸ”π’„πŸ + πŸ’π’„) βˆ’ 𝟐(πŸ‘π’„ + πŸ’π’„πŸ ) + (𝒄 + 𝟐)(𝒄 βˆ’ πŸ’)
2. (πŸπŸ”π’†πŸ‘ + πŸπŸπ’†) βˆ’ (πŸ’π’†πŸ’ βˆ’ πŸπ’†πŸ‘ ) + 𝒆(𝟐 + 𝒆)
3. βˆ’πŸ‘π’˜πŸ (π’˜ + 𝟏𝟏) βˆ’ (πŸπ’˜ + πŸ‘)(πŸ‘π’˜ βˆ’ πŸ“)
4. πŸπŸ“π’… + πŸπŸ‘π’…πŸ βˆ’ πŸπŸ•π’… + πŸπŸŽπ’…πŸ βˆ’ (πŸπ’… + πŸ’π’…πŸ )
Station 2 key
1. πŸπŸ”π’„πŸ + πŸ’π’„ βˆ’ πŸ”π’„ βˆ’ πŸ–π’„πŸ + π’„πŸ βˆ’ πŸ’π’„ + πŸπ’„ βˆ’ πŸ– = πŸ—π’„πŸ βˆ’ πŸ’π’„ βˆ’ πŸ– ; 2nd degree trinomial
2. πŸπŸ”π’†πŸ‘ + πŸπŸπ’† βˆ’ πŸ’π’†πŸ’ + πŸπ’†πŸ‘ + πŸπ’† + π’†πŸ = βˆ’πŸ’π’†πŸ’ + πŸπŸ–π’†πŸ‘ + π’†πŸ + πŸπŸ’π’† ; 4th degree polynomial
3. βˆ’πŸ‘π’˜πŸ‘ βˆ’ πŸ‘πŸ‘π’˜πŸ βˆ’ (πŸ”π’˜πŸ βˆ’ πŸπŸŽπ’˜ + πŸ—π’˜ βˆ’ πŸπŸ“) = βˆ’πŸ‘π’˜πŸ‘ βˆ’ πŸ‘πŸ—π’˜πŸ + π’˜ + πŸπŸ“ ; 3rd degree poly.
4. πŸπŸ“π’… + πŸπŸ‘π’…πŸ βˆ’ πŸπŸ•π’… + πŸπŸŽπ’…πŸ βˆ’ (πŸπ’… + πŸ’π’…πŸ ) = πŸπŸ—π’…πŸ βˆ’ πŸ’π’… ; 2nd degree binomial
Station 3 – GCF Factoring
Factor the following using the GCF. Use standard form.
1. πŸπŸŽπ’„πŸ‘ + πŸ’π’„πŸ βˆ’ πŸ”π’„πŸ’
2. πŸπŸ’π’‚πŸ‘ π’ƒπŸ + πŸπŸ”π’‚πŸ’ π’ƒπŸ‘ + πŸ–π’‚πŸ π’ƒπŸ“
3. βˆ’πŸπŸŽπ’™πŸ’ + πŸπŸ“π’™πŸ” βˆ’ πŸ’πŸŽπ’™πŸ‘
4. πŸπŸπ’šπŸ βˆ’ πŸπŸ”π’™π’šπŸ‘ + πŸπŸŽπ’™πŸ π’šπŸ’
Station 3 key
1. πŸπ’„πŸ (βˆ’πŸ‘π’„πŸ + πŸ“π’„ + 𝟐) = βˆ’πŸπ’„πŸ (πŸ‘π’„πŸ βˆ’ πŸ“π’„ βˆ’ 𝟐)
2. πŸ–π’‚πŸ π’ƒπŸ (π’ƒπŸ‘ + πŸπ’‚πŸ 𝒃 + πŸ‘π’‚)
3. βˆ’πŸ“π’™πŸ‘ (βˆ’πŸ‘π’™πŸ‘ + πŸ’π’™ + πŸ–) = πŸ“π’™πŸ‘ (πŸ‘π’™πŸ‘ βˆ’ πŸ’π’™ βˆ’ πŸ–)
4. πŸπ’šπŸ (πŸ” βˆ’ πŸπŸ‘π’™π’š + πŸ“π’™πŸ π’šπŸ ) = πŸπ’šπŸ (πŸ“π’™πŸ π’šπŸ βˆ’ πŸπŸ‘π’™π’š + πŸ”)
Station 4 - Multiplying polynomials
Find the product:
1. (πŸ•π’‚ + πŸπ’ƒ)(βˆ’πŸ’π’‚ βˆ’ πŸ‘π’ƒ)
4. (πŸ“π’— + 𝟐)(πŸ“π’— βˆ’ 𝟐)
2. (πŸπŸπ’š βˆ’ πŸ‘)(πŸ“π’˜ + πŸ’)
5. (πŸπŸŽπ’‘ βˆ’ πŸ’)𝟐
3. (πŸ‘π’˜ βˆ’ πŸ’)(πŸ“π’˜πŸ βˆ’ πŸπ’˜ + 𝟏𝟎)
6. (πŸ—π’š βˆ’ πŸ“)(πŸπ’š βˆ’ πŸ•)
Station 4 key
1.
βˆ’πŸπŸ–π’‚πŸ βˆ’ πŸπŸ—π’‚π’ƒ βˆ’ πŸ”π’ƒπŸ
4.
πŸπŸ“π’—πŸ βˆ’ πŸ’
2.
πŸ”πŸŽπ’˜π’š + πŸ’πŸ–π’š βˆ’ πŸπŸ“π’˜ βˆ’ 𝟏𝟐
5.
πŸπŸŽπŸŽπ’‘πŸ βˆ’ πŸ–πŸŽπ’‘ + πŸπŸ”
3.
πŸπŸ“π’˜πŸ‘ βˆ’ πŸπŸ”π’˜πŸ + πŸ‘πŸ–π’˜ βˆ’ πŸ’πŸŽ
6.
πŸπŸ–π’šπŸ βˆ’ πŸ“πŸ‘π’š βˆ’ πŸ‘πŸ“
Station 5 – Applications of perimeter and area
For the rectangle below, write the simplified expressions that represent the perimeter and area
of the rectangle.
4π‘₯ βˆ’ 1
2π‘₯ + 3
Station 5 key
Perimeter:
𝟐(πŸπ’™ + πŸ‘) + 𝟐(πŸ’π’™ βˆ’ 𝟏) = πŸ’π’™ + πŸ” + πŸ–π’™ βˆ’ 𝟐 = πŸπŸπ’™ + πŸ’
Area:
(πŸπ’™ + πŸ‘)(πŸ’π’™ βˆ’ 𝟏) = πŸ–π’™πŸ βˆ’ πŸπ’™ + πŸπŸπ’™ βˆ’ πŸ‘ = πŸ–π’™πŸ + πŸπŸŽπ’™ βˆ’ πŸ‘
Station 6 – Factoring
Factor the following trinomials:
1. π’˜πŸ βˆ’ πŸπŸ“π’˜ + πŸ’πŸ’
5. πŸπŸπ’™πŸ βˆ’ πŸ•π’™ βˆ’ 𝟏𝟎
2. π’‘πŸ + πŸπŸ‘π’‘ + 𝟐𝟐
6. πŸπ’™πŸ βˆ’ πŸπŸ—π’™ + πŸ‘πŸ—
3. π’“πŸ + πŸπŸ“π’“ + πŸ“πŸŽ
7. πŸ‘π’™πŸ βˆ’ πŸ‘π’™ βˆ’ πŸ”
4. π’’πŸ βˆ’ πŸπŸ“π’’ βˆ’ πŸ‘πŸ’
8. πŸ“π’™πŸ βˆ’ πŸ”π’™ + 𝟏
Station 6 key
1. (π’˜ βˆ’ 𝟏𝟏)(π’˜ βˆ’ πŸ’)
5. (πŸ’π’™ βˆ’ πŸ“)(πŸ‘π’™ + 𝟐)
2. (𝒑 + 𝟐𝟐)(𝒑 + 𝟏)
6. (πŸπ’™ βˆ’ πŸπŸ‘)(𝒙 βˆ’ πŸ‘)
3. (𝒓 + 𝟏𝟎)(𝒓 + πŸ“)
7. (πŸ‘π’™ + πŸ‘)(𝒙 βˆ’ 𝟐)
4. (𝒒 βˆ’ πŸπŸ•)(𝒒 + 𝟐)
8. (πŸ“π’™ βˆ’ 𝟏)(𝒙 βˆ’ 𝟏)
Station 7 - Dividing Polynomials
Using long division, divide 2x3 – 9x2 + 15 by 2x – 5.
Show all work and express any remainder as a fraction.
Station 7 key
Remember that since there is no β€œx term” you must fill in with Ox
so the answer is
Station 8 – Finding area using polynomial rules
Write a simplified expression to represent the shaded area below:
πŸ‘π’™ + πŸ“
πŸ’π’™ βˆ’ 𝟐
πŸπ’™ + 𝟏
Station 8 key
Area of rectangle
(πŸ‘π’™ + πŸ“)(πŸ’π’™ βˆ’ 𝟐) = πŸπŸπ’™πŸ βˆ’ πŸ”π’™ + πŸπŸŽπ’™ βˆ’ 𝟏𝟎 = πŸπŸπ’™πŸ + πŸπŸ’π’™ βˆ’ 𝟏𝟎
Area of triangle
(πŸ’π’™ βˆ’ 𝟐)(πŸπ’™ + 𝟏) πŸ–π’™πŸ βˆ’ πŸ’π’™ + πŸ’π’™ βˆ’ 𝟐 πŸ–π’™πŸ βˆ’ 𝟐
=
=
= πŸ’π’™πŸ βˆ’ 𝟏
𝟐
𝟐
𝟐
Now subtract and simplify
(πŸπŸπ’™πŸ + πŸπŸ’π’™ βˆ’ 𝟏𝟎) βˆ’ (πŸ’π’™πŸ βˆ’ 𝟏) = πŸ–π’™πŸ + πŸπŸ’π’™ βˆ’ πŸ—