Lesson Plans: Introduction Lesson Plan 1: The X Factor - Trinomials and Algebra Tiles Lesson Plan 2: Curses and Re-Curses! It's Happening Again.

Lesson Plan 1: The X Factor - Trinomials and Algebra Tiles

Supplies:

Teachers will need the following:
Students will need the following:
• A set of algebra tiles - click here for a printable page
• Graphing calculator
Steps

Introductory Activity:

1. Ask a student to define the word "factor." Elicit that "factor," when used as a verb, means to "to rewrite a number or expression as a product of two or more numbers or expressions."

2. Have students, in pairs or small groups, talk about ways to factor 30. Students should come up with four ways: 1 x 30, 2 x 15, 3 x 10, and 5 x 6. Note that students might also include methods that break 30 into more than 2 factors, such as 5 x 3 x 2 or 10 x 3 x 1.

3. Explain that the factors of 30 are "numerical factors" and that during today's lesson, students will look at algebraic factors.

Learning Activities:

1. Give some examples of algebraic expressions that can be written in factored form:
• 3x2 - 6x = 3x (x - 2)
• a2 + 7a + 12 = (a + 4)(a + 3)
2. Use the distributive property to multiply the factors and obtain a product, and then have students verify that both of these equations are true.

3. Reinforce the definition of factoring by asking, "Which side is in factored form?" Students should conclude that the right side is in factored form, because factoring means to rewrite an expression as a product or as a multiplication problem.

4. In groups, have students use algebra tiles to write the first trinomial on Factoring Gift in factored form. That is, explain that you would like them to find two binomials that, when multiplied, give this trinomial: x2 + 6x + 8. Reinforce that the algebra tiles must be arranged to form a rectangle with no gaps. It's important to note that students may arrange the tiles in any formation that results in a rectangle (for example: below, left). However, students must be able to identify the side length of the rectangle, which may be easier if the rectangle is arranged in a systematic manner (below, right). Students should understand that they can find the area of the rectangle by using the area formula, namely A = lw = (x + 4)(x + 2), or by adding the individual pieces inside. These pieces are
x2 + x + x + x + x + x + x + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = x2+ 6x + 8.

Because the two expressions measure the same area, they must be equivalent.

5. Once students have used the tiles to find the factored form (x + 2)(x + 4), have them verify the product using the distributive property.

6. Have students graph the trinomial x2 + 6x + 8 and its factored form, (x + 2)(x + 4), on the graphing calculator. Students should see that the graphs are identical.

7. Ask students to identify the values of x at which the graph crosses the x-intercept. Using the TRACE feature, students should identify (-2, 0) and (-4, 0) as the x-intercepts.

8. Give students 30 seconds to determine where the y-intercept occurs. Using the TRACE feature, students should identify (0, 8) as the y-intercept.

9. Have students use the same process for the second and third trinomials on the Factoring Gift handout: x2 + 7x + 6 and x2 + 8x + 12.
• Factor the trinomial using algebra tiles, and write it in factored form.
• Verify the product with the distributive property.
• Identify the x-intercepts.
• Identify the y-intercept.
On the chalkboard or overhead projector, create a table that lists all of the information that has so far been obtained:

 Polynomial Factored Form x-intercepts y-intercept x2 + 6x + 8 (x + 2)(x + 4) (-2, 0) and (-4, 0) (0, 8) x2 + 7x + 6 (x + 1)(x + 6) (-1, 0) and (-6, 0) (0, 6) x2 + 8x + 12 (x + 2)(x + 6) (-2, 0) and (-6, 0) (0, 12)

10. Give students one minute to discuss, in pairs or small groups, any patterns or relationships that exist in the table. Encourage them to look for several different patterns, and explain that there are many to find. Students should identify the following relationships:

• The y coordinate of the y-intercept is equal to the constant term in the trinomial; that is, if the polynomial is x2 + bx + c, the y-intercept occurs at (0, c).
• The x coordinates of the x-intercepts are equal to the opposite of the constant terms when the trinomial is written in factored form; that is, if the polynomial can be expressed as (x + m)(x + n), the x-intercepts occur at (-m, 0) and (-n, 0) because x + m = 0 or x + n = 0.
• If a trinomial x2 + bx + c can be written as (x + m)(x + n), then b = m + n and c = m x n.
Note: The final pattern listed above, that b = m + n and c = m x n, is one of the keystones of this lesson. Students must realize that this relationship always holds, and that it is the key to factoring trinomials. If students have not identified this relationship by this point of the lesson, have them continue to factor trinomials from the Factoring Gift handout until they identify this pattern.

11. Using the patterns identified in the table, students should factor the fourth trinomial from the Factoring Gift handout, x2 + 7x + 10, without algebra tiles or the graphing calculator. At this point, students should have discovered that they must find two numbers for which the product is 10 and the sum is 7, resulting in (x + 2)(x + 5).

12. Ask students to factor x2 + 4x + 6. To factor this trinomial, students must identify two numbers that have a product of 6 and a sum of 4. Because no real numbers exist for which this is true, students should conclude that this trinomial cannot be factored. Define such a polynomial as a "prime trinomial."

13. Have students graph x2 + 4x + 6 on graphing calculator. To make sure students understand this visual representation, which shows why this expression can't be factored, point out or elicit that the graph does not cross the x-axis, so it has no x-intercepts and consequently no real factors.

Culminating Activity/Assessment:

In their math journals, have students identify all positive values of C such that x2 + 6x + C is a factorable trinomial. Students should identify at least three values for C, namely 5, 8, and 9. Students may also notice that C = 0 is a solution, but it produces a slightly different form than this lesson addresses.
• 9 x2 + 6x + 9 = (x + 3)(x + 3)
• 8 x2 + 6x + 8 = (x + 2) (x + 4)
• 5 x2 + 6x + 5 = (x + 1)(x + 5)
• 0 x2 + 6x = x (x + 6)