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### 1.1 ss factoring the difference of two squares

• 1. FACTORS OF POLYNOMIALS Factoring the Difference of Two Squares
• 2. Jessebel G. Bautista Antonio J. Villegas Voc’l High School My Profile
• 3. OBJECTIVES • Identify the perfect squares • Factor the difference of two squares • Factor with accuracy
• 4. Factoring a polynomial involves writing it as a product of two or more polynomials. It reverses the process of polynomial multiplication. Every polynomial that is a difference of squares can be factored by applying the following formula: a2 – b2 = (a + b) (a – b)
• 5. Every polynomial that is a difference of squares can be factored by applying the following formula: a2 – b2 = (a + b) (a – b) Note that a and b in the pattern can be any algebraic expression. For example, for a = x , and b=2 we get the following: x2 – 22 = (x – 2) (x + 2) The polynomial x2 – 4 is now expressed in factored form, (x+2)(x−2). We can expand the right-hand side of this equation to justify the factorization: (x + 2) (x – 2) = x(x – 2) + 2(x – 2) = x2 – 2x + 2x – 4 = x2 – 4 Now that we understand the pattern, let's use it to factor a few more polynomials.
• 6. Both x2 and 16 are perfect square since x2 = (x)2 and 16 = (4)2. In other words. x2 – 16 = (x)2 – (4)2 EXAMPLE 1: Factoring x2 - 16 Since the two squares are being subtracted, we can see that this polynomial represents a difference of squares. We can use the difference of squares pattern to factor this expression: a2 – b2 = (a + b) (a – b) In our case, a = x and b = 4. Therefore, our polynomial factors as follows: (x)2 – (4)2 = (x – 4) (x + 4) We can check our work by ensuring the product of these two factors is x2 – 16 .
• 7. Both 9x2 and 100 are perfect square since 9x2 = (3x)2 and 100 = (10)2. In other words. 9x2 – 100 = (3x)2 – (10)2 EXAMPLE 2: Factoring 9x2 - 100 Since the two squares are being subtracted, we can see that this polynomial represents a difference of squares. We can use the difference of squares pattern to factor this expression: a2 – b2 = (a + b) (a – b) In our case, a = 3x and b = 10. Therefore, our polynomial factors as follows: (3x)2 – (10)2 = (3x – 10) (3x + 10) We can check our work by ensuring the product of these two factors is 9x2 – 100 .
• 8. Both 36m2 and 49n4 are perfect square since 36m2 = (6m)2 and 49n4 = (7n2)2. In other words. 36m2 – 49n4 = (6m)2 – (7n2)2 EXAMPLE 3: Factoring 36m2 – 49n4 Since the two squares are being subtracted, we can see that this polynomial represents a difference of squares. We can use the difference of squares pattern to factor this expression: a2 – b2 = (a + b) (a – b) In our case, a = 6m and b = 7n2. Therefore, our polynomial factors as follows: (6m)2 – (7n2)2 = (6m – 7n2) (6m + 7n2) We can check our work by ensuring the product of these two factors is 36m2 – 49n4 .
• 9. 1. 4x2 – 25 2. 9y2 – 100 3. 49m2 – 121n2 4. 144 – 16n2 5. m2n2 – 196 TO DO … Factor the following: 6. 169a2 – b2c2 7. 9x2 – 81 8. 289m2 – 256n2 9. 16b2 – 9c2 10. 121c4 – 169d2
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