Special Product
MATH 102 – College Algebra
Chapter 2 SPECIAL PRODUCTS AND FACTORING
TYPES OF SPECIAL PRODUCTS Some products of polynomials can be solved without applying the distributive property. These cases of products of polynomials have been classified because of the special forms of the factors and can be obtained by applying the Special Products Formulas. Type 1: Product of Two Binomials
(ax + by) (cx + dy) = acx + (ad + bc)xy + bdy
2
2
The product of a binomial by another binomial is obtained as follows: • • • Examples: 1. (3x – 2y)(4x + 3y) 2. (m3 – 5) (m3 – 2) 3. (3a – 5x)(6b + 7y) = = = = = = 3(4)x2 + [(3)(3)+(-2)(4)]xy + (-2)(3)y2 12x2 + xy – 6y2 (m3m3) + (- 5 – 2)m3 + 10 m6 - 7m3 + 10 (3a)(6b) + (3a)(7y) + (-5x)(6b) + (-5x)(7y) 18ab + 21ay – 30bx – 35xy The first term of the product is the product of the first terms of the multiplicand and the multiplier; The second or the middle term of the product is obtained by taking the cross product or adding the product of the extreme terms to the product of the adjacent terms. The last term of the product is the product of the last terms of the multiplicand and the multiplier.
Type 2: Square of Binomials
(x + y) = x + 2xy + y 2 2 2 (x – y) = x – 2xy + y
2
2
2
•
The result of a square of a binomial is a perfect trinomial square which comprises the following terms: o square of the first term; o twice the product of the first term and the second term; o square of the second term.
Examples: 1. (2x – 3y)2 = = 4x2 + 2(2x)(-3y) + 9y2 4x2 -12xy + 9y2 3. (3ab – 4m)2 = (3ab)2+2(3ab)(-4m)+(-4m)2 = 9a2b2 – 24abm + 16m2 4. (4x – 3y + 2)2 = [(4x – 3y) + 2]2 = (4x– y)2+2(4x–3y)(2)+ 22 = [16x2+2(4x)(-3y)+(-3y)2] + 4(4x – 3y) + 4 = 16x2–24xy+9y2+1x–12y+4
2. (5x + 2)2
= 25x2 + 2(5x)(2) + 4 = 25x2 + 20x + 4
1
Prepared by Mrs. Koni Gutierrez Cruz Assistant Professor I
| Bataan Peninsula State University
MATH 102 – College Algebra
Type 3: Product of the Sum and Difference of the Same Two
Chapter 2 SPECIAL PRODUCTS AND FACTORING
TYPES OF SPECIAL PRODUCTS Some products of polynomials can be solved without applying the distributive property. These cases of products of polynomials have been classified because of the special forms of the factors and can be obtained by applying the Special Products Formulas. Type 1: Product of Two Binomials
(ax + by) (cx + dy) = acx + (ad + bc)xy + bdy
2
2
The product of a binomial by another binomial is obtained as follows: • • • Examples: 1. (3x – 2y)(4x + 3y) 2. (m3 – 5) (m3 – 2) 3. (3a – 5x)(6b + 7y) = = = = = = 3(4)x2 + [(3)(3)+(-2)(4)]xy + (-2)(3)y2 12x2 + xy – 6y2 (m3m3) + (- 5 – 2)m3 + 10 m6 - 7m3 + 10 (3a)(6b) + (3a)(7y) + (-5x)(6b) + (-5x)(7y) 18ab + 21ay – 30bx – 35xy The first term of the product is the product of the first terms of the multiplicand and the multiplier; The second or the middle term of the product is obtained by taking the cross product or adding the product of the extreme terms to the product of the adjacent terms. The last term of the product is the product of the last terms of the multiplicand and the multiplier.
Type 2: Square of Binomials
(x + y) = x + 2xy + y 2 2 2 (x – y) = x – 2xy + y
2
2
2
•
The result of a square of a binomial is a perfect trinomial square which comprises the following terms: o square of the first term; o twice the product of the first term and the second term; o square of the second term.
Examples: 1. (2x – 3y)2 = = 4x2 + 2(2x)(-3y) + 9y2 4x2 -12xy + 9y2 3. (3ab – 4m)2 = (3ab)2+2(3ab)(-4m)+(-4m)2 = 9a2b2 – 24abm + 16m2 4. (4x – 3y + 2)2 = [(4x – 3y) + 2]2 = (4x– y)2+2(4x–3y)(2)+ 22 = [16x2+2(4x)(-3y)+(-3y)2] + 4(4x – 3y) + 4 = 16x2–24xy+9y2+1x–12y+4
2. (5x + 2)2
= 25x2 + 2(5x)(2) + 4 = 25x2 + 20x + 4
1
Prepared by Mrs. Koni Gutierrez Cruz Assistant Professor I
| Bataan Peninsula State University
MATH 102 – College Algebra
Type 3: Product of the Sum and Difference of the Same Two