Multiplication of Two or more Binomials

Multiplication of Two or more Monomials	Multiplication of Two or more Binomials	Multiplication of Two or more Trinomials	Multiplication of Monomial & Binomial	Multiplication of Monomial & Trinomial
Multiplication of Binomial & Trinomial

Before you understand this concept, you are advised to read: What are Binomials ? What are Like Terms ? How to Subtract Like Terms ? What are Terms of Polynomial ? How to add Like Terms ? Let's understand how to multiply two or more binomials with the help of following examples: Example 1: Multiply (a + 5) and (b + 6) Solution: In the given question we have two binomials: First Binomial = (a + 5) Second Binomial = (b + 6) Write in the multiplication expression and we get: (a + 5) (b + 6) Use Distributive Law and multiply each term of first binomial with every term of second binomial & we get: = a(b + 6) + 5(b + 6) = (a X b) + (a X 6) + (5 X b) + (5 X 6) = ab + a6 + 5b + 30 = ab + 6a + 5b + 30 (because a6 = 6a) Rearrange the terms and we get: = 6a + 5b + ab + 30 Hence, (a + 5) (b + 6) = 6a + 5b + ab + 30 Example 2: Multiply (x - 5) and (y - 2) Solution: In the given question we have two binomials: First Binomial = (x - 5) Second Binomial = (y - 2) Write in the multiplication expression and we get: (x - 5) (y - 2) Use Distributive Law and multiply each term of first binomial with every term of second binomial & we get: = x(y - 2) - 5(y - 2) = (x X y) - (x X 2) - (5 X y) - (5 x -2) = xy - 2x - 5y - (-10) = xy - 2x - 5y + 10 = -2x - 5y + xy + 10 Hence, (x - 3) (y - 2) = (-2x - 5y + xy + 10) Example 3: Multiply (p + 4) and (q - 3) Solution: In the given question we have two binomials: First Binomial = (p + 4) Second Binomial = (q - 3) Write in the multiplication expression and we get: (p + 4) (q - 3) Use Distributive Law and multiply each term of first binomial with every term of second binomial & we get: = p(q - 3) + 4(q - 3) = (p X q) - (p X 3) + (4 X q) + (4 X -3) = pq - 3p + 4q + (-12) = pq - 3p + 4q - 12 = - 3p + 4q + pq - 12 Hence, (p + 4) (q - 3) = (- 3p + 4q + pq - 12) Example 4: Multiply (a + b) and (2a + 3b) Solution: In the given question we have two binomials: First Binomial = (a + b) Second Binomial = (2a + 3b) Write in the multiplication expression and we get: (a + b) (2a + 3b) Use Distributive Law and multiply each term of first binomial with every term of second binomial & we get: = a(2a + 3b) + b(2a + 3b) = (a X 2a) + (a X 3b) + (b X 2a) + (b X 3b) = 2a2 + 3ab + 2ab + 3b2 Since 3ab and 2ab are like terms, so we add them and we get: = 2a2 + 5ab + 3b2 Or we can write it as : = 2a2 + 3b2 + 5ab Hence, (a + b) (2a + 3b) = (2a2 + 3b2 + 5ab) Example 5: Multiply (5 + 2x), (3 - x) and (2 + 2x) Solution: In the given question, we have three binomials: First Binomial = (5 + 2x) Second Binomial = (3 - x) Third Binomial = (2 + 2x) Write in the multiplication expression and we get: (5 + 2x) (3 - x) (2 + 2x) Use Distributive Law and multiply each term of first binomial with every term of second binomial and keep third binomial as such & this is done in the following steps: = [ 5(3 - x) + 2x(3 - x) ] (2 + 2x) = [ (5 X 3) - (5 X x) + (2x X 3) + (2x X -x) ] (2 + 2x) = [ 15 - 5x + 6x + (-2x2) ] (2 + 2x) = [ 15 - 5x + 6x - 2x2 ] (2 + 2x) Add Like terms (-5x + 6x = x), so we get: = [15 + x - 2x2] (2 + 2x) Now again use Distributive Law and multiply each term of polynomial [15 + x - 2x2] with every term of third binomial (2 + 2x) & this is done in the following steps: = 15(2 + 2x) + x(2 + 2x) - 2x2(2 + 2x) = (15 X 2) + (15 X 2x) + (x X 2) + (x X 2x) - (2x2 X 2) - (2x2 X 2x) = 30 + 30x + 2x + 2x2 - 4x2 - 4x3 Rearrange the terms and we get: = -4x3 + 2x2 - 4x2 + 30x + 2x + 30 Add Like terms (2x2 - 4x2 = -2x2) & (30x + 2x = 32x), so we get: = - 4x3 - 2x2 + 32x + 30 Hence, (5 + 2x) (3 - x) (2 + 2x) = (- 4x2 - 2x2 + 32x + 30)