Evaluating Expressions w/ the Distributive Law MPM2D: Principles of - PDF document

p o l y n o m i a l s p o l y n o m i a l s Evaluating Expressions w/ the Distributive Law MPM2D: Principles of Mathematics Consider the expression 4 × 7. Multiplying, 4 × 7 = 28. Now consider the expression 4(2 + 5). Evaluating inside of the brackets first gives Distributive Law 4(2 + 5) = 4 × 7 = 28 as before. Products of Two Binomials It is also possible (but not recommended) to use the Distributive Law to evaluate the expression. J. Garvin 4(2 + 5) = 4 · 2 + 4 · 5 = 8 + 20 = 28 J. Garvin — Distributive Law Slide 1/15 Slide 2/15 p o l y n o m i a l s p o l y n o m i a l s Evaluating Expressions w/ the Distributive Law Evaluating Expressions w/ the Distributive Law Now consider the expression (1 + 3)(2 + 5). Example Use the Distributive Law to evaluate (2 + 6)(4 + 1), and Evaluating inside of the brackets, as we should, gives (1 + 3)(2 + 5) = 4 · 7 = 28 as expected. verify the solution by evaluating within the brackets first. How could the Distributive Law be used here? Multiply each term in the first pair of brackets by each term To evaluate the expression using the Distributive Law, each in the second. term in the first pair of brackets is multiplied by each term in the second pair. (2 + 6)(4 + 1) = 2 · 4 + 2 · 1 + 6 · 4 + 6 · 1 = 8 + 2 + 24 + 6 (1 + 3)(2 + 5) = 1 · 2 + 1 · 5 + 3 · 2 + 3 · 5 = 40 = 2 + 5 + 6 + 15 Evaluating inside of the brackets first, = 28 (2 + 6)(4 + 1) = 8 · 5 = 40. J. Garvin — Distributive Law J. Garvin — Distributive Law Slide 3/15 Slide 4/15 p o l y n o m i a l s p o l y n o m i a l s Products of Two Binomials Products of Two Binomials Clearly, this is more work than necessary for evaluating an Using the Distributive Law as before, multiply each term in expression, so when is this useful? the first pair of brackets by each term in the second. Consider the expression ( x + 1)( x + 3). ( x + 1)( x + 3) = x · x + 3 · x + 1 · x + 1 · 3 Within the first pair of brackets, x and 1 are unlike terms, so = x 2 + 3 x + x + 3 they cannot be simplified. = x 2 + 4 x + 3 The same is true of x and 3 in the second pair of brackets. The new expression is a quadratic expression, and has the Both x + 1 and x + 3 are binomials – they are polynomials general form ax 2 + bx + c for some real values a , b amd c . that contain 2 terms. In this case, the quadratic expression we obtained is a The Distributive Law allows us to rewrite the product of two trinomial , since it contains three terms. binomials as a single expression instead. J. Garvin — Distributive Law J. Garvin — Distributive Law Slide 5/15 Slide 6/15

p o l y n o m i a l s p o l y n o m i a l s Products of Two Binomials Products of Two Binomials Example Example Expand and simplify ( x + 2)( x + 5). Expand and simplify ( x − 3)( x − 6). ( x − 3)( x − 6) = x 2 − 6 x − 3 x + 18 ( x + 2)( x + 5) = x · x + 5 · x + 2 · x + 2 · 5 = x 2 + 5 x + 2 x + 10 = x 2 − 9 x + 18 = x 2 + 7 x + 10 In this example, the sum of two negative numbers is negative, while the product of two negative numbers is positive. Note that all signs are positive in the final expression, since both the sum and the product of two positive numbers are positive. J. Garvin — Distributive Law J. Garvin — Distributive Law Slide 7/15 Slide 8/15 p o l y n o m i a l s p o l y n o m i a l s Products of Two Binomials Products of Two Binomials Example Example Expand and simplify ( x + 7)( x − 5). Expand and simplify ( x − 3) 2 . Using the definition of exponentiation, a 2 = a · a . Therefore, ( x − 3) 2 = ( x − 3)( x − 3). ( x + 7)( x − 5) = x 2 − 5 x + 7 x − 35 = x 2 + 2 x − 35 ( x − 3)( x − 3) = x 2 − 3 x − 3 x + 9 = x 2 − 6 x + 9 In this example, the product of a positive number and a negative number is negative. Note that ( x − 3) 2 � = x 2 − 9, since the middle term is absent. The sum of the two numbers (one positive, one negative) is This is a common error. positive, because the positive number has a greater magnitude. J. Garvin — Distributive Law J. Garvin — Distributive Law Slide 9/15 Slide 10/15 p o l y n o m i a l s p o l y n o m i a l s Products of Two Binomials Products of Two Binomials Example Example Expand and simplify (2 x + 1)(3 x − 4). Expand and simplify (3 x − 2)(5 x − 1). Even with coefficients, follow the same procedure as before. (3 x − 2)(5 x − 1) = 15 x 2 − 3 x − 10 x + 2 (2 x + 1)(3 x − 4) = 2 x · 3 x + 2 x · ( − 4) + 3 x · 1 + 1 · ( − 4) = 6 x 2 − 8 x + 3 x − 4 = 15 x 2 − 13 x + 2 = 6 x 2 − 5 x − 4 J. Garvin — Distributive Law J. Garvin — Distributive Law Slide 11/15 Slide 12/15

p o l y n o m i a l s p o l y n o m i a l s Products of Two Binomials Products of Two Binomials Example Example Expand and simplify (2 a + b )( a − 3 b ). Expand and simplify (4 p + 5 q )(2 p + 3 q ). Even with multiple variables, follow the same procedure. (4 p + 5 q )(2 p + 3 q ) = 8 p 2 + 12 pq + 10 pq + 15 q 2 (2 a + b )( a − 3 b ) = 2 a · a + 2 a · ( − 3 b ) + b · a + b · ( − 3 b ) = 2 a 2 − 6 ab + ab − 3 b 2 = 8 p 2 + 22 pq + 15 q 2 = 2 a 2 − 5 ab − 3 b 2 Note that the middle term of the trinomial involves both a and b , whereas the other terms only contain one variable. J. Garvin — Distributive Law J. Garvin — Distributive Law Slide 13/15 Slide 14/15 p o l y n o m i a l s Questions? J. Garvin — Distributive Law Slide 15/15