Match Equivalent Expressions: (a±b)(c±4) Expansions

Join expressions of equal value

  1. (a−b)(c−4) (a-b)(c-4)

  2. (a+b)(c+4) (a+b)(c+4)

  3. (a−b)(c+4) (a-b)(c+4)

  4. (a+b)(c−4) (a+b)(c-4)

    a.ac−4a+bc−4b ac-4a+bc-4b

    b.ac+4a−bc−4b ac+4a-bc-4b

    c.ac−4a−bc+4b ac-4a-bc+4b

    d.ac+4a+bc+4b ac+4a+bc+4b

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:00 Open brackets
00:03 Open brackets properly, multiply each factor by each factor
00:14 Calculate the multiplications, and group the factors
00:21 This is the simplification for 1, let's continue to 2
00:27 Open brackets properly, multiply each factor by each factor
00:32 Calculate the multiplications, and group the factors
00:38 This is the simplification for 2, let's continue to 3
00:42 Open brackets properly, multiply each factor by each factor
00:50 Calculate the multiplications, and group the factors
00:53 This is the simplification for 3, let's continue to 4
01:00 Open brackets properly, multiply each factor by each factor
01:03 Calculate the multiplications, and group the factors
01:10 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

Join expressions of equal value

  1. (a−b)(c−4) (a-b)(c-4)

  2. (a+b)(c+4) (a+b)(c+4)

  3. (a−b)(c+4) (a-b)(c+4)

  4. (a+b)(c−4) (a+b)(c-4)

    a.ac−4a+bc−4b ac-4a+bc-4b

    b.ac+4a−bc−4b ac+4a-bc-4b

    c.ac−4a−bc+4b ac-4a-bc+4b

    d.ac+4a+bc+4b ac+4a+bc+4b

2

Step-by-step solution

We use all the exercises of the extended distributive property:(a+b)×(c+d)=ac+ad+bc+bd (a+b)\times(c+d)=ac+ad+bc+bd

1.(a−b)(c−4)=ac−4a−bc+4b (a-b)(c-4)=ac-4a-bc+4b

2.(a+b)(c+4)=ac+4a+bc+4b (a+b)(c+4)=ac+4a+bc+4b

3.(a−b)(c+4)=ac+4a−bc−4b (a-b)(c+4)=ac+4a-bc-4b

4.(a+b)(c−4)=ac−4a+bc−4b (a+b)(c-4)=ac-4a+bc-4b

3

Final Answer

1-c, 2-d, 3-b, 4-a

Practice Quiz

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\( (3+20)\times(12+4)= \)

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