# 06 - Distribution

1. The distributive/distribution law can also be stated like this: a × (b + c) = a × b + b × a

False

True

2. The Boolean distributivity law is similar to distributivity in normal mathematics and has to do with expanding or simplifying ___________

full stops

AND gates

brackets

NOT gates

3. In practice, distribution is often carried out when either an 'AND' or an 'OR' is outside the brackets and the other is inside.

TRUE

FALSE

4. You can think of the distributive law as a law which deals with either 'multiplying out' or factoring. E.g in Maths: x(y+z) could give you:

xyz

xz+y

xy+xz

xy

5. Complete the following equation: e AND (f OR g) = (e AND f) OR (e AND g) and e OR (f AND g) = (e OR f) AND _________

eg+eg

(eg)

(e OR g)

(e AND g)

6. The distributive law would allow us to arrive at what output, from this expression: (x+y).(x+z)=

x.x+y

xyz

x+(yz)

xxyy

7. Using the distributive law, "simplify" the following expression: (A + B)(A + C)

A.1 + B.C

A.1 + B.C + C

A+B+C

A.A + A.C + A.B + B.C

8. Using the distributive law it is also possible to simplify (A + B).(A + C) to A(1 + B) + B.C

TRUE

FALSE

9. The expression (A + B)(A + C) can be simplified to ____________using the distributive law.

A(B.C)

A + (B.C)

A.B+C

ABC

10. Using the distributive law: X + Y Z = (X + Y) + (X. Z)

FALSE

TRUE

11. Using the distributive law: X. (Y + Z) = X Y + Z

TRUE

FALSE

12. Using the distributive law: X. (Y + Z) = X Y + X Z

TRUE

FALSE

13. Read the excerpt on removing factors below and fill in the blanks for the boolean equivalent
```Using the distributive law, you can also remove factors
(common variables), like in normal algebra.

For example: if you had (3 * 4) + (3 * 2), this could be
factorised into 3(4 + 2).

Note that both answers give 18.

The Boolean equivalent of this is:

______________________________________?```

(A AND B) OR (A AND C) ? A OR B OR C

(A AND B) OR (A AND C) ? A AND (B OR C)

(A AND B) OR (A AND C) ? ABC

(A AND B) OR (A AND C) ? A AND B

14. A(B + C) =

A.B + A.C

A.B. A.C

A+.B + A+C

A.BC

15. A + (B.C) =

(A + B).(A + C)

(A. B).(A.C)

(A. B)+(A.C)

(AB+AC)