The **distributive property of multiplication over addition** is one of the fundamental properties of algebra that shows how multiplication and addition interact. This property states:
**a(b + c) = ab + ac**
This means: when you multiply a number by a sum, you can multiply the number by each term in the sum separately and then add the results.
**Why This Matters:**
#### Case 1: When One Number in a Product is Increased by 1
**Problem:** Consider 23 × 27. By how much does the product increase if the first number (23) is increased by 1?
**Solution using the distributive property:**
If we increase 23 to 24 in the product 23 × 27:
Using a(b + c) = ab + ac:
23(27) + 1(27) = 23 × 27 + 27
The product increases by 27 (which is the second number).
**General Statement:**
a(b + 1) = ab + a
**The increase = a** (the first number)
Similarly, if we increase b to b + 1:
(a)(b + 1) = ab + a
The increase is a, the first number.
**Visualization using Arrays:**
When we increase b by 1, we add one more column of height a dots to the original array. This represents a × 1 = a additional dots.
#### Case 2: When One Number is Increased and the Other Decreased
**Problem:** What happens when 23 is increased by 1 and 27 is decreased by 1?
**Solution:**
(23 + 1)(27 - 1) = (23 + 1) × 27 - (23 + 1) × 1
= 23 × 27 + 1 × 27 - 23 - 1
= 23 × 27 + 27 - 23 - 1
**The change = 27 - 23 - 1 = 3**
In this case, we see that the change depends on which number is larger.
**General Statement for (a + 1)(b - 1):**
(a + 1)(b - 1) = ab + b - a - 1
**The change = b - a - 1**
**Important Observation:** The product does NOT always increase. For example:
#### Case 3: When Both Numbers are Increased by 1
**Problem:** What happens when both 23 and 27 are increased by 1?
**Solution:**
(23 + 1)(27 + 1) = (23 + 1) × 27 + (23 + 1) × 1
= 23 × 27 + 1 × 27 + 23 + 1
= 23 × 27 + 27 + 23 + 1
**The increase = 27 + 23 + 1 = 51**
**General Statement for (a + 1)(b + 1):**
(a + 1)(b + 1) = ab + a + b + 1
**The increase = a + b + 1** (the sum of both numbers plus 1)
**Verification:**
If a = 5, b = 7:
**When both numbers change by any amounts:**
If the initial numbers are a and b, and they become (a + m) and (b + n):
**(a + m)(b + n) = ab + an + bm + mn**
**Breaking this down:**
**Why this works:**
Consider the rectangular area interpretation: a rectangle with sides (a + m) and (b + n) can be divided into four smaller rectangles:
1. One with sides a and b (area = ab)
2. One with sides a and n (area = an)
3. One with sides m and b (area = bm)
4. One with sides m and n (area = mn)
The total area is ab + an + bm + mn.
**Proof using distributive property:**
(a + m)(b + n) = (a + m)b + (a + m)n
= ab + mb + an + mn
**Application of Identity 1:**
Example 1: Increase one number by 2 and decrease the other by 3.
Using m = 2, n = -3:
(a + 2)(b - 3) = (a + 2)(b + (-3))
= ab + a(-3) + b(2) + 2(-3)
= ab - 3a + 2b - 6
Example 2: Decrease both numbers, one by 3 and another by 4.
Using m = -3, n = -4:
(a - 3)(b - 4) = (a - 3)(b + (-4))
= ab + a(-4) + b(-3) + (-3)(-4)
= ab - 4a - 3b + 12
#### (a + u)(b - v) Form
Using Identity 1 with m = u and n = -v:
**(a + u)(b - v) = ab - av + ub - uv**
**Proof:**
(a + u)(b - v) = (a + u)b - (a + u)v
= ab + ub - av - uv
#### (a - u)(b + v) Form
Using Identity 1 with m = -u and n = v:
**(a - u)(b + v) = ab + av - ub - uv**
#### (a - u)(b - v) Form
Using Identity 1 with m = -u and n = -v:
**(a - u)(b - v) = ab - av - ub + uv**
**Remember:** The signs in the products follow the rules of integer multiplication:
**The distributive property also applies when brackets contain more than two terms.**
**Example 1:** Expand 3a/2 (a - b + 1/5)
Solution:
3a/2 (a - b + 1/5) = 3a/2 × a - 3a/2 × b + 3a/2 × 1/5
Simplifying each term:
Therefore:
3a/2 (a - b + 1/5) = 3a²/2 - 3ab/2 + 3a/10
**Cannot be simplified further** because no two terms are like terms (terms with exactly the same letter-numbers).
**Example 2:** Expand (a + b)(a + b)
Using the distributive property:
(a + b)(a + b) = (a + b)a + (a + b)b
= a² + ba + ab + b²
Combining like terms (ba and ab are like terms since ba = ab):
= a² + 2ab + b²
**Key Point:** Terms like 3a² and 5ab CANNOT be combined because they have different letter-numbers (a² vs ab).
**Like terms** are terms that have exactly the same letter-numbers (including their exponents).
Examples of like terms:
Examples of unlike terms:
**Only like terms can be combined** by adding or subtracting their coefficients.
**Example 3:** Expand (a + b)(a² + 2ab + b²)
Using the distributive property:
(a + b)(a² + 2ab + b²) = (a + b)a² + (a + b)(2ab) + (a + b)b²
Expanding each part:
= a × a² + b × a² + a × 2ab + b × 2ab + a × b² + b × b²
= a³ + a²b + 2a²b + 2ab² + ab² + b³
Combining like terms:
Final answer:
(a + b)(a² + 2ab + b²) = a³ + 3a²b + 3ab² + b³
The distributive property holds for all integers, including negative integers.
**Verification:** Check (a + 1)(b - 1) = ab + b - a - 1 with a = -5, b = 8:
An **algebraic identity** is a mathematical statement showing that two algebraic expressions are equal for all values of the variables.
**Examples of identities from this section:**
1. a(b + c) = ab + ac (Basic distributive property)
2. (a + m)(b + n) = ab + an + bm + mn (General identity for product change)
3. (a + 1)(b - 1) = ab + b - a - 1
4. (a + u)(b - v) = ab - av + ub - uv
**Key difference:** An identity is true for ALL values of variables, while an equation may be true for only specific values.
The distributive property can be used to develop quick mental multiplication tricks.
**Principle:** 11 = 10 + 1
For any number dcba (a 4-digit number where d, c, b, a are the digits):
dcba × 11 = dcba × (10 + 1)
= dcba × 10 + dcba
= (shift left by one place) + (original number)
**Digit-by-digit addition process:**
When we add:
```
d c b a 0
+ d c b a
-----------
```
Reading from right to left:
**Examples:**
Example 1: 3874 × 11
```
3 8 7 4 0
+ 3 8 7 4
-----------
4 2 6 1 4
```
Step-by-step:
Answer: 42614
Example 2: 94 × 11
```
9 4 0
+ 9 4
-------
1 0 3 4
```
Step-by-step:
Answer: 1034
**General Rule for multiplying by 11:**
1. Write the first digit (with any carry)
2. For each internal position, add the digit above it and the digit to its left
3. Write the last digit
**Practice problems:**
(i) 94 × 11 = 1034
(ii) 495 × 11 = 5445
(iii) 3279 × 11 = 36069
(iv) 4791256 × 11 = 52703816
**Principle:** 101 = 100 + 1
For a 4-digit number dcba:
dcba × 101 = dcba × (100 + 1)
= dcba × 100 + dcba
= (shift left by two places) + (original number)
```
d c b a 0 0
+ d c b a
---------------
d c (b+d) (a+c) b a
```
**Examples:**
Example 1: 3874 × 101
```
3 8 7 4 0 0
+ 3 8 7 4
---------------
3 9 0 5 7 4
```
Step-by-step:
Answer: 390574
Example 2: 89 × 101 = 8989
**General Rule for multiplying by 101:**
1. Write the rightmost two digits of the original number
2. Add the hundreds digit and thousands digit (if exists), write the result (with carry)
3. Continue adding pairs two places apart
4. Write the leftmost digit
**For 1001 = 1000 + 1:**
A 3-digit number abc × 1001 = abc × 1000 + abc
The addition:
```
a b c 0 0 0
+ a b c
-----------
a b (c+a) b c
```
**Example:** 265 × 1001
```
2 6 5 0 0 0
+ 2 6 5
-----------
2 6 7 6 5
```
Wait, let me recalculate: 5 + 2 = 7, so it's 265265? Let me verify with actual multiplication:
265 × 1001 = 265 × 1000 + 265 = 265000 + 265 = 265265
So the pattern shows digits are repeated with the sum of the overlapping digits.
**For 10001 = 10000 + 1:**
Similar principle applies, adding digits four places apart.
**Practice problems:**
(i) 89 × 101 = 8989
(ii) 949 × 101 = 95849
(iii) 265831 × 1001 = 265831265831
(iv) 1111 × 1001 = 1112111
**Principle:** 99 = 100 - 1 and 999 = 1000 - 1
For dcba × 99 = dcba × (100 - 1):
= dcba × 100 - dcba
= (shift left by two places) - (original number)
**Example:** 9734 × 99
First method:
9734 × 99 = 9734 × 100 - 9734
= 973400 - 9734
Subtraction:
```
973400
--------
963666
```
**Or:** We can think of 99 as (100 - 1) applied directly:
For a 4-digit number dcba:
99 × dcba = (100 - 1) × dcba
Using (a - 1) × b = ab - b:
= 100 × dcba - dcba
= dcba00 - dcba
**Quick trick:** In 99 × dcba, think of it as (dcba - 1) in the thousands-hundreds place and (100 - ba) in the tens-units place.
**Example:** 23478 × 999
= 23478 × (1000 - 1)
= 23478000 - 23478
= 23454522
**General approach for such fast multiplications:**
Always decompose the multiplier into a power of 10 plus or minus a small number, then use the distributive property to split the multiplication into simpler parts.
These fast multiplication methods were extensively discussed by:
Brahmagupta's Verse 12.55 states: "The multiplier is broken up into two or more parts whose sum is equal to it; the multiplicand is then multiplied by each of these and the results added."
---
#### Geometric Visualization
Consider a square with side length (a + b). We can divide this square into four regions:
1. A square of side a with area a²
2. A square of side b with area b²
3. Two rectangles of dimensions a × b, each with area ab
Total area:
(a + b)² = a² + b² + ab + ab = a² + 2ab + b²
**Real-life context:** Suppose a vegetable garden has been extended. The original square garden had side 60 m (area = 3600 m²). It was extended by 5 m on two adjacent sides. The extension includes:
#### Algebraic Derivation
Using the distributive property:
(a + b)² = (a + b)(a + b)
= (a + b) × a + (a + b) × b
= a² + ab + ab + b²
= a² + 2ab + b²
Or, applying the distributive property directly to each factor:
(a + b)(a + b) = a × a + a × b + b × a + b × b
= a² + ab + ab + b²
= a² + 2ab + b²
This is one of the most important algebraic identities and should be memorized.
**Components of the expansion:**
**Verification with numbers:**
(10 + 3)² = 13² = 169
Using identity: 10² + 2(10)(3) + 3² = 100 + 60 + 9 = 169 ✓
**Another example with variables:**
(x + y)² = x² + 2xy + y²
#### Example 1: Find (m + 3)²
(m + 3)² = m² + 2(m)(3) + 3²
= m² + 6m + 9
#### Example 2: Expand (6 + p)²
(6 + p)² = 6² + 2(6)(p) + p²
= 36 + 12p + p²
#### Example 3: Expand (6x + 5)²
**Method 1 - Using Identity:**
(6x + 5)² = (6x)² + 2(6x)(5) + 5²
= 36x² + 60x + 25
**Method 2 - Using Distributive Property:**
(6x + 5)² = (6x + 5)(6x + 5)
= (6x)(6x) + (6x)(5) + (5)(6x) + (5)(5)
= 36x² + 30x + 30x + 25
= 36x² + 60x + 25
Both methods give the same answer.
#### Example 4: Expand (3j + 2k)²
(3j + 2k)² = (3j)² + 2(3j)(2k) + (2k)²
= 9j² + 12jk + 4k²
**Is (a + b)² always greater than a² + b²?**
From the identity: (a + b)² = a² + 2ab + b²
Comparing with a² + b²:
(a + b)² - (a² + b²) = 2ab
**Conclusion:** (a + b)² is greater than a² + b² if and only if ab > 0 (both have same sign and are non-zero).
**Hint:** Decompose numbers into sums or differences of simpler numbers.
#### Example 1: Find 104²
104 = 100 + 4
104² = (100 + 4)²
= 100² + 2(100)(4) + 4²
= 10000 + 800 + 16
= 10816
#### Example 2: Find 37²
37 = 30 + 7
37² = (30 + 7)²
= 30² + 2(30)(7) + 7²
= 900 + 420 + 49
= 1369
Alternatively: 37 = 40 - 3 (might be harder, uses difference formula)
#### Geometric Visualization
Consider a square of side 60. We want the area of a smaller square of side (60 - 5) = 55 inside it.
Area of square with side 55:
= Area of big square - Area of two rectangles + Area of small square added back
When we remove two rectangles of dimensions 60 × 5 from the big square, we remove the small square (5 × 5) twice. So we add it back once:
(60 - 5)² = 60² - (60 × 5) - (5 × 60) + 5²
= 3600 - 300 - 300 + 25
= 3025
**Real-life context:** A square plot of land measures 80 m on each side. A square section measuring 6 m on each side is cut off from one corner. What is the remaining length of one side? (80 - 6)² = 74² = 5476 m² (if we're talking about a remaining square area, though the remaining shape wouldn't be square).
Better example: A square wall of side 50 m is reduced by 3 m on each side uniformly. The new square area = (50 - 3)² = 47² = 2209 m².
#### Algebraic Derivation
Using the distributive property:
(a - b)² = (a - b)(a - b)
= a(a - b) - b(a - b)
= a² - ab - ab + b²
= a² - 2ab + b²
Or, thinking of (a - b) as (a + (-b)):
(a + (-b))² = a² + 2(a)(-b) + (-b)²
= a² - 2ab + b²
This is another crucial algebraic identity to memorize.
**Components of the expansion:**
**Verification:**
(10 - 3)² = 7² = 49
Using identity: 10² - 2(10)(3) + 3² = 100 - 60 + 9 = 49 ✓
#### Example 1: Find 99²
99 = 100 - 1
99² = (100 - 1)²
= 100² - 2(100)(1) + 1²
= 10000 - 200 + 1
= 9801
#### Example 2: Find 58²
58 = 60 - 2
58² = (60 - 2)²
= 60² - 2(60)(2) + 2²
= 3600 - 240 + 4
= 3364
#### Example 3: Expand (b - 6)²
(b - 6)² = b² - 2(b)(6) + 6²
= b² - 12b + 36
#### Example 4: Expand (-2a + 3)²
Method 1 - Rewrite as (3 - 2a)²:
(3 - 2a)² = 3² - 2(3)(2a) + (2a)²
= 9 - 12a + 4a²
= 4a² - 12a + 9
Method 2 - Treat (-2a + 3) as (-2a + 3):
Using (a + b)² form with a = -2a and b = 3:
(-2a)² + 2(-2a)(3) + 3² = 4a² - 12a + 9
#### Example 5: Expand (7y - 3z/4)²
(7y - 3z/4)² = (7y)² - 2(7y)(3z/4) + (3z/4)²
= 49y² - (42yz)/4 + 9z²/16
= 49y² - (21yz)/2 + 9z²/16
**Observation:** These two expansions are very similar, differing only in the sign of the middle term.
**Important note:** Both formulas have a² and b² as positive terms. Only the middle term changes sign.
---
**Given patterns:**
**The pattern:** 2(a² + b²) = (a + b)² + (a - b)²
**Verification:**
Let's expand the right side:
(a + b)² + (a - b)² = (a² + 2ab + b²) + (a² - 2ab + b²)
= a² + a² + b² + b² + 2ab - 2ab
= 2a² + 2b²
= 2(a² + b²) ✓
This confirms the pattern!
**Finding examples:**
Take any two natural numbers. For example, a = 7, b = 4:
2(7² + 4²) = 2(49 + 16) = 2(65) = 130
(7 + 4)² + (7 - 4)² = 11² + 3² = 121 + 9 = 130 ✓
Another: a = 10, b = 6:
2(10² + 6²) = 2(100 + 36) = 2(136) = 272
(10 + 6)² + (10 - 6)² = 16² + 4² = 256 + 16 = 272 ✓
**Why this pattern matters:**
This identity shows that twice the sum of squares of any two numbers can always be written as the sum of two perfect squares. This has applications in number theory and geometry.
**Given patterns:**
**The algebraic pattern:** a² - b² = (a + b)(a - b)
Q1. What is the distributive property of multiplication over addition?
Answer: A — The distributive property states that a(b + c) = ab + ac, which allows us to multiply the number outside the bracket with each term inside.
Q2. By how much does the product ab increase when a is increased by 1?
Answer: B — Using the distributive property: a(b + 1) = ab + a. When we compare with the original product ab, the increase is a. Wait—the increase is a, so the answer should be A. Actually, when the first number a is increased by 1 to become (a+1), we get (a+1)b = ab + b, so the increase is b.
Q3. When both numbers a and b in the product ab are increased by 1, what is the increase in the product?
Answer: B — Expanding (a + 1)(b + 1) = ab + b + a + 1, so comparing with original product ab, the increase is a + b + 1.
Q4. A shopkeeper has a shop with dimensions 23 metres by 27 metres. If he increases both dimensions by 1 metre each, what will be the increase in the shop's area?
Answer: B — Using the identity (a + 1)(b + 1) = ab + a + b + 1, the increase is 23 + 27 + 1 = 51 square metres.
Q5. What is the expansion of (a + 2)(b + 3) using Identity 1: (a + m)(b + n) = ab + mb + an + mn?
Answer: A — Using Identity 1 with m = 2 and n = 3: (a + 2)(b + 3) = ab + 2b + a(3) + 2(3) = ab + 2b + 3a + 6 = ab + 3a + 2b + 6.
Q6. Which of the following is NOT a correct identity?
Answer: D — For (a - 1)(b - 1), we expand as ab - a - b + 1 (not ab - a - b - 1), because the product of the two negative terms (-1)(-1) = +1.
Q7. In a factory, a rectangular container has dimensions (x + 5) metres by (x + 3) metres. Which expression correctly represents its area?
Answer: A — Using (a + m)(b + n) = ab + mb + an + mn with a = x, m = 5, b = x, n = 3: (x + 5)(x + 3) = x² + 3x + 5x + 15 = x² + 8x + 15.
Q8. If one number is decreased by 2 and another number b is increased by 3, what is the increase in their product using Identity 1?
Answer: A — Using (a + m)(b + n) = ab + mb + an + mn with m = -2 and n = 3: the increase is mb + an + mn = (-2)b + a(3) + (-2)(3) = -2b + 3a - 6 = 3a - 2b - 6.
Q9. The distributive property works for which types of numbers?
Answer: C — The study material explicitly states that integers satisfy the distributive property, and this property holds when letter-numbers are replaced by any integers, including negative ones.
Q10. Two algebraic expressions are equal identities if and only if they:
Answer: B — An identity is defined as a mathematical statement where two algebraic expressions take on the same values for any replacement of their letter-numbers with numbers.
State the distributive property of multiplication over addition for three numbers a, b, and c.
a(b + c) = ab + ac, meaning the number outside the bracket multiplies with each term inside.
By how much does the product ab increase if b is increased by 1?
The product increases by a, because a(b + 1) = ab + a.
What is the expansion of (a + 1)(b + 1)?
(a + 1)(b + 1) = ab + a + b + 1, so the product increases by a + b + 1.
Write Identity 1: the expansion of (a + m)(b + n).
(a + m)(b + n) = ab + mb + an + mn, where each term of the first bracket multiplies each term of the second.
What is the increase in product (a + m)(b + n) compared to ab?
The increase is an + bm + mn, obtained from Identity 1 by subtracting ab.
Define like terms in algebra.
Terms are like terms if they have exactly the same letter-numbers; for example, 3ab and 5ab are like terms and combine to 8ab.
Expand 3a²(a − b + 1/5) using distributivity.
3a²(a − b + 1/5) = 3a³ − 3a²b + 3a/10 by multiplying 3a² with each term inside.
Expand (a − u)(b − v) using the distributive property.
(a − u)(b − v) = ab − ub − av + uv, where signs follow integer multiplication rules.
What is an algebraic identity?
An identity is a statement that two algebraic expressions are equal for all values of the variables, such as a(b + 8) = ab + 8a.
Expand (a + b)(a + b) and simplify.
(a + b)(a + b) = a² + 2ab + b² after combining like terms ba and ab into 2ab.
State the distributive property of multiplication over addition. [1 mark]
Write the property as a(b + c) = ... or use the phrase 'a single number multiplies each term inside the brackets'.
Expand 2a(3b − 5) and simplify. Show your working. [2 marks]
Apply distributivity: 2a multiplies 3b and −5 separately. Then simplify each term using rules of multiplication.
Using Identity 1, find (a + 3)(b − 2). Verify your answer by taking m = 3 and n = −2 in the formula (a + m)(b + n) = ab + mb + an + mn. [3 marks]
First substitute into Identity 1: ab + 3b + a(−2) + 3(−2). Then expand by distributivity and simplify with correct signs; verify both methods give the same result.
Expand (x + 2)(x² + 3x + 5) using the distributive property. Combine like terms and write the final simplified expression. Explain how you identified which terms were like terms. [5 marks]
Distribute (x + 2) to each term in the second bracket: x times each term, then 2 times each term. Identify like terms (same letter-numbers, such as x² terms, x terms) and add their coefficients. Write the expanded form in descending order of exponents.
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