(x + a)(x + b) Identity
The identity (x + a)(x + b) = x² + (a + b)x + ab is the fourth standard algebraic identity taught in Class 8. It is also called Identity IV. This identity gives the expanded form of the product of two binomials that share the same variable x but have different constant terms a and b.
Unlike identities I, II, and III where both binomials are closely related (like a + b and a + b, or a + b and a - b), this identity is more general. The values a and b can be any numbers — positive, negative, or zero. In fact, identities I, II, and III are all special cases of this identity.
This identity is particularly useful for factorising quadratic expressions of the form x² + px + q, where you need to find two numbers whose sum is p and whose product is q. This skill becomes essential in Class 9 and 10 when solving quadratic equations.
What is (x + a)(x + b) Identity?
Definition: The (x + a)(x + b) identity states:
(x + a)(x + b) = x² + (a + b)x + ab
In words: the product of two binomials (x + a) and (x + b) equals x-squared plus the sum of a and b times x plus the product of a and b.
Special cases:
- When a = b: (x + a)(x + a) = x² + 2ax + a² = (x + a)². This is Identity I.
- When b = -a: (x + a)(x - a) = x² + (a - a)x + a(-a) = x² - a². This is Identity III.
- When a and b are both negative: (x - p)(x - q) = x² + (-p - q)x + pq = x² - (p + q)x + pq.
(x + a)(x + b) Identity Formula
The Identity:
(x + a)(x + b) = x² + (a + b)x + ab
Three parts of the expansion:
- First term: x² — the square of the common variable
- Middle term: (a + b)x — the sum of the constants times x
- Last term: ab — the product of the two constants
Reverse application (Factorisation):
To factorise x² + px + q:
- Find two numbers a and b such that a + b = p and ab = q.
- Then x² + px + q = (x + a)(x + b).
Derivation and Proof
Algebraic Derivation:
Expand (x + a)(x + b) using the distributive property:
- (x + a)(x + b) = x(x + b) + a(x + b)
- = x² + xb + ax + ab
- = x² + (a + b)x + ab
The two middle terms (ax and bx) combine since they both have x as a factor.
Verification with numbers:
Let x = 5, a = 3, b = 4:
- LHS: (5 + 3)(5 + 4) = 8 x 9 = 72
- RHS: 5² + (3 + 4) x 5 + 3 x 4 = 25 + 35 + 12 = 72
- LHS = RHS. Verified.
Verification with negative values:
Let x = 10, a = -3, b = 7:
- LHS: (10 - 3)(10 + 7) = 7 x 17 = 119
- RHS: 10² + (-3 + 7) x 10 + (-3)(7) = 100 + 40 - 21 = 119
- LHS = RHS. Verified.
Types and Properties
Problems using the (x + a)(x + b) identity fall into these categories:
1. Expansion of binomial products:
- Expand expressions like (x + 3)(x + 7), (y - 2)(y + 8), (m - 5)(m - 9).
- Apply the identity directly: x² + (a + b)x + ab.
2. Numerical computation:
- Find products like 104 x 106, 97 x 103, or 55 x 57 by writing them as (100 + 4)(100 + 6), (100 - 3)(100 + 3), or (56 - 1)(56 + 1).
3. Factorisation of quadratic trinomials:
- Factorise expressions like x² + 9x + 20 by finding two numbers with sum 9 and product 20 (answer: 4 and 5, so the factorisation is (x + 4)(x + 5)).
4. Mixed sign problems:
- When one or both constants are negative: (x - 3)(x + 5) = x² + 2x - 15.
- The middle term coefficient is a + b (which can be positive, negative, or zero).
- The last term is ab (positive if a and b have the same sign, negative if opposite signs).
5. Comparison with other identities:
- Recognise when Identity IV reduces to Identity I (a = b), Identity II (b = -a), or Identity III (b = -a).
Solved Examples
Example 1: Example 1: Both constants positive
Problem: Expand (x + 4)(x + 9).
Solution:
Using the identity with a = 4, b = 9:
- x² + (4 + 9)x + (4)(9)
- = x² + 13x + 36
Answer: (x + 4)(x + 9) = x² + 13x + 36.
Example 2: Example 2: One constant negative
Problem: Expand (x + 6)(x - 2).
Solution:
Using the identity with a = 6, b = -2:
- x² + (6 + (-2))x + (6)(-2)
- = x² + 4x - 12
Answer: (x + 6)(x - 2) = x² + 4x - 12.
Example 3: Example 3: Both constants negative
Problem: Expand (x - 3)(x - 8).
Solution:
Using the identity with a = -3, b = -8:
- x² + (-3 + (-8))x + (-3)(-8)
- = x² - 11x + 24
Answer: (x - 3)(x - 8) = x² - 11x + 24.
Example 4: Example 4: Factorisation — finding a and b
Problem: Factorise x² + 11x + 28.
Solution:
Find two numbers whose sum = 11 and product = 28.
- Try: 4 + 7 = 11 and 4 x 7 = 28. Both conditions satisfied!
Therefore: x² + 11x + 28 = (x + 4)(x + 7).
Answer: x² + 11x + 28 = (x + 4)(x + 7).
Example 5: Example 5: Factorisation with negative middle term
Problem: Factorise x² - 9x + 18.
Solution:
Find two numbers whose sum = -9 and product = 18.
- Since the product is positive and the sum is negative, both numbers must be negative.
- Try: (-3) + (-6) = -9 and (-3)(-6) = 18. Both match!
Therefore: x² - 9x + 18 = (x - 3)(x - 6).
Answer: x² - 9x + 18 = (x - 3)(x - 6).
Example 6: Example 6: Factorisation with negative last term
Problem: Factorise x² + 3x - 28.
Solution:
Find two numbers whose sum = 3 and product = -28.
- Since the product is negative, one number is positive and one is negative.
- Try: 7 + (-4) = 3 and 7 x (-4) = -28. Both match!
Therefore: x² + 3x - 28 = (x + 7)(x - 4).
Answer: x² + 3x - 28 = (x + 7)(x - 4).
Example 7: Example 7: Numerical computation
Problem: Find 106 x 108 using the identity.
Solution:
Write 106 = 100 + 6 and 108 = 100 + 8.
(100 + 6)(100 + 8) = 100² + (6 + 8) x 100 + 6 x 8
= 10000 + 1400 + 48
= 11448
Answer: 106 x 108 = 11,448.
Example 8: Example 8: Numerical computation with negative
Problem: Find 95 x 103 using the identity.
Solution:
Write 95 = 100 - 5 and 103 = 100 + 3.
(100 - 5)(100 + 3) = 100² + (-5 + 3) x 100 + (-5)(3)
= 10000 - 200 - 15
= 9785
Answer: 95 x 103 = 9,785.
Example 9: Example 9: Variable other than x
Problem: Expand (p + 10)(p - 4).
Solution:
Using the identity with x = p, a = 10, b = -4:
- p² + (10 + (-4))p + (10)(-4)
- = p² + 6p - 40
Answer: (p + 10)(p - 4) = p² + 6p - 40.
Example 10: Example 10: Verifying by substitution
Problem: Verify the identity (x + a)(x + b) = x² + (a + b)x + ab for x = 2, a = -1, b = 5.
Solution:
LHS: (2 + (-1))(2 + 5) = (1)(7) = 7
RHS: 2² + (-1 + 5)(2) + (-1)(5) = 4 + 8 - 5 = 7
LHS = RHS = 7. Verified.
Answer: The identity is verified.
Real-World Applications
The (x + a)(x + b) identity has many applications in algebra and arithmetic:
- Factorisation of Quadratics: This identity is the primary tool for factorising quadratic trinomials of the form x² + bx + c. This is essential in Class 9 and 10 for solving quadratic equations.
- Mental Arithmetic: Products like 102 x 105 = 10000 + 700 + 10 = 10710 can be computed quickly without long multiplication.
- Solving Equations: To solve x² + 5x + 6 = 0, factorise as (x + 2)(x + 3) = 0, giving x = -2 or x = -3.
- Understanding Sign Patterns: The identity teaches how the signs of a and b affect the middle and last terms. This pattern recognition is critical for quick factorisation.
- Competition Mathematics: Problems involving products of consecutive or near-consecutive numbers often use this identity.
- Foundation for FOIL: The FOIL method (First, Outer, Inner, Last) for multiplying binomials is essentially this identity in procedural form. Understanding Identity IV gives a deeper understanding of why FOIL works.
- Polynomial Division: Factorising using this identity helps simplify polynomial fractions and perform polynomial long division.
Key Points to Remember
- (x + a)(x + b) = x² + (a + b)x + ab — Identity IV.
- The coefficient of x is the SUM of a and b: (a + b).
- The constant term is the PRODUCT of a and b: ab.
- a and b can be positive, negative, or zero.
- This identity is the most general of the four Class 8 identities.
- When a = b, it reduces to Identity I: (x + a)² = x² + 2ax + a².
- When b = -a, it reduces to Identity III: (x + a)(x - a) = x² - a².
- For factorisation: find two numbers with the required sum and product.
- If the last term is positive, both numbers have the same sign. If negative, they have opposite signs.
- This identity is the foundation for factorising quadratic expressions in higher classes.
Practice Problems
- Expand (x + 5)(x + 8) using the identity.
- Expand (y - 6)(y + 3).
- Expand (m - 4)(m - 7).
- Factorise: x² + 12x + 35.
- Factorise: x² - 2x - 15.
- Factorise: y² - 13y + 40.
- Find 104 x 102 using the identity.
- Find 97 x 105 using the identity.
Frequently Asked Questions
Q1. What is the (x + a)(x + b) identity?
(x + a)(x + b) = x² + (a + b)x + ab. The product of two binomials with common variable x equals x² plus the sum of the constants times x plus the product of the constants.
Q2. How is this identity used for factorisation?
To factorise x² + px + q, find two numbers a and b such that a + b = p and a x b = q. Then x² + px + q = (x + a)(x + b).
Q3. Can a and b be negative?
Yes. Both a and b can be any real numbers — positive, negative, or zero. For example, (x - 3)(x + 5) uses a = -3 and b = 5.
Q4. How is this identity related to Identity I?
When a = b, (x + a)(x + a) = x² + 2ax + a² = (x + a)², which is Identity I.
Q5. How is this identity related to Identity III?
When b = -a, (x + a)(x - a) = x² + 0 - a² = x² - a², which is Identity III (difference of squares).
Q6. What determines the sign of the last term (ab)?
If a and b have the same sign (both positive or both negative), ab is positive. If they have opposite signs, ab is negative.
Q7. What determines the sign of the middle term coefficient (a + b)?
It depends on which of a or b has the larger absolute value and whether they are positive or negative. It can be positive, negative, or even zero.
Q8. What is the FOIL method?
FOIL stands for First, Outer, Inner, Last — it is a procedural way to multiply two binomials. For (x + a)(x + b): First = x.x = x², Outer = x.b = bx, Inner = a.x = ax, Last = a.b = ab. Combining gives x² + (a+b)x + ab, which is this identity.
Q9. Can this identity be used for expressions like (2x + 3)(2x + 5)?
Yes. Let the common part be 2x instead of x. Then a = 3, b = 5, and the result is (2x)² + (3+5)(2x) + 15 = 4x² + 16x + 15.
Q10. How do you factorise when the last term is negative?
When the last term is negative (e.g., x² + 2x - 35), find two numbers whose product is -35 and sum is 2. Since the product is negative, the numbers have opposite signs: 7 and -5. So x² + 2x - 35 = (x + 7)(x - 5).
