The identity (a + b)2 = a2 + 2ab + b2 can also be used to find factors of some algebraic expressions. Consider the algebraic expression x2 + 4x + 4.
Answer: x2 + 4x + 4 = (x + 2)2, so (x + 2) is a factor.
Step-by-step solution
Given: x2 + 4x + 4
To find: Its factors
To find: Its factors
Idea: Read the identity backwards: a2 + 2ab + b2 = (a + b)2. If the first and last terms are perfect squares and the middle term is twice the product of their square roots, the expression is a perfect square.
- First term: x2 = (x)2. Last term: 4 = 22. So try a = x, b = 2.½ mark
- Check the middle term: 2ab = 2(x)(2) = 4x ✓ — it matches.½ mark
- So x2 + 4x + 4 = x2 + 2(x)(2) + 22 = a2 + 2ab + b2½ mark
- = (a + b)2 = (x + 2)2. Therefore (x + 2) is a factor of x2 + 4x + 4.½ mark
x² + 4x + 4 = (x + 2)² = (x + 2)(x + 2); (x + 2) is a factor.
Check: Expand back: (x + 2)2 = x2 + 4x + 4 ✓. Also with x = 3: 9 + 12 + 4 = 25 = 52 ✓.
Answer to write in the exam
x2 + 4x + 4 = x2 + 2(x)(2) + 22 [a2 + 2ab + b2 = (a + b)2]
∴ x2 + 4x + 4 = (x + 2)2 = (x + 2)(x + 2)
Common mistakes that cost marks
- Writing (x + 4)2. The last term is 4 = 22, so b = 2, not 4.
- Not checking the middle term. x2 + 5x + 4 also has square end terms, but it is not a perfect square because 5x ≠ 2(x)(2).
- Stopping at (x + 2)2 when asked for factors: the factors are (x + 2) and (x + 2).
How this can come in the exam
MCQ (1 mark)
A factor of y2 + 10y + 25 is
- y + 25
- y + 5
- y − 5
- y + 10
Show answer
(B) y + 5
y2 + 2(y)(5) + 52 = (y + 5)2.
Short answer (2 marks)
Factorise x2 + 14x + 49.
Show answer
x2 = (x)2, 49 = 72, 14x = 2(x)(7) (1 mark). So x2 + 14x + 49 = (x + 7)2 (1 mark).Try one yourself
Factorise m2 + 6m + 9.
Show answer
m2 + 2(m)(3) + 32 = (m + 3)2.