Answer
(a) \( (a - 2b - 4x)(a - 2b + 4x) \)
(b) \( (c - a + b)(c + a - b) \)
(c) \( (a + b - 3x)(a + b + 3x) \)
(d) \( (a - x - y)(a + x + y) \)
(e) \( (4x² - a + 3b)(4x² + a - 3b) \)
(f) \( (5y² - 3m + 1)(5y² + 3m - 1) \)
Solution
We will factorise each expression step‐by‐step.
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Example (a)
Factorise
a² – 4ab + 4b² – 16x²
Step 1. Notice that the first three terms form a perfect square:
a² – 4ab + 4b² = (a – 2b)²
Step 2. Then your expression becomes a difference of two squares:
(a – 2b)² – (4x)²
Step 3. Factor using the difference‐of‐squares formula:
X² – Y² = (X – Y)(X + Y)
Thus,
(a – 2b – 4x)(a – 2b + 4x)
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Example (b)
Factorise
c² – a² + 2ab – b²
Step 1. Group the last three terms:
a² – 2ab + b² = (a – b)²
So rewrite the expression as:
c² – (a – b)²
Step 2. Again a difference of two squares:
= (c – (a – b)) (c + (a – b))
which simplifies to
= (c – a + b)(c + a – b)
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Example (c)
Factorise
a² + 2ab + b² – 9x²
Step 1. Recognize a² + 2ab + b² = (a + b)², so:
= (a + b)² – (3x)²
Step 2. Factor the difference of two squares:
= (a + b – 3x)(a + b + 3x)
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Example (d)
Factorise
a² – 2xy – y² – x²
Step 1. Notice that the last three terms can be regrouped:
x² + 2xy + y² = (x + y)²
Thus, write:
a² – (x + y)²
Step 2. Factor as a difference of squares:
= (a – (x + y))(a + (x + y))
= (a – x – y)(a + x + y)
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Example (e)
Factorise
16x⁴ – a² + 6ab – 9b²
Step 1. Group the last three terms. Note that:
a² – 6ab + 9b² = (a – 3b)²
So write:
16x⁴ – (a – 3b)²
Step 2. Recognise 16x⁴ = (4x²)², then use difference of two squares:
= (4x² – (a – 3b)) (4x² + (a – 3b))
= (4x² – a + 3b)(4x² + a – 3b)
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Example (f)
Factorise
25y⁴ – 1 – 9m² + 6m
Step 1. Rearrange the terms to group two pairs:
(25y⁴ – 1) – (9m² – 6m)
Step 2. Notice that 25y⁴ – 1 is a difference of two squares,
25y⁴ – 1 = (5y²)² – 1² = (5y² – 1)(5y² + 1)
However, the second group does not factor nicely by itself.
A better idea is to rewrite the last three terms as a complete square:
9m² – 6m + 1 = (3m – 1)² so that
–(9m² – 6m + 1) = –(3m – 1)²
Step 3. Now write the whole expression as:
25y⁴ – (3m – 1)²
Step 4. Then factor as a difference of squares:
= (5y² – (3m – 1)) (5y² + (3m – 1))
= (5y² – 3m + 1)(5y² + 3m – 1)
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Example (g)
Factorise
9r² – 4 + 12a – 9a²
Step 1. Rearrange the terms to group those in a:
9r² – 9a² + 12a – 4
Step 2. Notice that 9a² – 12a + 4 = (3a – 2)². Thus write:
9r² – (3a – 2)²
Step 3. Factor as a difference of two squares:
= (3r – (3a – 2)) (3r + (3a – 2))
= (3r – 3a + 2)(3r + 3a – 2)
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Example (h)
Factorise
1 – 81a² – 18ab – b²
Step 1. Group the last three terms. Notice that:
81a² + 18ab + b² = (9a + b)²
Step 2. Then the expression becomes:
1 – (9a + b)²
Step 3. Factor as a difference of two squares:
= (1 – (9a + b))(1 + (9a + b))
= (1 – 9a – b)(1 + 9a + b)
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Example (i)
Factorise
20ab + 4a² + 25b² – 9x⁴
Step 1. Rearrange the terms so that the a–b part is together:
4a² + 20ab + 25b² = (2a + 5b)²
Step 2. Then the expression becomes:
(2a + 5b)² – 9x⁴
and 9x⁴ = (3x²)²
Step 3. Factor by difference of squares:
= (2a + 5b – 3x²)(2a + 5b + 3x²)
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Example (j)
Factorise
24xy – 3z² + 3y² + 48x²
Step 1. Group the terms in x and y:
48x² + 24xy + 3y²
Notice that this is a perfect square since
48x² + 24xy + 3y² = 3(16x² + 8xy + y²) = 3(4x + y)²
Step 2. Now the whole expression becomes:
3(4x + y)² – 3z²
Step 3. Factor out the common factor 3 and use the difference of squares:
= 3[(4x + y)² – z²]
= 3[(4x + y) – z][(4x + y) + z]
= 3(4x + y – z)(4x + y + z)
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Summary of Answers
(a) (a – 2b – 4x)(a – 2b + 4x)
(b) (c – a + b)(c + a – b)
(c) (a + b – 3x)(a + b + 3x)
(d) (a – x – y)(a + x + y)
(e) (4x² – a + 3b)(4x² + a – 3b)
(f) (5y² – 3m + 1)(5y² + 3m – 1)
Reviewed and approved by the UpStudy tutoring team
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Explain
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Simplify this solution