Exercise: 14-A
Q1: 6xy − 4xz
Step 1: Identify the Greatest Common Factor (G.C.F.) of coefficients 6 and 4 → G.C.F. = 2.
Step 2: Identify the common variables in both terms: 6xy and 4xz share the factor x.
Step 3: Therefore, the overall common factor is 2x.
Step 4: Factor out 2x: 6xy – 4xz = 2x(3y) – 2x(2z).
Step 5: Write the expression as a product: 6xy – 4xz = 2x(3y – 2z).
Step 6 (Check): Expand to verify → 2x(3y – 2z) = 6xy – 4xz ✔️
Answer: 2x(3y – 2z)
Q2: 9bx² + 15b²x
Step 1: Look at the coefficients 9 and 15.
G.C.F. of 9 and 15 = 3.
Step 2: Look at the variables.
First term = 9bx², second term = 15b²x.
Common variable factor = bx.
Step 3: Therefore, overall common factor = 3bx.
Step 4: Factorise:
9bx² + 15b²x = 3bx(3x) + 3bx(5b).
Step 5: Write as product:
9bx² + 15b²x = 3bx(3x + 5b).
Step 6 (Check): Expand:
3bx(3x + 5b) = 9bx² + 15b²x ✔️
Answer: 3bx(3x + 5b)
Q3: \(4a^2b − 6ab^2 + 8ab\)
Step 1: Consider coefficients 4, 6, 8. Their G.C.F. is 2.
Step 2: Each term has variables a and b in common ⇒ common variable factor is ab.
Step 3: Therefore, overall common factor = 2ab.
Step 4: Factor out \(2ab: 4a^2b − 6ab^2 + 8ab = 2ab(2a) − 2ab(3b) + 2ab(4)\).
Step 5: Write compactly: \(4a^2b − 6ab^2 + 8ab = 2ab(2a − 3b + 4)\).
Step 6 (Check): Expand \(2ab(2a − 3b + 4) = 4a^2b − 6ab^2 + 8ab\) ✔️
Answer: \(2ab(2a − 3b + 4)\)
Q4: 5x² + 15x³ − 20x
Step 1: Look at the coefficients 5, 15, 20. Their G.C.F. = 5.
Step 2: For variables, each term contains at least one factor of x. Common variable factor = x.
Step 3: So, overall common factor = 5x.
Step 4: Factor out 5x:
5x² ÷ 5x = x,
15x³ ÷ 5x = 3x²,
−20x ÷ 5x = −4.
Step 5: Write in product form: 5x² + 15x³ − 20x = 5x(x + 3x² − 4).
Step 6 (Check): Expand 5x(x + 3x² − 4) = 5x² + 15x³ − 20x ✔️
Answer: 5x(x + 3x² − 4)
Q5: 2x³ + x² − 4x
Step 1: Observe the coefficients 2, 1, and −4. Their G.C.F. = 1.
Step 2: For variables, each term has at least one factor of x. Common variable factor = x.
Step 3: So, overall common factor = x.
Step 4: Factor out x:
2x³ ÷ x = 2x²,
x² ÷ x = x,
−4x ÷ x = −4.
Step 5: Write in product form: 2x³ + x² − 4x = x(2x² + x − 4).
Step 6 (Check): Expand x(2x² + x − 4) = 2x³ + x² − 4x ✔️
Answer: x(2x² + x − 4)
Q6: a³b − a²b² − b³
Step 1: The three terms are: a³b, −a²b², and −b³.
Step 2: Find the common factors of variables.
• First term = a³b → contains a² and b
• Second term = −a²b² → contains a² and b
• Third term = −b³ → contains only b
The only common factor among all terms = b.
Step 3: Factor out b:
a³b ÷ b = a³
−a²b² ÷ b = −a²b
−b³ ÷ b = −b²
Step 4: So, expression becomes:
a³b − a²b² − b³ = b(a³ − a²b − b²).
Step 5 (Check): Expand b(a³ − a²b − b²) = a³b − a²b² − b³ ✔️
Answer: b(a³ − a²b − b²)
Q7: 28a²b²c − 42ab²c²
Step 1: Look at the coefficients 28 and 42.
G.C.F. of 28 and 42 = 14.
Step 2: Now check the variables:
• First term = 28a²b²c → contains a², b², c
• Second term = −42ab²c² → contains a, b², c²
Common variable factors = a, b², c.
Step 3: Therefore, overall common factor = 14abc²? Let’s check carefully.
– From a² and a → common factor = a
– From b² and b² → common factor = b²
– From c and c² → common factor = c
✅ So, common factor = 14ab²c.
Step 4: Divide each term by 14ab²c:
28a²b²c ÷ 14ab²c = 2a
−42ab²c² ÷ 14ab²c = −3c
Step 5: So the factorised form is:
28a²b²c − 42ab²c² = 14ab²c(2a − 3c).
Step 6 (Check): Expand 14ab²c(2a − 3c) = 28a²b²c − 42ab²c² ✔️
Answer: 14ab²c(2a − 3c)
Q8: 15x²y − 6xy² + 9y³
Step 1: Look at the coefficients 15, −6, 9.
G.C.F. of 15, 6, and 9 = 3.
Step 2: Now check the variables:
• First term = 15x²y → contains x² and y
• Second term = −6xy² → contains x and y²
• Third term = 9y³ → contains y³
The common variable factor is y.
Step 3: Therefore, overall common factor = 3y.
Step 4: Divide each term by 3y:
15x²y ÷ 3y = 5x²
−6xy² ÷ 3y = −2xy
9y³ ÷ 3y = 3y²
Step 5: So the factorised form is:
15x²y − 6xy² + 9y³ = 3y(5x² − 2xy + 3y²).
Step 6 (Check): Expand 3y(5x² − 2xy + 3y²) = 15x²y − 6xy² + 9y³ ✔️
Answer: 3y(5x² − 2xy + 3y²)
Q9: 10x²y + 15xy² − 25x²y²
Step 1: Look at the coefficients 10, 15, −25.
G.C.F. of 10, 15, and 25 = 5.
Step 2: Now check the variables:
• First term = 10x²y → contains x² and y
• Second term = 15xy² → contains x and y²
• Third term = −25x²y² → contains x² and y²
Common variable factor = xy.
Step 3: Therefore, overall common factor = 5xy.
Step 4: Divide each term by 5xy:
10x²y ÷ 5xy = 2x
15xy² ÷ 5xy = 3y
−25x²y² ÷ 5xy = −5xy
Step 5: So the factorised form is:
10x²y + 15xy² − 25x²y² = 5xy(2x + 3y − 5xy).
Step 6 (Check): Expand 5xy(2x + 3y − 5xy) = 10x²y + 15xy² − 25x²y² ✔️
Answer: 5xy(2x + 3y − 5xy)
Q10: 18x²y − 24xyz
Step 1: Look at the coefficients 18 and 24.
G.C.F. of 18 and 24 = 6.
Step 2: Now check the variables:
• First term = 18x²y → contains x² and y
• Second term = −24xyz → contains x, y, z
Common variable factor = xy.
Step 3: Therefore, overall common factor = 6xy.
Step 4: Divide each term by 6xy:
18x²y ÷ 6xy = 3x
−24xyz ÷ 6xy = −4z
Step 5: So the factorised form is:
18x²y − 24xyz = 6xy(3x − 4z).
Step 6 (Check): Expand 6xy(3x − 4z) = 18x²y − 24xyz ✔️
Answer: 6xy(3x − 4z)
Q11: 27a³b³ − 45a⁴b²
Step 1: Look at the coefficients 27 and 45.
G.C.F. of 27 and 45 = 9.
Step 2: Now check the variables:
• First term = 27a³b³ → contains a³ and b³
• Second term = −45a⁴b² → contains a⁴ and b²
Common variable factor = a³ and b².
Step 3: Therefore, overall common factor = 9a³b².
Step 4: Divide each term by 9a³b²:
27a³b³ ÷ 9a³b² = 3b
−45a⁴b² ÷ 9a³b² = −5a
Step 5: So the factorised form is:
27a³b³ − 45a⁴b² = 9a³b²(3b − 5a).
Step 6 (Check): Expand 9a³b²(3b − 5a) = 27a³b³ − 45a⁴b² ✔️
Answer: 9a³b²(3b − 5a)
Q12: 4(a + b) − 6(a + b)²
Step 1: Look at the coefficients 4 and −6.
G.C.F. of 4 and 6 = 2.
Step 2: Now check the variable factors:
• First term = 4(a + b) → contains (a + b)
• Second term = −6(a + b)² → contains (a + b)²
Common factor = (a + b).
Step 3: Therefore, overall common factor = 2(a + b).
Step 4: Divide each term by 2(a + b):
4(a + b) ÷ 2(a + b) = 2
−6(a + b)² ÷ 2(a + b) = −3(a + b)
Step 5: So the factorised form is:
4(a + b) − 6(a + b)² = 2(a + b)[2 − 3(a + b)].
Step 6 (Check): Expand 2(a + b)[2 − 3(a + b)]
= 4(a + b) − 6(a + b)² ✔️
Answer: 2(a + b)[2 − 3(a + b)]
Q13: 2a(3x + 5y) − 5b(3x + 5y)
Step 1: Both terms have a common factor (3x + 5y).
Step 2: Take out (3x + 5y) as the common factor.
Step 3: Divide each term by (3x + 5y):
• 2a(3x + 5y) ÷ (3x + 5y) = 2a
• −5b(3x + 5y) ÷ (3x + 5y) = −5b
Step 4: So the factorised form is:
2a(3x + 5y) − 5b(3x + 5y) = (3x + 5y)(2a − 5b).
Step 5 (Check): Expand (3x + 5y)(2a − 5b)
= 2a(3x + 5y) − 5b(3x + 5y) ✔️
Answer: (3x + 5y)(2a − 5b)
Q14: 2x(p² + q²) + 4y(p² + q²)
Step 1: Both terms contain a common factor (p² + q²).
Step 2: Take out (p² + q²) as the common factor.
Step 3: Divide each term by (p² + q²):
• 2x(p² + q²) ÷ (p² + q²) = 2x
• 4y(p² + q²) ÷ (p² + q²) = 4y
Step 4: So the factorised form is:
2x(p² + q²) + 4y(p² + q²) = (p² + q²)(2x + 4y) = 2(p² + q²)(x + 2y).
Step 5 (Check): Expand 2(p² + q²)(x + 2y)
= 2x(p² + q²) + 4y(p² + q²) ✔️
Answer: 2(p² + q²)(x + 2y)
Q15: 8(3a − 2b)² − 10(3a − 2b)
Step 1: Look at the coefficients 8 and 10.
G.C.F. of 8 and 10 = 2.
Step 2: Now check the variable factor:
• First term = 8(3a − 2b)² → contains (3a − 2b)²
• Second term = −10(3a − 2b) → contains (3a − 2b)
Common factor = (3a − 2b).
Step 3: Therefore, overall common factor = 2(3a − 2b).
Step 4: Divide each term by 2(3a − 2b):
• 8(3a − 2b)² ÷ 2(3a − 2b) = 4(3a − 2b)
• −10(3a − 2b) ÷ 2(3a − 2b) = −5
Step 5: So the factorised form is:
8(3a − 2b)² − 10(3a − 2b) = 2(3a − 2b)[4(3a − 2b) − 5] = 2(3a − 2b)(12a − 8b − 5).
Step 6 (Check): Expand 2(3a − 2b)(12a − 8b − 5)
= 8(3a − 2b)² − 10(3a − 2b) ✔️
Answer: 2(3a − 2b)(12a − 8b − 5)
Q16: Factorize \(x(x+y)^3 – 3x^2y(x+y)\)
Step 1: Identify the common factors in both terms.
Both terms have \(x\) and \((x+y)\) as factors:
\[
x(x+y)^3 – 3x^2y(x+y) = x(x+y) \left[(x+y)^2 – 3xy\right]
\]Step 2: Expand \((x+y)^2\):
\[
(x+y)^2 = x^2 + 2xy + y^2
\]
So,
\[
(x+y)^2 – 3xy = x^2 + 2xy + y^2 – 3xy = x^2 – xy + y^2
\]Step 3: Write the final factored form:
\[
x(x+y)(x^2 – xy + y^2)
\]Answer: \(x(x+y)(x^2 – xy + y^2)\)
Q17: a(3x − 2y) + b(2y − 3x)
Step 1: Notice that (2y − 3x) is the negative of (3x − 2y).
Because: (2y − 3x) = −(3x − 2y).
Step 2: Rewrite the expression:
a(3x − 2y) + b(2y − 3x)
= a(3x − 2y) + b[−(3x − 2y)]
= a(3x − 2y) − b(3x − 2y).
Step 3: Now both terms have a common factor (3x − 2y).
Step 4: Take out (3x − 2y):
= (3x − 2y)(a − b).
Step 5 (Check): Expand (3x − 2y)(a − b)
= a(3x − 2y) − b(3x − 2y)
= a(3x − 2y) + b(2y − 3x) ✔️
Answer: (3x − 2y)(a − b)
Q18: \(a(b-c)^2 – d(c-b)\)
Step 1: Note that \(c – b = -(b – c)\), so replace \(c – b\) by \(-(b – c)\):
\[
a(b – c)^2 – d(c – b) = a(b – c)^2 – d(-1)(b – c) \\
= a(b – c)^2 + d(b – c)
\]Step 2: Factor out the common factor \((b – c)\):
\[
= (b – c)\left[a(b – c) + d\right] \\
= (b – c)\left[a(b – c) + d\right]
\]Step 3: Expand \(a(b – c)\) if needed:
\[
= (b – c)\left[ab – ac + d\right]
\]
Step 4 (Check): Expand (b – c)\left[ab – ac + d\right]
\(a(b-c)^2 – d(c-b)\)✔️
Answer: \((b – c)(ab – ac + d)\)
Q19: a(4a − b) + 2b(4a − b) − c(4a − b)
Step 1: Observe the terms:
• First term = a(4a − b)
• Second term = 2b(4a − b)
• Third term = −c(4a − b)
Clearly, (4a − b) is common in all terms.
Step 2: Factor out (4a − b):
= (4a − b)[a + 2b − c].
Step 3 (Check): Expand back:
(4a − b)(a + 2b − c)
= a(4a − b) + 2b(4a − b) − c(4a − b) ✔️
Answer: (4a − b)(a + 2b − c)
Q20: m(mx+ny)² + mn(mx+ny) + n(mx+ny)
Step 1: Observe each term:
• First term = m(mx+ny)²
• Second term = mn(mx+ny)
• Third term = n(mx+ny)
Common factor is clearly (mx+ny).
Step 2: Factor out (mx+ny):
= (mx+ny)[ m(mx+ny) + mn + n ].
Step 3: Simplify the bracket:
Inside = m(mx+ny) + mn + n
= m²x + mny + mn + n.
Step 4: Final factorised form is:
(mx + ny)(m²x + mny + mn + n).
Answer: (mx + ny)(m²x + mny + mn + n)
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