Q1: The common factor of x2y2 and x3y3 is (a) x2y2 (b) x3y3 (c) x2y3 (d) x3y2
Solution:
Ans: (a) x2y2 = x × x × y × y x3y3 = x × x × x × y × y × y
Hence, Common Factor of x2y2 and x3y3 is x2y2
Q2: The common factor of x3y2 and x4y is (a) x43y2 (b) x4y (c) x3y2 (d) x3y
Solution:
Ans: (d) x3y2 = x × x × x × y × y x4y = x × x × x × x × y
Hence, Common Factor of x3y2 and x4y is x3y .
Q3: The common factor of a2m4 and a4m2 is (a) a4m4 (b) a2m2 (c) a2m4 (d) a4m2
Solution:
Ans: (b) a2m4 = a × a × m × m × m × m a4m2 = a × a × a × a × m × m
Hence, Common Factor of a2m4 and a4m2 is a2m2 .
Q4: The common factor of 2x, 3x3, 4 is (a) 1 (b) 2 (c) 3 (d) 4
Solution:
Ans: (a) 2x = 2 × x 3x3 = 3 × x × x × x 4 = 2 × 2 Hence the common factor is 1
Q5: The common factor of 10ab, 30bc, 50ca is (a) 10 (b) 30 (c) 50 (d) abc
Solution:
Ans: (a) 10ab = 2 × 5 × a × b 30bc = 2 × 3 × 5 × b × c 50ca = 2 × 5 × 5 × c × a
Hence the common factor is 10.
Q6: The common factor of 14a2b and 35a4b2 is (a) a4b2 (b) 35a4b2 (c) 14a2b (d) 7a2b
Solution:
Ans: (d) 14a2b = 2 × 7 × a × a × b 35a4b2 = 5 × 7 × a × a × a × a × b × b
Hence the common factor of 14a2b and 35a4b2 is 7a2b.
Q7: The common factor of 72x3y4z4, 120z2d4x4 and 96y3z4d4 is (a) 96z3 (b) 120z3 (c) 72z3 (d) 24z2
Solution:
Ans: (d) 72x3y4z4 = 2 × 2 × 2 × 3 × 3 × x × x × x × y × y × y × y × z × z × z × z. 120z2d4x4 = 2 × 2 × 2 × 3 × 5 × z × z × d × d × d × d × x × x × x × x 96y3z4d4 = 2 × 2 × 2 × 2 × 2 × 3 × y × y × z × z × z × z × d × d × d × d Hence common factor is 2 x 2 x 2 x 3 = 24z2
Q8: The factorisation of 12a2b + 15ab2 is a) 3ab (4a + 5b) (b) 3a2b (4a + 5b) (c) 3ab2 (4a + 5b) (d) 3a2b2 (4a + 5b)
Solution:
Ans: (a) 12a2b + 15ab2 = 3ab(4a + 5b)
12a2b=2×2×3×a×a×b 15ab2=3×5×a×b×b
Common factors: 3×a×b
Hence factorisation of 12a2b + 15ab2 is 3ab(4a+5b)
Q9: The factorisation of x2yz + xy2z + xyz2 is (a) xyz(x + y + z) (b) x2yz(x + y + z) (c) xy2z(x + y + z) (d) xyz2(x + y + z)
Solution:
Ans: (a) x2yz + xy2z + xyz2 = xyz (x + y + z)
Hence factorisation of x2yz + xy2z + xyz2 is xyz(x + y + z).
Q10: The factorisation of a (x + y + z) + b(x + y + z) + c(x + y + z) is (a) (a + b + c)(x + y + z) (b) (ab + bc + ca)(x + y + z) (c) (xy + yz + zx)(a + b + c) (d) none of these
Solution:
Ans: (a) a(x + y + z) + b(x + y + z) + c(x + y + z) = (x + y + z) (a + b + c).
Hence factorisation of a (x + y + z) + b(x + y + z) + c(x + y + z) is (a + b + c)(x + y + z).
1. How can factorisation help in simplifying algebraic expressions?
Ans. Factorisation helps in simplifying algebraic expressions by breaking down the expression into its factors, making it easier to work with and solve. It also helps in finding common factors and simplifying complex expressions.
2. What are the different methods of factorisation in algebra?
Ans. Some common methods of factorisation in algebra include grouping, finding common factors, using the distributive property, and using special factorisation formulas such as the difference of squares or perfect square trinomials.
3. How can factorisation be used to solve algebraic equations?
Ans. Factorisation can be used to solve algebraic equations by factoring out common factors or applying factorisation methods to simplify the equation. This makes it easier to find the solutions or roots of the equation.
4. Can factorisation be used in real-life applications outside of algebra?
Ans. Yes, factorisation can be used in various real-life applications such as in economics, physics, engineering, and computer science. It helps in simplifying complex problems and finding optimal solutions.
5. What are the benefits of mastering factorisation in algebra?
Ans. Mastering factorisation in algebra can help improve problem-solving skills, enhance understanding of algebraic concepts, and make it easier to work with complex equations and expressions. It is a fundamental skill that is essential for further studies in mathematics.
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