Factorization MCQ Test — Class 8 CBSE

Solution:

Factorization is essentially the reverse process of multiplication.

When we factorize an algebraic expression, we express it as a product of two or more simpler expressions, which are called its factors.

Solution:

Identify the terms: 6xy and -9x.

Find the GCF of the numerical coefficients (6 and 9) which is 3. Find the GCF of the variables (xy and x) which is x. So, the overall GCF is 3x.

Divide each term by 3x: (6xy) ÷ (3x) = 2y and (-9x) ÷ (3x) = -3.

Write the expression as the product of the GCF and the remaining terms: 3x(2y - 3).

Solution:

Check Step 1: The GCF of 12 and 18 is 6. The GCF of a²b and ab² is ab. So, the GCF is 6ab. Step 1 is correct.

Check Step 2: (12a²b) ÷ (6ab) = 2a, which is correct. (18ab²) ÷ (6ab) = 3b, which is also correct. Step 2 is correct.

Check Step 3: Since the original expression had a '+' sign between the terms, the terms inside the bracket should also be added. 6ab(2a + 3b) is correct. Step 3 is correct.

All steps in Rahul's solution are completely correct.

Solution:

Group the terms in pairs that have a common factor: (ab + bc) + (ax + cx).

Factor out the common factor from each group: b(a + c) + x(a + c).

Now, (a + c) is a common binomial factor. Factor it out from the entire expression: (a + c)(b + x).

Solution:

The given expression is 7p²q - 14pq².

The greatest common factor (GCF) of 7p²q and 14pq² is 7pq.

Priya's factorization: 7pq(p - 2q) is correct and complete because 7pq is the GCF, and the expression inside the bracket cannot be further factorized by common factors.

Rohan's factorization: 7(p²q - 2pq²) is partially correct because 7 is a common factor, but it is not the *greatest* common factor. The expression inside the bracket (p²q - 2pq²) can still be factorized by taking out pq.

Therefore, only Priya's factorization is correct and complete.

Solution:

Identify the terms: 5x³, -10x², and 25x.

Find the GCF of the numerical coefficients (5, -10, and 25) which is 5.

Find the GCF of the variables (x³, x², and x) which is x.

The overall GCF of the expression is 5x. Divide each term by 5x: (5x³) ÷ (5x) = x², (-10x²) ÷ (5x) = -2x, and (25x) ÷ (5x) = 5.

Write the factored form: 5x(x² - 2x + 5).

Solution:

We are looking for two expressions that multiply to x² - 8x + 15. We know one factor is (x - 3).

Let the other factor be (x + k). So, (x - 3)(x + k) = x² - 8x + 15.

Expand the left side: x² + kx - 3x - 3k = x² - 8x + 15. This simplifies to x² + (k - 3)x - 3k = x² - 8x + 15.

By comparing the constant terms, -3k = 15, which implies k = -5. By comparing the coefficient of x, k - 3 = -8, which also implies k = -5. Thus, the other factor is (x - 5).

Solution:

Observe the given expression: 5(x - 2y) - 3z(x - 2y).

Notice that the binomial (x - 2y) is a common factor in both terms.

Factor out the common binomial (x - 2y) from both terms: (x - 2y) [5 - 3z].

The factored form is (x - 2y)(5 - 3z).

Solution:

The given expression is 49x² - 16y².

Recognize that 49x² can be written as (7x)² and 16y² can be written as (4y)².

This expression matches the algebraic identity for the difference of two squares: a² - b² = (a - b)(a + b).

Here, a = 7x and b = 4y. Substitute these values into the identity: (7x - 4y)(7x + 4y).

Solution:

Option A (x² + 5x + 6) is a quadratic trinomial, typically factorized by splitting the middle term or using identities like (x+a)(x+b).

Option B (3ab - 6ac + 5b - 10c) has four terms. We can group (3ab - 6ac) and (5b - 10c) to get 3a(b - 2c) + 5(b - 2c), which then factors as (b - 2c)(3a + 5). This is suitable for factorization by grouping.

Option C (4y² - 9) is a difference of squares (2y)² - 3², which is factorized using the a² - b² identity.

Option D (7m - 14n) is a binomial, factorized by taking out the common monomial factor 7: 7(m - 2n).

Therefore, 3ab - 6ac + 5b - 10c is the most appropriate for factorization by grouping terms.