RBSE Solutions Class 8 Maths Chapter 10 Factorization Important Questions

Rajasthan Board RBSE Class 8 Maths Chapter 10 Factorization Additional Questions

I. Objective Type Questions

 

Question 1. Factors of expression \( x^2 + (a + b) x + ab \) are
(a) \( (x + a)(x - b) \)
(b) \( (x - a) (x + b) \)
(c) \( (x + a)(x + b) \)
(d) \( (x - a) (x - b) \)
Answer: (c) (x + a)(x + b)
In simple words: To find the factors, group the terms and take out common factors. This shows how a polynomial can be broken down into simpler parts.

🎯 Exam Tip: Remember the common factorization identity \( x^2 + (a+b)x + ab = (x+a)(x+b) \). This makes solving faster.

 

Question 2. Square of (2x + 3) is
(a) \( 4x^2 + 6x + 9 \)
(b) \( 4x^2 + 2x + 9 \)
(c) \( 4x^2 + 12x + 9 \)
(d) \( 4x^2 + 9 \)
Answer: (c) \( 4x^2 + 12x + 9 \)
In simple words: Squaring means multiplying the expression by itself. You can use the formula for \( (a+b)^2 \). This is an algebraic expansion.

🎯 Exam Tip: Always use the identity \( (a+b)^2 = a^2 + 2ab + b^2 \) for squaring binomials. Don't forget the middle term \( 2ab \).

 

Question 3. Square of (6x + 1) is
(a) \( 36x^2 + 1 \)
(b) \( 36x^2 + 6x + 1 \)
(c) \( 36x^2 + 6 \)
(d) \( 36x^2 + 12x + 1 \).
Answer: (d) \( 36x^2 + 12x + 1 \)
In simple words: To find the square of \( (6x+1) \), you multiply it by itself. This results in three terms: the square of the first term, twice the product of both terms, and the square of the second term.

🎯 Exam Tip: Be careful with the multiplication. Remember that \( (a+b)^2 = a^2 + 2ab + b^2 \), so \( (6x)^2 + 2(6x)(1) + (1)^2 \) gives the correct answer.

 

Question 4. The product of expression (2a - 3) (2a + 3) is
(a) \( 4a^2 + 2a + 9 \)
(b) \( 4a^2 - 9 \)
Answer: (b) \( 4a^2 - 9 \)
In simple words: When you multiply \( (2a-3) \) by \( (2a+3) \), it's like using the special formula \( (A-B)(A+B) = A^2-B^2 \). This gives a simplified two-term answer.

🎯 Exam Tip: Recognize the difference of squares identity: \( (a-b)(a+b) = a^2 - b^2 \). Applying this identity directly simplifies the calculation.

 

Question 5.
(a) y
(b) 6y
(c) 6x
(d) xy
Answer: (c) 6x
In simple words: This question asks to identify a specific term or result from a set of options. Without the full question, the answer is chosen from the given choices.

🎯 Exam Tip: For MCQ questions, always read all options carefully before selecting your answer, even if the question text is incomplete.

 

Question 6. Factor of \( 2x^3 + x^2 + 2x + 1 \) is
(a) \( (2x + 1)(x^2 + 1) \)
(b) \( (x^2 + 2)(x + 1) \)
(c) \( (x + 2)(x^2 + 1) \)
(d) \( (x^2 + 1)(x + 1) \)
Answer: (a) \( (2x + 1)(x^2 + 1) \)
In simple words: To find the factors, group the terms and take out common factors from each pair. This helps to simplify the expression into a product of two binomials.

🎯 Exam Tip: When factorizing expressions with four terms, try grouping the first two terms and the last two terms. Look for a common binomial factor.

 

Question 7. Factor of \( 4x^2 + 8xy + 4y^2 \) is
(a) \( (2x + 2y)^2 \)
(b) \( (2x - 2y)^2 \)
(c) \( (2x + y)^2 \)
(d) \( (x + 2y)^2 \)
Answer: (a) \( (2x + 2y)^2 \)
In simple words: This expression is a perfect square trinomial. It matches the pattern \( a^2 + 2ab + b^2 \), which factors into \( (a+b)^2 \).

🎯 Exam Tip: Always look for perfect square trinomials in factorization problems. They have a distinct form and simplify quickly to a squared binomial.

 

Question 8. The factors of expression \( a^2 + 2ab + b^2 \) are
(a) \( (a + b)(a - b) \)
(b) \( (a + b)^2 \)
(c) \( (a - b)^2 \)
(d) \( (a^2 + b^2)^2 \)
Answer: (b) \( (a + b)^2 \)
In simple words: The expression \( a^2 + 2ab + b^2 \) is a well-known algebraic identity, meaning it always equals \( (a+b)^2 \). This is one of the fundamental identities used in algebra.

🎯 Exam Tip: This is a fundamental algebraic identity. Recognizing it directly can save a lot of time in factorization problems.

 

Question 9. The factors of expression \( a^2 – b^2 \) are
(a) \( (a^2 - b^2) \)
(b) \( (a^2 – b^2) (a + b) \)
(c) \( (a + b) (a + b) \)
(d) \( (a + b) (a – b) \)
Answer: (d) \( (a + b) (a – b) \)
In simple words: The expression \( a^2 - b^2 \) is called a difference of squares. It can always be factored into two parts: one part where 'a' and 'b' are added, and another where 'b' is subtracted from 'a'.

🎯 Exam Tip: Remember the difference of squares identity: \( a^2 - b^2 = (a-b)(a+b) \). This is another essential identity for factorization.

II. Fill In The Blanks

 

Question 1. The equation which is true for all value of variables is called _______.
Answer: Identity
In simple words: An identity is like a special math rule that works for any numbers you put into it, always staying true. It's a statement that is always correct.

🎯 Exam Tip: Clearly define "identity" as an equation true for all variable values, distinct from a standard equation that is true only for specific values.

 

Question 2. \( (x + a) (x + b) = x^2 + (....) x + (ab) \)
Answer: \( (a + b) \)
In simple words: When you multiply two binomials like \( (x+a) \) and \( (x+b) \), the middle term in the expanded form is the sum of 'a' and 'b' multiplied by 'x'. This is a standard algebraic expansion.

🎯 Exam Tip: Memorize the expansion: \( (x+a)(x+b) = x^2 + (a+b)x + ab \). This saves time and prevents errors in polynomial multiplication.

 

Question 3. \( a^2 - b^2 = (...) x (a - b) \)
Answer: \( (a + b) \)
In simple words: The difference of two squares, \( a^2 - b^2 \), can be broken down into two factors: \( (a-b) \) and \( (a+b) \). This is a very common factorization rule.

🎯 Exam Tip: Always recall the identity \( a^2 - b^2 = (a+b)(a-b) \). It is a key tool for factoring many expressions.

 

Question 4. \( (x - 1) (x + 1) \) is equal to _______
Answer: \( x^2 - 1 \)
In simple words: This is a special case of the difference of squares formula. When you multiply \( (x-1) \) by \( (x+1) \), you get the square of 'x' minus the square of '1'.

🎯 Exam Tip: Recognize \( (x-1)(x+1) \) as \( (x^2-1^2) \). This is a frequent application of the difference of squares identity, so remember it.

 

Question 5. The value of \( 3.5 \times 3.5 – 2.5 \times 2.5 \) is _______
Answer: 6
In simple words: You can calculate this by recognizing it as a difference of squares, \( a^2 - b^2 \). This makes the calculation simpler by changing it to \( (a-b)(a+b) \).

🎯 Exam Tip: Convert the decimal multiplication into a difference of squares: \( (3.5)^2 - (2.5)^2 = (3.5-2.5)(3.5+2.5) \). This method is much faster than squaring each number separately.

 

Question 2. Expansion of \( { \left( x+\frac { 1 }{ x } \right) }^{ 2 } \) is \( {x}^{2}+\frac { 1 }{ {x}^{2} }+2 \)
Answer: True
In simple words: When you square the expression \( (x + \frac{1}{x}) \), you apply the formula for \( (a+b)^2 \). This results in \( x^2 + 2(x)(\frac{1}{x}) + (\frac{1}{x})^2 \), which simplifies to \( x^2 + 2 + \frac{1}{x^2} \).

🎯 Exam Tip: Remember the special case where \( b = \frac{1}{a} \) in the \( (a+b)^2 \) identity. This simplifies the \( 2ab \) term to just 2.

 

Question 3. Factors of \( x^2 - 7x + 12 \) are \( (x + 3) (x + 4) \).
Answer: False
In simple words: To factorize \( x^2 - 7x + 12 \), you need two numbers that multiply to 12 and add up to -7. These numbers are -3 and -4, so the correct factors are \( (x-3)(x-4) \).

🎯 Exam Tip: Always double-check your factorization by multiplying the factors back. If you don't get the original expression, your factors are incorrect.

 

Question 4. If x = 2 and y = 1 then value of \( x^2 + 4xy + 4y^2 \) is 0.
Answer: True
In simple words: If you substitute \( x=2 \) and \( y=1 \) into the expression \( x^2 + 4xy + 4y^2 \), you should get the value 0. This checks if the given statement holds true for the specific values.

🎯 Exam Tip: Always substitute the given values into the expression carefully, following the order of operations, to determine if the statement is true or false.

IV. Matching Type Questions

 

Question. Match the following.

Answer:
1. \( \Leftrightarrow \) (a) 11021
2. \( \Leftrightarrow \) (d) \( 11 (2y - 3z) \)
3. \( \Leftrightarrow \) (b) 2y
4. \( \Leftrightarrow \) (c) - 4y
In simple words: We need to match each expression in Part 1 with its correct simplified form or factor in Part 2. This involves performing multiplication, factorization, finding common factors, and division.

🎯 Exam Tip: For matching questions, work out each item in Part 1 and then find its corresponding match in Part 2. Use algebraic identities where possible to speed up calculations.

V. Very Short Answer Type Questions

 

Question 1. Factorize \( 3x^3 + 3x^2 + x + 1 \).
Answer:
\( 3x^3 + 3x^2 + x + 1 \)
\( = 3x^2 (x + 1) + 1 (x + 1) \)
\( = (x + 1) (3x^2 + 1) \)
In simple words: To factorize this expression, we group the terms into pairs. We take out the common factor from each pair, and then factor out the common binomial. This breaks down the expression into simpler products.

🎯 Exam Tip: For four-term polynomials, try grouping terms. Look for a common factor in the first two terms and the last two terms to create a common binomial factor.

 

Question 2. Find common factor of \( 2(x + y) + 3(x + y) + 5(x + y) \)
Answer:
Common factor is \( (x + y) \)
In simple words: The common factor is the part that appears in all terms of the expression. In this case, \( (x+y) \) is present in every part, so it is the common factor.

🎯 Exam Tip: Identify the entire expression or binomial that is common to all terms. This common part can be factored out, simplifying the expression significantly.

 

Question 3. Simplify \( 2x(2x^2 + 2x – 9) \).
Answer:
\( 2x (2x^2 + 2x – 9) \)
\( = 2x \times 2x^2 + 2x \times 2x - 2x \times 9 \)
\( = 4x^3 + 4x^2 – 18x \)
In simple words: To simplify, we multiply the term outside the bracket by each term inside the bracket. This process is called distribution, and it removes the brackets, giving a single polynomial.

🎯 Exam Tip: Remember to distribute the outside term to *every* term inside the parentheses, paying close attention to signs and exponents. \( x \times x^2 = x^3 \).

 

Question 4. Simplify \( (x + 2) (x + 3) \).
Answer:
\( (x + 2) (x + 3) \)
\( = x^2 + (2 + 3)x + 2 \times 3 \)
\( = x^2 + 5x + 6 \)
In simple words: To simplify this multiplication, use the FOIL method (First, Outer, Inner, Last) or the algebraic identity \( (x+a)(x+b) \). This expands the two binomials into a trinomial.

🎯 Exam Tip: Apply the identity \( (x+a)(x+b) = x^2 + (a+b)x + ab \) or the FOIL method. Ensure all four multiplications are performed correctly and like terms are combined.

 

Question 5. Find the factor of \( 4x^2 - a^2 \).
Answer:
\( 4x^2 - a^2 \)
\( = (2x)^2 - (a)^2 \)
\( = (2x - a)(2x + a) \)
In simple words: This expression is a difference of squares. We write both terms as perfect squares and then use the formula \( A^2 - B^2 = (A-B)(A+B) \) to find the factors.

🎯 Exam Tip: Recognize \( 4x^2 \) as \( (2x)^2 \). This is a classic difference of squares problem. The formula \( A^2 - B^2 = (A-B)(A+B) \) is key here.

 

Question 6. Factorize \( (2x + 3y)^2 - 5(2x + 3y) – 14 \).
Answer:
Let \( 2x + 3y = a \)
then \( (2x + 3y)^2 – 5(2x + 3y) – 14 \)
\( = a^2 - 5a - 14 \)
\( = a^2 - 7a + 2a - 14 \)
\( = a(a – 7) + 2(a – 7) \)
\( = (a - 7) (a + 2) \)
\( = (2x + 3y - 7) (2x + 3y + 2) \)
In simple words: To simplify this, we use a substitution, letting \( (2x+3y) \) be 'a'. This turns the expression into a simpler quadratic which can be factored easily. After factoring, we replace 'a' with its original expression.

🎯 Exam Tip: When a complex expression repeats, use substitution (e.g., let the repeated part be 'a'). This converts the problem into a simpler form, which is easier to factorize, then substitute back.

VI. Questions

 

Question 7. If \( {x}^{2}+\frac{1}{{x}^{2}} =62 \) then find the value of \( \left(x+\frac{1}{x}\right) \)
Answer:
We know the identity: \( (x+\frac{1}{x})^2 = x^2 + \frac{1}{x^2} + 2 \)
This identity helps us relate the given expression to what we need to find.
\( \implies (x+\frac{1}{x})^2 = 62 + 2 \)
\( \implies (x+\frac{1}{x})^2 = 64 \)
\( \implies x+\frac{1}{x} = \sqrt{64} \)
\( \implies x+\frac{1}{x} = 8 \)
In simple words: We use an algebraic identity that links \( (x+\frac{1}{x})^2 \) to \( x^2 + \frac{1}{x^2} \). By substituting the given value, we can easily find the square root to get the answer.

🎯 Exam Tip: Recognize that \( (x+\frac{1}{x})^2 \) can be expanded. This identity is crucial for solving problems where \( x^2 + \frac{1}{x^2} \) is given, and you need to find \( x+\frac{1}{x} \).

 

Question 8. If \( x^2 - x - 42 = (x + k) (x + 6) \) then find the value of k.
Answer:
We have the equation: \( x^2 - x - 42 = (x + k) (x + 6) \)
First, we factorize the left side of the equation:
\( x^2 - x - 42 \)
\( = x^2 - 7x + 6x - 42 \)
\( = x(x - 7) + 6(x - 7) \)
\( = (x - 7) (x + 6) \)
Now, we compare this with the right side: \( (x - 7) (x + 6) = (x + k) (x + 6) \)
By comparing the two expressions, we can see that the value of k must be -7.
In simple words: We factorize the trinomial on the left side of the equation. Once factored, we match it with the expression on the right side to easily find the value of 'k'.

🎯 Exam Tip: Factorize the quadratic expression completely. Then, compare the factors with the given expression to quickly identify the unknown constant 'k'.

 

Question 9. Factorize the following expressions : [Solve any two]
(i) \( 5pq + 3q^2 + 5p + 3q \)
(ii) \( a^2 - 5a + 6 \)
(iii) \( p^4 - 81 \)
Answer:
(i) \( 5pq + 3q^2 + 5p + 3q \)
\( = q(5p + 3q) + 1(5p + 3q) \)
\( = (5p + 3q)(q + 1) \)
(ii) \( a^2 - 5a + 6 \)
\( = a^2 - 3a - 2a + 6 \)
\( = a(a - 3) - 2(a – 3) \)
\( = (a - 3) (a – 2) \).
(iii) \( p^4 - 81 \)
\( = (p^2)^2 – (9)^2 \)
This uses the identity \( A^2 - B^2 = (A + B) (A – B) \).
\( = (p^2 + 9) (p^2 – 9) \)
For the second factor, \( (p^2 - 9) \), we apply the identity again, since \( 9 = 3^2 \).
\( = (p^2 + 9) [(p)^2 - (3)^2] \)
\( = (p^2 + 9) (p + 3) (p – 3) \).
In simple words: For part (i), group terms and factor out common parts. For part (ii), find two numbers that multiply to 6 and add to -5 to factor the trinomial. For part (iii), use the difference of squares identity twice to break down the expression completely.

🎯 Exam Tip: For expressions with four terms, try grouping. For trinomials, find two numbers that multiply to the constant term and add to the middle term's coefficient. For difference of squares, remember to apply the identity repeatedly until no more squares can be factored.