RBSE Solutions Class 8 Maths Chapter 9 Algebraic Expressions Important Questions

Rajasthan Board RBSE Class 8 Maths Chapter 9 Algebraic Expressions Additional Questions

I. Objective Type Questions

Question 1. The product of xy, yz and zx is
(a) 2xyz
(b) \( x^2y^2z^2 \)
(c) \( xy + yz + zx \)
(d) \( x^3y^3z^3 \)
Answer: (b) \( x^2y^2z^2 \)
In simple words: When you multiply \( xy \), \( yz \), and \( zx \) together, you combine all the variables. Each variable, \( x, y, \) and \( z \), appears twice, so their powers become 2.

🎯 Exam Tip: Remember that when multiplying variables, you add their exponents. For example, \( x \times x = x^2 \).

Question 2. Coefficient of \( x^5 \) in \( \frac{8}{5}x^5 \) is
(a) \( \frac{8}{5} \)
(b) 5
(c) 8
(d) \( \frac{5}{8} \)
Answer: (a) \( \frac{8}{5} \)
In simple words: The coefficient is the number that is multiplied by the variable part. In this case, \( \frac{8}{5} \) is multiplied by \( x^5 \).

🎯 Exam Tip: The coefficient includes the sign and any fraction or decimal part attached to the variable.

Question 3. Coefficient of yz in \( \frac{15}{16} xyz \) is
(a) \( \frac{15}{12} \)
(b) \( \frac{12x}{16} \)
(c) \( 16y \)
(d) \( \frac{15}{16}x \)
Answer: (d) \( \frac{15}{16}x \)
In simple words: To find the coefficient of \( yz \), we look at what's left after taking \( yz \) out of the term. What remains is \( \frac{15}{16}x \).

🎯 Exam Tip: When asked for the coefficient of a part of a term, remove that part and the rest of the term (including numbers and other variables) is the coefficient.

Question 4. The number of terms in the expression \( 8x^2 + 2xy + 3x^2 + 2y^2 + 2x^2 \) is
(a) 3
(b) 2
(c) 5
(d) 4
Answer: (a) 3
In simple words: First, combine all the terms that are alike. We have \( 8x^2, 3x^2, \) and \( 2x^2 \), which add up to \( 13x^2 \). Then we are left with \( 13x^2 + 2xy + 2y^2 \), which has 3 different terms.

🎯 Exam Tip: Always simplify the expression by combining like terms before counting the number of terms. Like terms have the same variables raised to the same powers.

Question 6. The product of \( 3x(x + y) \) is
(a) \( 3x^2 + 3xy \)
(b) \( 3x^2 + y \)
(c) \( x^2 + 3xy \)
(d) \( 3x + 3y \)
Answer: (a) \( 3x^2 + 3xy \)
In simple words: To multiply, you give \( 3x \) to both \( x \) and \( y \) inside the bracket. So, \( 3x \times x \) becomes \( 3x^2 \), and \( 3x \times y \) becomes \( 3xy \).

🎯 Exam Tip: Remember to apply the distributive property correctly, multiplying the term outside the parenthesis by every term inside it.

Question 7. Subtracting \( 2a + 3b \) from \( 5a + 2b \), we get
(a) \( 3a + 2b \)
(b) \( 2a + 3b \)
(c) \( 3a - b \)
(d) \( 5a - 3b \)
Answer: (c) \( 3a - b \)
In simple words: When you subtract, change the signs of the terms being subtracted, then combine the \( a \) terms and the \( b \) terms separately. So, \( 5a - 2a = 3a \) and \( 2b - 3b = -b \).

🎯 Exam Tip: "Subtract A from B" means B - A. Be careful with signs, especially when subtracting negative terms.

Question 8. The product of \( pq + qr + 2p \) and 0 is
(a) 0
(b) 1
(c) pqr
(d) \( pq + qr + rp \)
Answer: (a) 0
In simple words: Any number or expression multiplied by zero always results in zero. This is a basic rule of multiplication.

🎯 Exam Tip: This is a trick question! The "zero property of multiplication" is a fundamental concept – anything times zero is zero.

Question 9. Like term of expression \( 7x^2y \) is
(a) \( 7xy \)
(b) \( -10x^2y \)
(c) 7
(d) \( 7x^2 \)
Answer: (b) \( -10x^2y \)
In simple words: Like terms have the same variables with the same powers. The numbers in front (coefficients) can be different. Here, \( -10x^2y \) has the same \( x^2y \) part as \( 7x^2y \).

🎯 Exam Tip: When identifying like terms, focus only on the variable part and its exponents; the coefficient (the number in front) can be any real number.

II. Fill In The Blanks

Question 1. The coefficient of \( x^4 \) in \( \frac{2}{8}x^4 \) is __
Answer: \( \frac{2}{8} \)
In simple words: The coefficient is the number multiplying the variable. For \( x^4 \), the number multiplied is \( \frac{2}{8} \).

🎯 Exam Tip: Always remember that the coefficient includes any fractional part attached to the variable term.

Question 2. The solution of \( 8xyz - 5xyz \) is __
Answer: \( 3xyz \)
In simple words: Since both terms have the exact same variables \( xyz \), we can just subtract the numbers in front. So, \( 8 - 5 = 3 \).

🎯 Exam Tip: You can only add or subtract like terms. If the variable parts are identical, simply perform the operation on their coefficients.

Question 3. The sum of \( 8x^2 + 2x \) and \( 4x + 2 \) is __
Answer: \( 8x^2 + 6x + 2 \)
In simple words: To find the sum, we combine the like terms. We add \( 2x \) and \( 4x \) to get \( 6x \). The \( 8x^2 \) and \( 2 \) have no like terms, so they stay as they are.

🎯 Exam Tip: When adding polynomials, group like terms together (terms with the same variable and exponent) and then add their coefficients.

Question 4. The product of \( 2x, 4x \) and \( \frac{2}{3}x \) is __
Answer: \( \frac{16x^3}{3} \)
In simple words: Multiply all the numbers together: \( 2 \times 4 \times \frac{2}{3} = \frac{16}{3} \). Then multiply all the \( x \) terms: \( x \times x \times x = x^3 \).

🎯 Exam Tip: When multiplying monomials, multiply the coefficients first, then multiply the variables, adding their exponents if they are the same base.

Question 5. \( x^2 + (a + b)x + ab = (x + a) ___ \)
Answer: \( (x + b) \)
In simple words: This is a standard algebraic identity. The expression \( x^2 + (a + b)x + ab \) can be factored into \( (x + a)(x + b) \).

🎯 Exam Tip: Recognize the standard factorization for quadratic trinomials in the form \( x^2 + (sum)x + (product) \).

III. True/False Type Questions

Question 3. \( (x + a)(x + b) = x^2 + (a + b)x + ab \) is an identity.
Answer: True
In simple words: An identity is an equation that is true for all possible values of the variables. This equation is a known algebraic identity because if you multiply out the left side, you get the right side.

🎯 Exam Tip: An identity holds true for any value of the variables, while an equation is true only for specific values.

Question 4. 1 and 100 are homogeneous expressions.
Answer: True
In simple words: A homogeneous expression means all terms have the same degree. Constant terms (like 1 or 100) are considered to have a degree of zero, making them homogeneous with each other.

🎯 Exam Tip: The degree of a constant term is 0. All constant terms are considered homogeneous expressions of degree 0.

IV. Matching Type Questions

Question 1. Match \( 4x \times 5y \times 7z \) with its correct pair.
(a) \( 140xyz \)
(b) \( 0 \)
(c) \( a^2 - b^2 \)
(d) \( 1 \)
Answer: (a) \( 140xyz \)
In simple words: To find the product, multiply the numbers \( 4 \times 5 \times 7 \) to get 140. Then, combine the variables \( x, y, \) and \( z \).

🎯 Exam Tip: When multiplying monomials, multiply coefficients and then list variables in alphabetical order.

Question 2. Match the coefficient of x in \( 1 + x + x^2 \) with its correct pair.
(a) \( 140xyz \)
(b) \( 0 \)
(c) \( a^2 - b^2 \)
(d) \( 1 \)
Answer: (d) \( 1 \)
In simple words: The term with \( x \) is just \( x \). When no number is written in front of a variable, the coefficient is always 1.

🎯 Exam Tip: A variable term like \( x \) implicitly has a coefficient of 1, just as \( -x \) has a coefficient of -1.

Question 3. Match the sum of \( ab - bc, bc - ca, ca - ab \) with its correct pair.
(a) \( 140xyz \)
(b) \( 0 \)
(c) \( a^2 - b^2 \)
(d) \( 1 \)
Answer: (b) \( 0 \)
In simple words: When you add all these terms, you will see that \( ab \) and \( -ab \) cancel each other out, \( -bc \) and \( bc \) cancel out, and \( -ca \) and \( ca \) cancel out. Everything cancels, leaving zero.

🎯 Exam Tip: Identify and group like terms (terms with the exact same variable part) and then sum their coefficients. Opposite terms will cancel out.

Question 4. Match \( (a + b)(a - b) \) with its correct pair.
(a) \( 140xyz \)
(b) \( 0 \)
(c) \( a^2 - b^2 \)
(d) \( 1 \)
Answer: (c) \( a^2 - b^2 \)
In simple words: This is a very common algebraic identity called the difference of squares. When you multiply a sum and a difference of the same two terms, the middle terms cancel out.

🎯 Exam Tip: Memorize the identity \( (a+b)(a-b) = a^2-b^2 \) as it is frequently used in algebraic simplification and factorization.

Question 1. Find the coefficient of \( x^2 \) in \( -\frac{3}{7}x^2y^2 \).
Answer: The coefficient of \( x^2 \) in \( -\frac{3}{7}x^2y^2 \) is \( -\frac{3}{7}y^2 \). When looking for the coefficient of a specific variable part, the remaining factors, including numbers and other variables, form the coefficient.
In simple words: The number and other letters multiplied with \( x^2 \) are \( -\frac{3}{7}y^2 \). So, this is the coefficient.

🎯 Exam Tip: Remember to include the sign and all other factors (numbers and variables) that are multiplied by the specified variable part when determining the coefficient.

Question 2. Write the number of terms in \( 5xy + 2xz + 3xy + x^2 + y^2 \).
Answer: First, we combine the like terms in the expression: \( 5xy + 2xz + 3xy + x^2 + y^2 \). The terms \( 5xy \) and \( 3xy \) are like terms. Adding them gives \( (5+3)xy = 8xy \). So, the simplified expression becomes \( 8xy + 2xz + x^2 + y^2 \). Each part of this new expression is a different term. Therefore, the number of terms in the simplified expression is 4.
In simple words: Combine \( 5xy \) and \( 3xy \) to get \( 8xy \). Then the terms are \( 8xy, 2xz, x^2, \) and \( y^2 \). There are 4 different terms.

🎯 Exam Tip: Always simplify an algebraic expression by combining all like terms before counting the number of terms.

Question 3. Find the sum \( \frac{5}{6}x + \frac{1}{6}x \).
Answer: To find the sum of \( \frac{5}{6}x + \frac{1}{6}x \), since both terms have the same variable part \( x \) and the same denominator, we can add their numerators directly: \( \frac{5x + x}{6} \). This simplifies to \( \frac{6x}{6} \). Finally, dividing \( 6x \) by 6 gives \( x \). So, the sum is \( x \). This operation is similar to adding regular fractions.
In simple words: Both terms have \( x \) and a common bottom number (6). So, add the top numbers: \( 5x + x = 6x \). Then divide by 6: \( \frac{6x}{6} = x \).

🎯 Exam Tip: When adding or subtracting fractions with the same denominator and variable, simply add or subtract the numerators and keep the common denominator and variable.

Question 4. Find the sum of \( (2x + 7), (4x - 2) \) and \( (6x + 4) \).
Answer: To find the sum, we add the expressions by combining like terms: \( (2x + 7) + (4x - 2) + (6x + 4) \) First, add the \( x \) terms: \( 2x + 4x + 6x = 12x \). Next, add the constant terms: \( 7 - 2 + 4 = 5 + 4 = 9 \). Combining these, the sum is \( 12x + 9 \). This method is often preferred for clarity when adding multiple expressions. \( \begin{array}{r} 2x + 7 \\ 4x - 2 \\ + \quad 6x + 4 \\ \hline 12x + 9 \end{array} \)
In simple words: Add all the numbers with \( x \) together: \( 2x + 4x + 6x = 12x \). Then, add all the plain numbers together: \( 7 - 2 + 4 = 9 \). The total sum is \( 12x + 9 \).

🎯 Exam Tip: Arrange the polynomials in columns with like terms aligned vertically for easier addition and subtraction, ensuring you account for all signs.

Question 5. Find the product of \( (x^2 + 2x) \) and \( (2x + 3) \).
Answer: Solution not provided in source.
In simple words: The steps to multiply these two expressions were not given in the provided content.

🎯 Exam Tip: To find the product of two binomials, use the distributive property, multiplying each term in the first expression by each term in the second, then combine like terms.

Question 6. Find the value of \( 3x^2 + 4xy + 2y^2 \) if \( x = 5 \) and \( y = 2 \).
Answer: We need to substitute \( x = 5 \) and \( y = 2 \) into the expression \( 3x^2 + 4xy + 2y^2 \). Substitute the values: \( = 3(5)^2 + 4(5)(2) + 2(2)^2 \) Calculate the squares: \( = 3(25) + 4(10) + 2(4) \) Perform the multiplications: \( = 75 + 40 + 8 \) Finally, add the numbers: \( = 123 \) So, the value of the expression is 123. Always follow the order of operations (PEMDAS/BODMAS).
In simple words: Put the numbers \( x=5 \) and \( y=2 \) into the expression. Work out the powers first, then multiply, then add. The final answer is 123.

🎯 Exam Tip: When substituting values into an expression, always use parentheses around the substituted numbers to avoid errors with signs and exponents.

Question 7. Find the product: \( (3x + 5)(5x - 3) \).
Answer: To find the product \( (3x + 5)(5x - 3) \), we use the distributive property (FOIL method). Multiply each term of the first binomial by each term of the second binomial: First terms: \( 3x \times 5x = 15x^2 \) Outer terms: \( 3x \times (-3) = -9x \) Inner terms: \( 5 \times 5x = 25x \) Last terms: \( 5 \times (-3) = -15 \) Now, add these results and combine like terms: \( 15x^2 - 9x + 25x - 15 \) Combine the \( x \) terms: \( -9x + 25x = 16x \) So, the final product is \( 15x^2 + 16x - 15 \). This method ensures all terms are correctly multiplied and combined.
In simple words: Multiply the first parts, then the outside parts, then the inside parts, then the last parts. After that, add the terms that are alike. The answer is \( 15x^2 + 16x - 15 \).

🎯 Exam Tip: The FOIL method (First, Outer, Inner, Last) is a helpful mnemonic for multiplying two binomials, but remember it's just a specific application of the distributive property.

Question 8. If \( x = \frac{1}{2}, y = \frac{2}{3} \) and \( z = \frac{1}{3} \) then find the value of \( \frac{1}{8}xyz \).
Answer: To find the value of \( \frac{1}{8}xyz \), we substitute the given values of \( x, y, \) and \( z \): \( \frac{1}{8}xyz = \frac{1}{8} \times \frac{1}{2} \times \frac{2}{3} \times \frac{1}{3} \) Multiply the numerators and denominators: Numerator: \( 1 \times 1 \times 2 \times 1 = 2 \) Denominator: \( 8 \times 2 \times 3 \times 3 = 144 \) So, the expression becomes \( \frac{2}{144} \). Now, simplify the fraction by dividing both the numerator and denominator by their greatest common divisor, which is 2: \( \frac{2 \div 2}{144 \div 2} = \frac{1}{72} \) Thus, the value of \( \frac{1}{8}xyz \) is \( \frac{1}{72} \). Multiplying fractions involves multiplying straight across.
In simple words: Put the numbers for \( x, y, \) and \( z \) into the expression. Multiply all the top numbers together, and all the bottom numbers together. Then, simplify the fraction you get. The answer is \( \frac{1}{72} \).

🎯 Exam Tip: Always simplify fractions to their lowest terms after performing multiplications to get the final answer. Look for common factors in the numerator and denominator.

VI. Short Answer Type Questions

Question 1. Simplify \( (x - \frac{1}{x})(x + \frac{1}{x})(x^2 + \frac{1}{x^2}) \).
Answer: We need to simplify the expression \( (x - \frac{1}{x})(x + \frac{1}{x})(x^2 + \frac{1}{x^2}) \). First, consider the product of the first two terms: \( (x - \frac{1}{x})(x + \frac{1}{x}) \). This is in the form of \( (a - b)(a + b) = a^2 - b^2 \), where \( a = x \) and \( b = \frac{1}{x} \). So, \( (x - \frac{1}{x})(x + \frac{1}{x}) = x^2 - (\frac{1}{x})^2 = x^2 - \frac{1}{x^2} \). Now, substitute this back into the original expression: \( (x^2 - \frac{1}{x^2})(x^2 + \frac{1}{x^2}) \) Again, this is in the form of \( (a - b)(a + b) = a^2 - b^2 \), where \( a = x^2 \) and \( b = \frac{1}{x^2} \). So, \( (x^2 - \frac{1}{x^2})(x^2 + \frac{1}{x^2}) = (x^2)^2 - (\frac{1}{x^2})^2 \). This simplifies to \( x^4 - \frac{1}{x^4} \). Applying identities step-by-step simplifies complex products efficiently.
In simple words: We use the identity \( (A - B)(A + B) = A^2 - B^2 \) two times. First, apply it to \( (x - \frac{1}{x})(x + \frac{1}{x}) \) to get \( x^2 - \frac{1}{x^2} \). Then apply the identity again to \( (x^2 - \frac{1}{x^2})(x^2 + \frac{1}{x^2}) \) to get \( x^4 - \frac{1}{x^4} \).

🎯 Exam Tip: Recognize and apply the difference of squares identity \( (a-b)(a+b) = a^2-b^2 \) repeatedly to simplify complex products of terms.

Question 2. If \( x + \frac{1}{x} = 7 \), then find the value of \( x^2 + \frac{1}{x^2} \).
Answer: We are given \( x + \frac{1}{x} = 7 \). To find \( x^2 + \frac{1}{x^2} \), we can square both sides of the given equation. \( (x + \frac{1}{x})^2 = 7^2 \) Using the identity \( (a + b)^2 = a^2 + 2ab + b^2 \), where \( a = x \) and \( b = \frac{1}{x} \): \( x^2 + 2(x)(\frac{1}{x}) + (\frac{1}{x})^2 = 49 \) The middle term simplifies: \( 2(x)(\frac{1}{x}) = 2 \). So, the equation becomes: \( x^2 + 2 + \frac{1}{x^2} = 49 \) Now, subtract 2 from both sides to isolate \( x^2 + \frac{1}{x^2} \): \( x^2 + \frac{1}{x^2} = 49 - 2 \)
\( \implies x^2 + \frac{1}{x^2} = 47 \). Squaring both sides is a common technique to find sums of squares.
In simple words: We are given \( x + \frac{1}{x} = 7 \). If we square both sides, we get \( (x + \frac{1}{x})^2 = 7^2 \). This expands to \( x^2 + 2(x)(\frac{1}{x}) + (\frac{1}{x})^2 = 49 \). The middle part becomes 2. So, \( x^2 + 2 + \frac{1}{x^2} = 49 \). Take 2 to the other side: \( x^2 + \frac{1}{x^2} = 49 - 2 = 47 \).

🎯 Exam Tip: When given an expression like \( x + \frac{1}{x} \) or \( x - \frac{1}{x} \) and asked for the value of \( x^2 + \frac{1}{x^2} \), square both sides of the given equation and use the appropriate algebraic identity.

Question 3. If \( 2(a^2 + b^2) = (a + b)^2 \) then prove that \( a = b \).
Answer: We are given the equation \( 2(a^2 + b^2) = (a + b)^2 \). We need to prove that \( a = b \). First, expand the right side of the equation using the identity \( (a + b)^2 = a^2 + 2ab + b^2 \): \( 2(a^2 + b^2) = a^2 + 2ab + b^2 \) Distribute the 2 on the left side: \( 2a^2 + 2b^2 = a^2 + 2ab + b^2 \) Now, move all terms to one side to set the equation to zero: \( 2a^2 - a^2 + 2b^2 - b^2 - 2ab = 0 \) Combine the like terms: \( a^2 + b^2 - 2ab = 0 \) Recognize that \( a^2 - 2ab + b^2 \) is the expansion of \( (a - b)^2 \): \( (a - b)^2 = 0 \) To make a square equal to zero, the base must be zero: \( a - b = 0 \)
\( \implies a = b \). This shows that the given condition implies \( a \) must be equal to \( b \).
In simple words: Start with the given equation. Expand the right side using \( (a+b)^2 = a^2 + 2ab + b^2 \). Then move all terms to one side. You will get \( a^2 - 2ab + b^2 = 0 \), which is the same as \( (a - b)^2 = 0 \). If a square is zero, the thing inside must be zero, so \( a - b = 0 \), which means \( a = b \).

🎯 Exam Tip: When proving identities or relationships, always expand and simplify both sides of the equation, then manipulate the terms to reach the desired conclusion, often by recognizing perfect square identities.

Question 4. What is the value of a if \( 2x^2 + x - a \) is equal to 5 and \( x = 0 \).
Answer: We are given the expression \( 2x^2 + x - a \). We know that when \( x = 0 \), the expression is equal to 5. Substitute \( x = 0 \) into the expression: \( 2(0)^2 + (0) - a = 5 \) Simplify the terms: \( 2(0) + 0 - a = 5 \) \( 0 + 0 - a = 5 \) \( -a = 5 \) To find the value of \( a \), multiply both sides by -1:
\( \implies a = -5 \). This direct substitution is key for finding unknown constants.
In simple words: Replace \( x \) with 0 in the expression \( 2x^2 + x - a \). We are told the answer is 5. So, \( 2(0)^2 + 0 - a = 5 \). This simplifies to \( -a = 5 \), which means \( a = -5 \).

🎯 Exam Tip: When an expression's value is given for a specific variable value, always substitute the given variable value directly into the expression to solve for any unknown constants.

Question 5. Simplify the expression \( 2(a^2 + ab) + 3 - ab \) and find its value when \( a = 5 \) and \( b = -3 \).
Answer: First, simplify the expression \( 2(a^2 + ab) + 3 - ab \). Distribute the 2 into the parenthesis: \( 2a^2 + 2ab + 3 - ab \) Combine the like terms \( 2ab \) and \( -ab \): \( 2ab - ab = ab \) So, the simplified expression is \( 2a^2 + ab + 3 \). Now, substitute \( a = 5 \) and \( b = -3 \) into the simplified expression: \( 2(5)^2 + (5)(-3) + 3 \) Calculate \( (5)^2 = 25 \): \( 2(25) + (-15) + 3 \) Perform the multiplications: \( 50 - 15 + 3 \) Perform the additions/subtractions from left to right: \( 35 + 3 = 38 \) Thus, the value of the expression is 38. Simplifying first often makes substitution easier.
In simple words: First, simplify the expression: \( 2a^2 + 2ab + 3 - ab \) becomes \( 2a^2 + ab + 3 \). Then, put \( a = 5 \) and \( b = -3 \) into this simpler expression. Calculate \( 2(5)^2 + (5)(-3) + 3 \), which works out to \( 50 - 15 + 3 = 38 \).

🎯 Exam Tip: Always simplify an algebraic expression as much as possible before substituting values to reduce the chances of calculation errors.

Question 6. Using proper identity, find the value of \( (1.2)^2 - (0.8)^2 \).
Answer: We need to find the value of \( (1.2)^2 - (0.8)^2 \) using a proper identity. This expression is in the form of the difference of squares identity, \( a^2 - b^2 = (a + b)(a - b) \). Here, \( a = 1.2 \) and \( b = 0.8 \). Substitute these values into the identity: \( (1.2)^2 - (0.8)^2 = (1.2 + 0.8)(1.2 - 0.8) \) Perform the additions and subtractions inside the parentheses: \( (1.2 + 0.8) = 2.0 \) \( (1.2 - 0.8) = 0.4 \) Now, multiply these two results: \( = (2.0)(0.4) \) \( = 0.8 \) So, the value of \( (1.2)^2 - (0.8)^2 \) is 0.8. Using identities can greatly speed up calculations.
In simple words: We use the special rule \( a^2 - b^2 = (a + b)(a - b) \). Here, \( a=1.2 \) and \( b=0.8 \). So, it becomes \( (1.2 + 0.8) \times (1.2 - 0.8) \). This is \( 2.0 \times 0.4 \), which equals \( 0.8 \).

🎯 Exam Tip: When evaluating expressions involving squares of decimals or larger numbers, the difference of squares identity often provides a much quicker and simpler method than direct squaring and subtraction.

Question 7. Using suitable identities, evaluate \( 201 \times 202 \).
Answer: To evaluate \( 201 \times 202 \) using suitable identities, we can rewrite the numbers in terms of a base value, like 200. So, \( 201 \) can be written as \( (200 + 1) \). And \( 202 \) can be written as \( (200 + 2) \). The product becomes \( (200 + 1)(200 + 2) \). This is in the form of the identity \( (x + a)(x + b) = x^2 + (a + b)x + ab \). Here, \( x = 200, a = 1, \) and \( b = 2 \). Substitute these values into the identity: \( (200)^2 + (1 + 2)(200) + (1 \times 2) \) Calculate the terms: \( (200)^2 = 40000 \) \( (1 + 2)(200) = 3 \times 200 = 600 \) \( (1 \times 2) = 2 \) Add these results: \( 40000 + 600 + 2 = 40602 \) Therefore, \( 201 \times 202 = 40602 \). Using identities simplifies mental math and reduces calculation errors.
In simple words: We can write \( 201 \) as \( (200 + 1) \) and \( 202 \) as \( (200 + 2) \). Then use the rule \( (x + a)(x + b) = x^2 + (a + b)x + ab \). So, \( 200^2 + (1+2)200 + (1 \times 2) \). This is \( 40000 + 600 + 2 \), which is \( 40602 \).

🎯 Exam Tip: When multiplying numbers close to a round figure (like 10, 100, 200), use the identity \( (x+a)(x+b) \) to make calculations simpler and faster.

Question 8. To get \( -x^2 - y^2 + 6xy + 20 \), what is to be subtracted from \( 3x^2 - 4y^2 + 5xy + 20 \)?
Answer: Let the expression to be subtracted be P. According to the problem, we have: \( (3x^2 - 4y^2 + 5xy + 20) - P = -x^2 - y^2 + 6xy + 20 \) To find P, we rearrange the equation: \( P = (3x^2 - 4y^2 + 5xy + 20) - (-x^2 - y^2 + 6xy + 20) \) Now, remove the parentheses, remembering to change the signs of all terms inside the second parenthesis due to the minus sign in front: \( P = 3x^2 - 4y^2 + 5xy + 20 + x^2 + y^2 - 6xy - 20 \) Group the like terms together: \( P = (3x^2 + x^2) + (-4y^2 + y^2) + (5xy - 6xy) + (20 - 20) \) Combine the like terms: \( P = 4x^2 - 3y^2 - xy + 0 \) So, the expression that needs to be subtracted is \( 4x^2 - 3y^2 - xy \). Careful handling of signs is crucial in subtraction.
In simple words: We want to find what to subtract from the first expression to get the second. This means we subtract the second expression from the first. So, \( (3x^2 - 4y^2 + 5xy + 20) - (-x^2 - y^2 + 6xy + 20) \). Change the signs of the second expression and then combine like terms. This gives \( 4x^2 - 3y^2 - xy \).

🎯 Exam Tip: When subtracting algebraic expressions, distribute the negative sign to *every* term within the parentheses being subtracted, then combine like terms carefully.

Question 9. Simplify \( (a + b)(a - b) - (a^2 + b^2) \).
Answer: We need to simplify the expression \( (a + b)(a - b) - (a^2 + b^2) \). First, use the identity for the product of the first two terms: \( (a + b)(a - b) = a^2 - b^2 \). Substitute this into the expression: \( (a^2 - b^2) - (a^2 + b^2) \) Now, remove the parentheses. Remember to change the signs of the terms inside the second parenthesis because of the minus sign in front: \( a^2 - b^2 - a^2 - b^2 \) Group the like terms: \( (a^2 - a^2) + (-b^2 - b^2) \) Combine the like terms: \( 0 + (-2b^2) \) So, the simplified expression is \( -2b^2 \). Applying identities and managing signs are essential steps.
In simple words: First, \( (a + b)(a - b) \) becomes \( a^2 - b^2 \) (using the difference of squares rule). So the expression is \( (a^2 - b^2) - (a^2 + b^2) \). Remove the brackets, changing signs after the minus sign: \( a^2 - b^2 - a^2 - b^2 \). The \( a^2 \) terms cancel out, and \( -b^2 - b^2 \) makes \( -2b^2 \).

🎯 Exam Tip: Always prioritize applying algebraic identities to simplify parts of an expression, then carefully handle the signs when removing parentheses, especially after a subtraction sign.

Question 10. Using identity \( a^2 - b^2 = (a + b)(a - b) \), find the product.
(i) \( (2a + 7)(2a - 7) \)
(ii) \( (p^2 + q^2)(p^2 - q^2) \)
Answer:
(i) For \( (2a + 7)(2a - 7) \), we use the identity \( (a + b)(a - b) = a^2 - b^2 \). Here, \( a = 2a \) and \( b = 7 \). So, \( (2a + 7)(2a - 7) = (2a)^2 - (7)^2 \). Calculate the squares: \( (2a)^2 = 4a^2 \) and \( (7)^2 = 49 \). Thus, \( (2a + 7)(2a - 7) = 4a^2 - 49 \). This is a direct application of the difference of squares.
(ii) For \( (p^2 + q^2)(p^2 - q^2) \), we again use the identity \( (A + B)(A - B) = A^2 - B^2 \). Here, \( A = p^2 \) and \( B = q^2 \). So, \( (p^2 + q^2)(p^2 - q^2) = (p^2)^2 - (q^2)^2 \). Calculate the powers: \( (p^2)^2 = p^{2 \times 2} = p^4 \) and \( (q^2)^2 = q^{2 \times 2} = q^4 \). Thus, \( (p^2 + q^2)(p^2 - q^2) = p^4 - q^4 \). The identity applies even with higher powers.
In simple words:
(i) For \( (2a + 7)(2a - 7) \), it's like \( (A+B)(A-B) \). So the answer is \( (2a)^2 - 7^2 \), which is \( 4a^2 - 49 \).
(ii) For \( (p^2 + q^2)(p^2 - q^2) \), it's also like \( (A+B)(A-B) \). So the answer is \( (p^2)^2 - (q^2)^2 \), which is \( p^4 - q^4 \).

🎯 Exam Tip: The difference of squares identity can be applied to terms that are themselves powers or products, as long as the pattern \( (first + second)(first - second) \) is maintained.

Question 11. Find the product using suitable identity.
(i) \( (5x - 3)(5x + 3) \)
(ii) \( 103 \times 99 \)
Answer:
(i) For \( (5x - 3)(5x + 3) \), we use the identity \( (a - b)(a + b) = a^2 - b^2 \). Here, \( a = 5x \) and \( b = 3 \). So, the product is \( (5x)^2 - (3)^2 \). Calculate the squares: \( (5x)^2 = 25x^2 \) and \( (3)^2 = 9 \). Thus, the product is \( 25x^2 - 9 \). This identity makes multiplication quick and easy.
(ii) For \( 103 \times 99 \), we can express these numbers using a common base, 100. \( 103 = (100 + 3) \) \( 99 = (100 - 1) \) So, the product is \( (100 + 3)(100 - 1) \). This can be seen as an application of the identity \( (x + a)(x + b) = x^2 + (a + b)x + ab \), where \( x = 100, a = 3, \) and \( b = -1 \). So, \( (100)^2 + (3 + (-1))(100) + (3 \times -1) \). Calculate the terms: \( (100)^2 = 10000 \) \( (3 + (-1))(100) = (2)(100) = 200 \) \( (3 \times -1) = -3 \) Add these values: \( 10000 + 200 - 3 = 10197 \). Rewriting numbers in terms of a convenient base is very helpful.
In simple words:
(i) For \( (5x - 3)(5x + 3) \), use the rule \( (A-B)(A+B) = A^2-B^2 \). This means \( (5x)^2 - 3^2 \), which is \( 25x^2 - 9 \).
(ii) For \( 103 \times 99 \), think of them as \( (100 + 3) \) and \( (100 - 1) \). Using the rule \( (x+a)(x+b) = x^2 + (a+b)x + ab \), you get \( 100^2 + (3 - 1)100 + (3 \times -1) \). This is \( 10000 + 200 - 3 \), which is \( 10197 \).

🎯 Exam Tip: For numerical products like \( 103 \times 99 \), consider if they can be written as \( (x+a)(x+b) \) or \( (x+a)(x-a) \) to use identities and simplify calculations.

Question 12. Simplify \( (x - 5)^2 + 10x \).
Answer: We need to simplify \( (x - 5)^2 + 10x \). First, expand \( (x - 5)^2 \) using the identity \( (a - b)^2 = a^2 - 2ab + b^2 \). Here, \( a = x \) and \( b = 5 \). So, \( (x - 5)^2 = x^2 - 2(x)(5) + (5)^2 = x^2 - 10x + 25 \). Now, substitute this expansion back into the original expression: \( (x^2 - 10x + 25) + 10x \) Combine the like terms \( -10x \) and \( +10x \): \( -10x + 10x = 0 \) So, the expression simplifies to \( x^2 + 25 \). Recognizing and applying the square identities is crucial.
In simple words: Expand \( (x - 5)^2 \) using the rule \( (A-B)^2 = A^2 - 2AB + B^2 \). This gives \( x^2 - 10x + 25 \). Now add \( 10x \) to it: \( x^2 - 10x + 25 + 10x \). The \( -10x \) and \( +10x \) cancel out, leaving \( x^2 + 25 \).

🎯 Exam Tip: When simplifying expressions that include squared binomials, always expand the squared term first using the appropriate identity before combining other terms.

Question 13. Using suitable identities, evaluate \( 72 \times 68 \).
Answer: To evaluate \( 72 \times 68 \) using suitable identities, we can express these numbers relative to a convenient base, such as 70. So, \( 72 \) can be written as \( (70 + 2) \). And \( 68 \) can be written as \( (70 - 2) \). The product becomes \( (70 + 2)(70 - 2) \). This is in the form of the difference of squares identity, \( (a + b)(a - b) = a^2 - b^2 \). Here, \( a = 70 \) and \( b = 2 \). Substitute these values into the identity: \( (70)^2 - (2)^2 \) Calculate the squares: \( (70)^2 = 4900 \) \( (2)^2 = 4 \) Perform the subtraction: \( 4900 - 4 = 4896 \) Thus, \( 72 \times 68 = 4896 \). This identity makes complex multiplication straightforward.
In simple words: We can write \( 72 \) as \( (70 + 2) \) and \( 68 \) as \( (70 - 2) \). This looks like \( (A+B)(A-B) \), which is \( A^2 - B^2 \). So, \( 70^2 - 2^2 \). This is \( 4900 - 4 \), which equals \( 4896 \).

🎯 Exam Tip: Look for numbers that are equidistant from a round number. This allows you to apply the difference of squares identity, simplifying the multiplication process significantly.