GSEB Class 8 Maths Solutions Chapter 9 Algebraic Expressions and Identities InText Question

Try These (Page 138)

Question 1. Give five examples of expressions containing one variable and five examples of expressions containing two variables?
Answer:
(a) Here are five examples of expressions containing one variable:
• \( x - 4 \)
• \( 5 - y \)
• \( 2x + 5 \)
• \( 10x + 4 \)
• \( 11 - z \)
(b) Here are five examples of expressions containing two variables:
• \( 5x + 2y - 10 \)
• \( 6x + z - 2 \)
• \( 15y - z \)
• \( 9x - 10z \)
• \( xy \)
In simple words: Expressions can have one letter (variable) like 'x' or 'y', or they can have two different letters like 'x' and 'y' together. We just need to give examples for each kind.

Exam Tip: Remember that a variable is a letter representing an unknown number. An expression with "one variable" means only one type of letter is used throughout (e.g., all 'x's or all 'y's). An expression with "two variables" means two different types of letters are used (e.g., 'x' and 'y').

Question 2. Show on the number line \( x \), \( x - 4 \), \( 2x + 1 \), \( 3x - 2 \).
Answer:
(i) Representation of \( x \) on the number line:

The point X represents the variable \( 'x' \).
(ii) Representation of \( x - 4 \) on the number line:

(iii) Representation of \( 2x + 1 \) on the number line:

(iv) Representation of \( 3x - 2 \) on the number line:

In simple words: To show an expression on a number line, we draw a line with numbers. We mark a point for 'O' (zero) and then another point to stand for the expression, like 'x' or 'x-4'. We label that point with the expression it represents.

Exam Tip: When representing expressions on a number line, draw a clear line with an origin (0) and indicate the direction with arrows. Mark the position of the expression relative to the origin or other known points, ensuring labels are distinct.

Try These (Page 138)

Question 1. Identify the coefficient of each term in the expression \( x^{2}y^{2} - 10x^{2}y + 5xy^{2} - 20 \)?
Answer:
The given expression is \( x^{2}y^{2} - 10x^{2}y + 5xy^{2} - 20 \). The terms involved are \( x^{2}y^{2} \), \( -10x^{2}y \), \( 5xy^{2} \), and \( -20 \).
The coefficient of \( x^{2}y^{2} \) is \( 1 \).
The coefficient of \( x^{2}y \) is \( -10 \).
The coefficient of \( xy^{2} \) is \( 5 \).
The coefficient of the constant term \( -20 \) is \( -20 \) itself.
In simple words: A coefficient is the number multiplied by the variables in a term. If there's no number written, it's a 1. For example, in \( x^{2}y^{2} \), the coefficient is 1. In \( -10x^{2}y \), it's -10. For a number without variables, like -20, that number is its own coefficient.

Exam Tip: Always remember that if a variable term appears without an explicit numerical factor (e.g., \( x^{2}y^{2} \)), its coefficient is 1. Pay close attention to negative signs, as they are part of the coefficient.

Try These (Page 138)

Question 1. Classify the following polynomials as monomials, binomials, trinomials: \( -z + 5 \), \( x + y + z \), \( y + z + 100 \), \( ab - ac \), \( 17 \).
Answer:

Exam Tip: To classify a polynomial, count the number of terms. Terms are separated by addition or subtraction signs. For example, \( x+y+z \) has three terms, making it a trinomial.

Question 2. Construct:
(a) 3 binomials with only \( x \) as a variable
(b) 3 binomials with \( x \) and \( y \) as variable
(c) 3 monomials with \( x \) and \( y \) as variables
(d) 2 polynomials with 4 or more terms
Answer:
(a) 3 binomials with only \( x \) as a variable are:
• \( 5x - 4 \)
• \( 3x + 2 \)
• \( 10 + 2x \)
(b) 3 binomials with \( x \) and \( y \) as variables are:
• \( 2x + 3y \)
• \( 5x - y \)
• \( x - 3y \)
(c) 3 monomials with \( x \) and \( y \) as variables are:
• \( xy \)
• \( -7xy \)
• \( 3x^{2}y \)
(d) 2 polynomials with 4 or more terms are:
• \( 2x^{3} - x^{2} + 6x - 8 \)
• \( 10 + 7x - 2x^{2} + 4x^{3} \)
In simple words: We need to create algebraic expressions based on the number of terms and the types of letters used. 'Binomials' have two terms, 'monomials' have one, and 'polynomials with 4 or more terms' have many parts joined by plus or minus signs.

Exam Tip: When constructing polynomials, pay attention to the number of terms (monomial, binomial, trinomial, or general polynomial) and the specific variables required for each part. Ensure that each example meets all given criteria.

Try These (Page 139)

Question 1. Write two terms which are like:
1. \( 7xy \)
2. \( 4mn^{2} \)
3. \( 21 \)
Answer:
1. Two terms like \( 7xy \) are: \( -3xy \) and \( 8xy \).
2. Two terms like \( 4mn^{2} \) are: \( 6mn^{2} \) and \( -2n^{2}m \).
3. Two terms like \( 21 \) are: \( 51 \) and \( -7 \).
In simple words: 'Like terms' means terms that have the exact same letters and the exact same powers on those letters. Only the numbers in front can be different. For example, \( 7xy \) and \( 8xy \) are like terms because both have \( xy \). Constants (plain numbers) are also like terms to each other.

Exam Tip: To identify like terms, check two things: first, the variables must be identical (e.g., \( xy \) is not like \( x^2y \)); second, the powers of each variable must also be identical. The order of variables (e.g., \( mn^2 \) vs. \( n^2m \)) does not matter as multiplication is commutative.

Try These (Page 143)

Question 1. Find \( 4x \times 5y \times 7z \). First find \( 4x \times 5y \) and multiply it by \( 7z \); or first find \( 5y \times 7z \) and multiply it by \( 4x \). Is the result the same? What do you observe? Does the order in which you carry out the multiplication matter?
Answer:
We have the expression \( 4x \times 5y \times 7z \).
First, let's find \( 4x \times 5y \) and then multiply by \( 7z \):
\( 4x \times 5y \times 7z = (4x \times 5y) \times 7z \)
\( = (4 \times 5 \times x \times y) \times 7z \)
\( = 20xy \times 7z \)
\( = (20 \times 7 \times x \times y \times z) \)
\( = 140xyz \)
Now, let's find \( 5y \times 7z \) first and then multiply by \( 4x \):
\( 4x \times 5y \times 7z = 4x \times (5y \times 7z) \)
\( = 4x \times (5 \times 7 \times y \times z) \)
\( = 4x \times 35yz \)
\( = (4 \times 35 \times x \times y \times z) \)
\( = 140xyz \)
We observe that the result is the same in both cases.
The product of monomials is associative, meaning the order in which we multiply the terms does not change the final product.
In simple words: When you multiply three things, it doesn't matter which two you multiply first. You'll always get the same answer. It's like a special rule for multiplication that means the grouping doesn't change anything.

Exam Tip: This question demonstrates the associative property of multiplication for monomials. Remember that for multiplication, you can group terms in any order without changing the outcome. Multiply the numerical coefficients first, then combine the variables.

Try These (Page 144)

Question 1. Find the product:
1. \( 2x \times (3x + 5xy) \)
2. \( a^{2} \times (2ab - 5c) \)
Answer:
1. \( 2x(3x + 5xy) \)
\( = (2x \times 3x) + (2x \times 5xy) \)
\( = (2 \times 3 \times x \times x) + (2 \times 5 \times x \times x \times y) \)
\( = 6x^{2} + 10x^{2}y \)
2. \( a^{2}(2ab - 5c) \)
\( = (a^{2} \times 2ab) + (a^{2} \times (-5c)) \)
\( = (1 \times 2 \times a^{2} \times a \times b) + (1 \times (-5) \times a^{2} \times c) \)
\( = 2a^{3}b - 5a^{2}c \)
In simple words: To find the product, we multiply the term outside the bracket by each term inside the bracket. We multiply the numbers first and then the letters, adding up the powers for the same letters.

Exam Tip: Use the distributive property for multiplying a monomial by a polynomial. Multiply the numerical coefficients and add the exponents of like variables. Remember to handle negative signs carefully.

Try These (Page 145)

Question 1. Find the product: \( (4p^{2} + 5p + 7) \times 3p \)
Answer:
\( (4p^{2} + 5p + 7) \times 3p \)
\( = (4p^{2} \times 3p) + (5p \times 3p) + (7 \times 3p) \)
\( = [(4 \times 3) \times p^{2} \times p] + [(5 \times 3) \times p \times p] + [(7 \times 3) \times p] \)
\( = 12p^{3} + 15p^{2} + 21p \)
In simple words: To multiply a polynomial by a single term (monomial), distribute the monomial to every term inside the polynomial. Multiply the numbers, and for letters, add their powers.

Exam Tip: Apply the distributive property here. Each term inside the parentheses must be multiplied by the monomial outside. Ensure you correctly apply exponent rules (\( p^2 \times p = p^3 \)).

Question 2. Simplify: \( 2x(x - 1) + 5 \) and find its value at \( x = -1 \).
Answer:
First, simplify the expression: \( 2x(x - 1) + 5 \)
\( = (2x \times x) + (2x \times (-1)) + 5 \)
\( = (2 \times 1 \times x \times x) + (2 \times (-1) \times x) + 5 \)
\( = 2x^{2} - 2x + 5 \)
Now, find its value at \( x = -1 \):
Substitute \( x = -1 \) into the simplified expression:
\( 2(-1)^{2} - 2(-1) + 5 \)
\( = 2(1) - (-2) + 5 \)
\( = 2 + 2 + 5 \)
\( = 9 \)
In simple words: First, we use the distribution rule to make the expression simpler. After that, we put the number \( -1 \) wherever we see 'x' in the simplified expression and then calculate the final answer.

Exam Tip: Always simplify the expression completely before substituting the given value of the variable. This reduces the chances of calculation errors. Be careful with signs, especially when squaring negative numbers or multiplying by negative numbers.

Question 3. Carry out the multiplication of the expressions in each of the following pairs?
1. \( 4p, q + r \)
2. \( ab, a - b \)
3. \( a + b, 7a^{2}b^{2} \)
4. \( a^{2} - 9, 4a \)
5. \( pq + qr + rp, 0 \)
Answer:
1. \( 4p \times (q + r) \)
\( = (4p \times q) + (4p \times r) \)
\( = (4 \times 1 \times p \times q) + (4 \times 1 \times p \times r) \)
\( = 4pq + 4pr \)
2. \( ab \times (a - b) \)
\( = (ab \times a) + (ab \times (-b)) \)
\( = (1 \times 1 \times a \times b \times a) + (1 \times (-1) \times a \times b \times b) \)
\( = a^{2}b - ab^{2} \)
3. \( (a + b) \times 7a^{2}b^{2} \)
\( = (a \times 7a^{2}b^{2}) + (b \times 7a^{2}b^{2}) \)
\( = (1 \times 7 \times a \times a^{2} \times b^{2}) + (1 \times 7 \times b \times a^{2} \times b^{2}) \)
\( = 7a^{3}b^{2} + 7a^{2}b^{3} \)
4. \( (a^{2} - 9) \times 4a \)
\( = (a^{2} \times 4a) + ((-9) \times 4a) \)
\( = (1 \times 4 \times a^{2} \times a) + ((-9) \times 4 \times a) \)
\( = 4a^{3} - 36a \)
5. \( (pq + qr + rp) \times 0 \)
\( = (pq \times 0) + (qr \times 0) + (rp \times 0) \)
\( = 0 + 0 + 0 \)
\( = 0 \)
In simple words: For each pair, we multiply the first expression by the second. If there are brackets, we use the distributive rule, multiplying the term outside by every term inside. Remember that anything multiplied by zero becomes zero.

Exam Tip: When multiplying expressions, apply the distributive property. Multiply coefficients together and add exponents for the same variables. Remember that multiplying any expression by zero always results in zero.

Question 1. Multiply the binomials.
1. \( (2x + 5) \) and \( (4x - 3) \)
2. \( (y - 8) \) and \( (3y - 4) \)
3. \( (2.5l - 0.5m) \) and \( (2.5l + 0.5m) \)
4. \( (a + 3b) \) and \( (x + 5) \)
5. \( (2pq + 3q^{2}) \) and \( (3pq - 2q^{2}) \)
6. \( (\frac{3}{4} a^{2} + 3b^{2}) \) and \( 4 (a^{2} - \frac{2}{3}b^{2}) \)
Answer:
1. Multiply \( (2x + 5) \) by \( (4x - 3) \):
\( (2x + 5) \times (4x - 3) \)
\( = 2x(4x - 3) + 5(4x - 3) \)
\( = (2x \times 4x) - (2x \times 3) + (5 \times 4x) - (5 \times 3) \)
\( = 8x^{2} - 6x + 20x - 15 \)
\( = 8x^{2} + (-6 + 20)x - 15 \)
\( = 8x^{2} + 14x - 15 \)
2. Multiply \( (y - 8) \) by \( (3y - 4) \):
\( (y - 8) \times (3y - 4) \)
\( = y(3y - 4) - 8(3y - 4) \)
\( = (y \times 3y) - (y \times 4) - (8 \times 3y) - (8 \times (-4)) \)
\( = 3y^{2} - 4y - 24y + 32 \)
\( = 3y^{2} - 28y + 32 \)
3. Multiply \( (2.5l - 0.5m) \) by \( (2.5l + 0.5m) \):
\( (2.5l - 0.5m) \times (2.5l + 0.5m) \)
\( = 2.5l(2.5l + 0.5m) - 0.5m(2.5l + 0.5m) \)
\( = (2.5l \times 2.5l) + (2.5l \times 0.5m) - (0.5m \times 2.5l) - (0.5m \times 0.5m) \)
\( = 6.25l^{2} + 1.25lm - 1.25lm - 0.25m^{2} \)
\( = 6.25l^{2} + (0)lm - 0.25m^{2} \)
\( = 6.25l^{2} - 0.25m^{2} \)
4. Multiply \( (a + 3b) \) by \( (x + 5) \):
\( (a + 3b) \times (x + 5) \)
\( = a(x + 5) + 3b(x + 5) \)
\( = (a \times x) + (a \times 5) + (3b \times x) + (3b \times 5) \)
\( = ax + 5a + 3bx + 15b \)
5. Multiply \( (2pq + 3q^{2}) \) by \( (3pq - 2q^{2}) \):
\( (2pq + 3q^{2}) \times (3pq - 2q^{2}) \)
\( = 2pq(3pq - 2q^{2}) + 3q^{2}(3pq - 2q^{2}) \)
\( = (2pq \times 3pq) - (2pq \times 2q^{2}) + (3q^{2} \times 3pq) - (3q^{2} \times 2q^{2}) \)
\( = 6p^{2}q^{2} - 4pq^{3} + 9pq^{3} - 6q^{4} \)
\( = 6p^{2}q^{2} + (-4 + 9)pq^{3} - 6q^{4} \)
\( = 6p^{2}q^{2} + 5pq^{3} - 6q^{4} \)
6. Multiply \( (\frac{3}{4} a^{2} + 3b^{2}) \) by \( 4(a^{2} - \frac{2}{3}b^{2}) \):
\( (\frac{3}{4} a^{2} + 3b^{2}) \times 4(a^{2} - \frac{2}{3}b^{2}) \)
First, distribute the 4 into the second binomial:
\( = (\frac{3}{4} a^{2} + 3b^{2}) \times (4a^{2} - 4 \times \frac{2}{3}b^{2}) \)
\( = (\frac{3}{4} a^{2} + 3b^{2}) \times (4a^{2} - \frac{8}{3}b^{2}) \)
Now, multiply the binomials:
\( = \frac{3}{4} a^{2} (4a^{2} - \frac{8}{3}b^{2}) + 3b^{2} (4a^{2} - \frac{8}{3}b^{2}) \)
\( = (\frac{3}{4} a^{2} \times 4a^{2}) - (\frac{3}{4} a^{2} \times \frac{8}{3}b^{2}) + (3b^{2} \times 4a^{2}) - (3b^{2} \times \frac{8}{3}b^{2}) \)
\( = (3a^{4}) - (2a^{2}b^{2}) + (12b^{2}a^{2}) - (8b^{4}) \)
\( = 3a^{4} - 2a^{2}b^{2} + 12a^{2}b^{2} - 8b^{4} \)
\( = 3a^{4} + (-2 + 12)a^{2}b^{2} - 8b^{4} \)
\( = 3a^{4} + 10a^{2}b^{2} - 8b^{4} \)
In simple words: To multiply two binomials, multiply each term in the first binomial by each term in the second binomial. Then, combine any terms that are alike. Make sure to be careful with fractions and negative signs during your calculations.

Exam Tip: When multiplying binomials or any polynomials, use the distributive property (often remembered as FOIL for binomials: First, Outer, Inner, Last). Remember to combine like terms after all multiplications are performed. Pay extra attention to signs and fractional coefficients.

Question 2. Find the product:
1. \( (5 - 2x)(3 + x) \)
2. \( (x + 7y)(7x - y) \)
3. \( (a^{2} + b)(a + b^{2}) \)
4. \( (p^{2} - q^{2})(2p + q) \)
Answer:
1. Find the product of \( (5 - 2x) \) and \( (3 + x) \):
\( (5 - 2x) \times (3 + x) \)
\( = 5(3 + x) - 2x(3 + x) \)
\( = (5 \times 3) + (5 \times x) - (2x \times 3) - (2x \times x) \)
\( = 15 + 5x - 6x - 2x^{2} \)
\( = -2x^{2} + (5 - 6)x + 15 \)
\( = -2x^{2} - x + 15 \)
2. Find the product of \( (x + 7y) \) and \( (7x - y) \):
\( (x + 7y) \times (7x - y) \)
\( = x(7x - y) + 7y(7x - y) \)
\( = (x \times 7x) - (x \times y) + (7y \times 7x) - (7y \times y) \)
\( = 7x^{2} - xy + 49xy - 7y^{2} \)
\( = 7x^{2} + (-1 + 49)xy - 7y^{2} \)
\( = 7x^{2} + 48xy - 7y^{2} \)
3. Find the product of \( (a^{2} + b) \) and \( (a + b^{2}) \):
\( (a^{2} + b) \times (a + b^{2}) \)
\( = a^{2}(a + b^{2}) + b(a + b^{2}) \)
\( = (a^{2} \times a) + (a^{2} \times b^{2}) + (b \times a) + (b \times b^{2}) \)
\( = a^{3} + a^{2}b^{2} + ab + b^{3} \)
4. Find the product of \( (p^{2} - q^{2}) \) and \( (2p + q) \):
\( (p^{2} - q^{2}) \times (2p + q) \)
\( = p^{2}(2p + q) - q^{2}(2p + q) \)
\( = (p^{2} \times 2p) + (p^{2} \times q) - (q^{2} \times 2p) - (q^{2} \times q) \)
\( = 2p^{3} + p^{2}q - 2pq^{2} - q^{3} \)
In simple words: To multiply these pairs of expressions, take each term from the first expression and multiply it by every term in the second expression. Then, combine any like terms to get the final simplified answer.

Exam Tip: Remember that product means multiplication. Be careful with signs when distributing and combine only terms that have the exact same variables raised to the exact same powers.

Question 3. Simplify:
1. \( (x^{2} - 5)(x + 5) + 25 \)
2. \( (a^{2} + 5)(b^{3} + 3) + 5 \)
3. \( (t + s^{2})(t^{2} - s) \)
4. \( (a + b)(c - d) + (a - b)(c + d) + 2(ac + bd) \)
5. \( (x + y)(2x + y) + (x + 2y)(x - y) \)
6. \( (x + y)(x^{2} - xy + y^{2}) \)
7. \( (1.5x - 4y)(1.5x + 4y + 3) - 4.5x + 12y \)
8. \( (a + b + c)(a + b - c) \)
Answer:
1. Simplify \( (x^{2} - 5)(x + 5) + 25 \):
\( (x^{2} - 5)(x + 5) + 25 \)
\( = x^{2}(x + 5) - 5(x + 5) + 25 \)
\( = (x^{2} \times x) + (x^{2} \times 5) - (5 \times x) - (5 \times 5) + 25 \)
\( = x^{3} + 5x^{2} - 5x - 25 + 25 \)
\( = x^{3} + 5x^{2} - 5x \)
2. Simplify \( (a^{2} + 5)(b^{3} + 3) + 5 \):
\( (a^{2} + 5)(b^{3} + 3) + 5 \)
\( = a^{2}(b^{3} + 3) + 5(b^{3} + 3) + 5 \)
\( = (a^{2} \times b^{3}) + (a^{2} \times 3) + (5 \times b^{3}) + (5 \times 3) + 5 \)
\( = a^{2}b^{3} + 3a^{2} + 5b^{3} + 15 + 5 \)
\( = a^{2}b^{3} + 3a^{2} + 5b^{3} + 20 \)
3. Simplify \( (t + s^{2})(t^{2} - s) \):
\( (t + s^{2})(t^{2} - s) \)
\( = t(t^{2} - s) + s^{2}(t^{2} - s) \)
\( = (t \times t^{2}) - (t \times s) + (s^{2} \times t^{2}) + (s^{2} \times (-s)) \)
\( = t^{3} - ts + s^{2}t^{2} - s^{3} \)
4. Simplify \( (a + b)(c - d) + (a - b)(c + d) + 2(ac + bd) \):
\( (a + b)(c - d) + (a - b)(c + d) + 2(ac + bd) \)
\( = [a(c - d) + b(c - d)] + [a(c + d) - b(c + d)] + 2ac + 2bd \)
\( = [ac - ad + bc - bd] + [ac + ad - bc - bd] + 2ac + 2bd \)
\( = ac - ad + bc - bd + ac + ad - bc - bd + 2ac + 2bd \)
Combine like terms:
\( = (ac + ac + 2ac) + (-ad + ad) + (bc - bc) + (-bd - bd + 2bd) \)
\( = 4ac + 0 + 0 + 0 \)
\( = 4ac \)
5. Simplify \( (x + y)(2x + y) + (x + 2y)(x - y) \):
\( (x + y)(2x + y) + (x + 2y)(x - y) \)
\( = [x(2x + y) + y(2x + y)] + [x(x - y) + 2y(x - y)] \)
\( = [(x \times 2x) + (x \times y) + (y \times 2x) + (y \times y)] + [(x \times x) - (x \times y) + (2y \times x) - (2y \times y)] \)
\( = [2x^{2} + xy + 2xy + y^{2}] + [x^{2} - xy + 2xy - 2y^{2}] \)
\( = 2x^{2} + 3xy + y^{2} + x^{2} + xy - 2y^{2} \)
Combine like terms:
\( = (2x^{2} + x^{2}) + (3xy + xy) + (y^{2} - 2y^{2}) \)
\( = 3x^{2} + 4xy - y^{2} \)
6. Simplify \( (x + y)(x^{2} - xy + y^{2}) \):
\( (x + y)(x^{2} - xy + y^{2}) \)
\( = x(x^{2} - xy + y^{2}) + y(x^{2} - xy + y^{2}) \)
\( = (x \times x^{2}) - (x \times xy) + (x \times y^{2}) + (y \times x^{2}) - (y \times xy) + (y \times y^{2}) \)
\( = x^{3} - x^{2}y + xy^{2} + yx^{2} - xy^{2} + y^{3} \)
Combine like terms:
\( = x^{3} + (-x^{2}y + yx^{2}) + (xy^{2} - xy^{2}) + y^{3} \)
\( = x^{3} + 0 + 0 + y^{3} \)
\( = x^{3} + y^{3} \)
7. Simplify \( (1.5x - 4y)(1.5x + 4y + 3) - 4.5x + 12y \):
\( (1.5x - 4y)(1.5x + 4y + 3) - 4.5x + 12y \)
Multiply the first two expressions:
\( = 1.5x(1.5x + 4y + 3) - 4y(1.5x + 4y + 3) - 4.5x + 12y \)
\( = (1.5x \times 1.5x) + (1.5x \times 4y) + (1.5x \times 3) - (4y \times 1.5x) - (4y \times 4y) - (4y \times 3) - 4.5x + 12y \)
\( = 2.25x^{2} + 6xy + 4.5x - 6xy - 16y^{2} - 12y - 4.5x + 12y \)
Combine like terms:
\( = 2.25x^{2} + (6xy - 6xy) + (4.5x - 4.5x) - 16y^{2} + (-12y + 12y) \)
\( = 2.25x^{2} + (0)xy + (0)x - 16y^{2} + (0)y \)
\( = 2.25x^{2} - 16y^{2} \)
8. Simplify \( (a + b + c)(a + b - c) \):
\( (a + b + c)(a + b - c) \)
Let \( A = (a + b) \). Then the expression becomes \( (A + c)(A - c) \).
Using the identity \( (X + Y)(X - Y) = X^{2} - Y^{2} \):
\( = (a + b)^{2} - c^{2} \)
Now expand \( (a + b)^{2} = a^{2} + 2ab + b^{2} \):
\( = a^{2} + 2ab + b^{2} - c^{2} \)
Alternatively, by direct multiplication:
\( = a(a + b - c) + b(a + b - c) + c(a + b - c) \)
\( = (a \times a) + (a \times b) - (a \times c) + (b \times a) + (b \times b) - (b \times c) + (c \times a) + (c \times b) - (c \times c) \)
\( = a^{2} + ab - ac + ba + b^{2} - bc + ca + cb - c^{2} \)
Combine like terms:
\( = a^{2} + b^{2} - c^{2} + (ab + ba) + (-ac + ca) + (-bc + cb) \)
\( = a^{2} + b^{2} - c^{2} + 2ab + 0 + 0 \)
\( = a^{2} + b^{2} - c^{2} + 2ab \)
In simple words: To simplify expressions, first perform all multiplications using the distributive property. Then, identify and combine any like terms by adding or subtracting their coefficients. Remember special identities like \( (A+B)(A-B) = A^2-B^2 \) or \( (A+B)^2 = A^2+2AB+B^2 \) if they apply, as they can save steps.

Exam Tip: For complex simplifications, break down the problem into smaller steps: expand products first, then collect and combine like terms. Be meticulous with signs and exponents, especially when dealing with multiple variables or terms.

Try These (Page 149)

Question 1. Verify Identity (IV), for \( a = 2, b = 3, x = 5 \).
Answer:
Identity (IV) is: \( (x + a)(x + b) = x^{2} + (a + b)x + ab \)
We are given \( a = 2, b = 3 \), and \( x = 5 \).
Let's verify the Left Hand Side (LHS):
LHS \( = (x + a)(x + b) \)
Substitute the values:
\( = (5 + 2)(5 + 3) \)
\( = (7)(8) \)
\( = 56 \)
Now, let's verify the Right Hand Side (RHS):
RHS \( = x^{2} + (a + b)x + ab \)
Substitute the values:
\( = (5)^{2} + (2 + 3) \times 5 + (2 \times 3) \)
\( = 25 + (5) \times 5 + 6 \)
\( = 25 + 25 + 6 \)
\( = 56 \)
Since LHS \( = \) RHS, the given identity is true for these particular values.
In simple words: To check if an identity works, we put the given numbers into both sides of the equation. If both sides give the same final answer, then the identity is correct for those numbers.

Exam Tip: When verifying an identity, always evaluate the Left Hand Side (LHS) and the Right Hand Side (RHS) separately. Show all substitution and calculation steps clearly. The identity is verified if LHS = RHS.

Question 2. Consider the special case of Identity (IV) with \( a = b \), what do you get? Is it related to Identity (I)?
Answer:
Identity (IV) is: \( (x + a)(x + b) = x^{2} + (a + b)x + ab \)
Consider the special case where \( a = b \). Let's replace \( b \) with \( a \) in Identity (IV):
\( (x + a)(x + a) = x^{2} + (a + a)x + (a \times a) \)
\( (x + a)^{2} = x^{2} + (2a)x + a^{2} \)
\( (x + a)^{2} = x^{2} + 2ax + a^{2} \)
This result is identical to Identity (I), which is \( (x + y)^{2} = x^{2} + 2xy + y^{2} \), by replacing \( y \) with \( a \).
Therefore, yes, it is related to Identity (I).
In simple words: If we make the 'a' and 'b' in Identity (IV) the same, it turns into Identity (I). This means Identity (I) is a specific type of Identity (IV) where the two extra numbers are equal.

Exam Tip: When exploring special cases of identities, perform the substitution carefully. Recognize common algebraic identities to quickly identify relationships between them. For \( (x+a)^2 \), remember the expansion is \( x^2 + 2ax + a^2 \).

Question 3. Consider the special case of Identity (IV) with \( a = -c \) and \( b = -c \). What do you get? Is it related to Identity (II)?
Answer:
Identity (IV) is: \( (x + a)(x + b) = x^{2} + (a + b)x + ab \)
Consider the special case where \( a = -c \) and \( b = -c \). Substitute these into Identity (IV):
\( (x + (-c))(x + (-c)) = x^{2} + ((-c) + (-c))x + ((-c) \times (-c)) \)
\( (x - c)(x - c) = x^{2} + (-2c)x + (c^{2}) \)
\( (x - c)^{2} = x^{2} - 2cx + c^{2} \)
This result is identical to Identity (II), which is \( (x - y)^{2} = x^{2} - 2xy + y^{2} \), by replacing \( y \) with \( c \).
Therefore, yes, it is related to Identity (II).
In simple words: If we replace 'a' and 'b' in Identity (IV) with the same negative letter, like \( -c \), the identity changes into Identity (II). This shows that Identity (II) is a specific version of Identity (IV) when both extra terms are negative and equal.

Exam Tip: Be very careful with signs when substituting negative values into identities. Remember that \( (-c) \times (-c) = c^2 \) and \( (-c) + (-c) = -2c \). This special case leads to the square of a binomial difference.

Question 4. Consider the special case of Identity (IV) with \( b = -a \). What do you get? It is related to Identity (III).
Answer:
Identity (IV) is: \( (x + a)(x + b) = x^{2} + (a + b)x + ab \)
Consider the special case where \( b = -a \). Substitute this into Identity (IV):
\( (x + a)(x + (-a)) = x^{2} + (a + (-a))x + (a \times (-a)) \)
\( (x + a)(x - a) = x^{2} + (a - a)x + (-a^{2}) \)
\( (x + a)(x - a) = x^{2} + (0)x - a^{2} \)
\( (x + a)(x - a) = x^{2} - a^{2} \)
This result is identical to Identity (III), which is \( (x + y)(x - y) = x^{2} - y^{2} \), by replacing \( y \) with \( a \).
Therefore, yes, it is related to Identity (III).
In simple words: When we change 'b' to be the negative of 'a' in Identity (IV), it transforms into Identity (III). This identity helps us quickly multiply two expressions that are the same except for a plus and minus sign.

Exam Tip: This special case highlights the "difference of squares" identity. Recognizing \( (x+a)(x-a) = x^2 - a^2 \) is a fundamental shortcut in algebraic simplification and factorization. Always remember that when a variable and its negative counterpart are added, they sum to zero.