Samacheer Kalvi Class 8 Maths Solutions Chapter 3 Algebra Exercise 3.1

Question 1. Complete the table.
Answer:

In simple words: To fill this table, you need to multiply the term in the first row by the term in the first column for each cell. Remember to multiply the numbers, and add the powers of the same variables. For example, \( x^2 \times x^4 = x^{2+4} = x^6 \).

🎯 Exam Tip: Always pay close attention to the signs (plus or minus) when multiplying terms, especially when dealing with negative numbers, to avoid calculation errors.

Question 2. Find the product of the terms.
(i) -2mn, (2m)², -3mn
(ii) 3x²y, -3xy³, x²y²
Answer:
(i) We need to multiply \( (-2mn) \times (2m)^2 \times (-3mn) \).
First, simplify \( (2m)^2 = 2^2 \times m^2 = 4m^2 \).
So, \( (-2mn) \times (4m^2) \times (-3mn) \)
Multiply the numerical coefficients: \( (-2) \times 4 \times (-3) = 24 \).
Multiply the variables: \( m \times m^2 \times m = m^{1+2+1} = m^4 \).
Multiply the variable \( n \): \( n \times n = n^{1+1} = n^2 \).
Therefore, the product is \( 24m^4n^2 \). When multiplying terms, combine like bases by adding their exponents.
(ii) We need to multiply \( (3x^2y) \times (-3xy^3) \times (x^2y^2) \).
Multiply the numerical coefficients: \( 3 \times (-3) \times 1 = -9 \).
Multiply the variable \( x \): \( x^2 \times x \times x^2 = x^{2+1+2} = x^5 \).
Multiply the variable \( y \): \( y \times y^3 \times y^2 = y^{1+3+2} = y^6 \).
Therefore, the product is \( -9x^5y^6 \).
In simple words: To find the product, multiply all the numbers together. Then, for each letter, add up all the small power numbers (exponents) it has. If there are negative signs, remember that two negative signs multiply to make a positive sign.

🎯 Exam Tip: Always remember that \( (a^m)^n = a^{m \times n} \) when a power is raised to another power, and \( a^m \times a^n = a^{m+n} \) when multiplying terms with the same base.

Question 3. If I = 4pq², b = -3p²q h = 2p³q³ then, find the value of I × b × h.
Answer:
Given values are:
\( l = 4pq^2 \)
\( b = -3p^2q \)
\( h = 2p^3q^3 \)
We need to find the value of \( l \times b \times h \).
Substitute the given values into the expression:
\( l \times b \times h = (4pq^2) \times (-3p^2q) \times (2p^3q^3) \)
First, multiply the numerical coefficients: \( 4 \times (-3) \times 2 = -24 \).
Next, multiply the variable \( p \) terms: \( p \times p^2 \times p^3 = p^{1+2+3} = p^6 \).
Then, multiply the variable \( q \) terms: \( q^2 \times q \times q^3 = q^{2+1+3} = q^6 \).
So, the final product is \( -24p^6q^6 \). This process of combining like terms is fundamental in algebraic simplification.
In simple words: To multiply these three terms, first multiply all the normal numbers. Then, for each letter like 'p', add all the small power numbers it has. Do the same for the letter 'q'. Remember to keep track of any minus signs.

🎯 Exam Tip: When multiplying multiple algebraic terms, always handle the coefficients (numbers) first, then each variable separately by adding their exponents, ensuring all parts are correctly combined.

Question 4. Expand
(i) 5x(2y - 3)
(ii) -2p(5p² – 3p + 7)
(iii) 3mn(m³n³ – 5m²n + 7mn²)
(iv) x²(x + y + z)+ y²(x+ y + z) + z²(x – y – z)
Answer:
(i) To expand \( 5x(2y - 3) \), we multiply \( 5x \) by each term inside the bracket:
\( 5x \times (2y - 3) = (5x \times 2y) - (5x \times 3) \)
\( = (5 \times 2)(x \times y) - (5 \times 3)x \)
\( = 10xy - 15x \). This is an example of the distributive property.
(ii) To expand \( -2p(5p^2 - 3p + 7) \), we multiply \( -2p \) by each term:
\( -2p \times (5p^2 - 3p + 7) = (-2p \times 5p^2) + (-2p \times -3p) + (-2p \times 7) \)
\( = (-10p^3) + (6p^2) + (-14p) \)
\( = -10p^3 + 6p^2 - 14p \).
(iii) To expand \( 3mn(m^3n^3 - 5m^2n + 7mn^2) \), we multiply \( 3mn \) by each term:
\( 3mn \times (m^3n^3 - 5m^2n + 7mn^2) = (3mn \times m^3n^3) + (3mn \times -5m^2n) + (3mn \times 7mn^2) \)
\( = (3m^{1+3}n^{1+3}) + (3 \times -5 m^{1+2}n^{1+1}) + (3 \times 7 m^{1+1}n^{1+2}) \)
\( = 3m^4n^4 - 15m^3n^2 + 21m^2n^3 \).
(iv) To expand \( x^2(x + y + z) + y^2(x + y + z) + z^2(x - y - z) \), we distribute each term:
\( x^2(x + y + z) = x^2 \times x + x^2 \times y + x^2 \times z = x^3 + x^2y + x^2z \)
\( y^2(x + y + z) = y^2 \times x + y^2 \times y + y^2 \times z = xy^2 + y^3 + y^2z \)
\( z^2(x - y - z) = z^2 \times x + z^2 \times (-y) + z^2 \times (-z) = xz^2 - yz^2 - z^3 \)
Now, add all the expanded terms together:
\( (x^3 + x^2y + x^2z) + (xy^2 + y^3 + y^2z) + (xz^2 - yz^2 - z^3) \)
\( = x^3 + y^3 - z^3 + x^2y + x^2z + xy^2 + y^2z + xz^2 - yz^2 \). This combines all terms after expansion.
In simple words: To expand means to multiply a term outside a bracket by every single term inside that bracket. After multiplying, combine any similar terms by adding or subtracting them. Remember to handle the signs carefully for each multiplication.

🎯 Exam Tip: When expanding expressions with multiple terms, use the distributive property systematically. Multiply each term in the first polynomial by every term in the second polynomial and then combine like terms for a simplified final answer.

Question 5. Find the product of
(i) (2x + 3)(2x - 4)
(ii) (y² – 4)(2y² + 3y)
(iii) (m² – n)(5m²n² – n²)
(iv) 3(x - 5) x 2(x – 1)
Answer:
(i) To find the product of \( (2x + 3)(2x - 4) \), we use the distributive property (FOIL method):
\( (2x + 3)(2x - 4) = 2x(2x - 4) + 3(2x - 4) \)
\( = (2x \times 2x) - (2x \times 4) + (3 \times 2x) - (3 \times 4) \)
\( = 4x^2 - 8x + 6x - 12 \)
Combine the like terms \( -8x \) and \( +6x \):
\( = 4x^2 + (-8 + 6)x - 12 \)
\( = 4x^2 - 2x - 12 \).
(ii) To find the product of \( (y^2 - 4)(2y^2 + 3y) \), we distribute each term:
\( (y^2 - 4)(2y^2 + 3y) = y^2(2y^2 + 3y) - 4(2y^2 + 3y) \)
\( = (y^2 \times 2y^2) + (y^2 \times 3y) - (4 \times 2y^2) - (4 \times 3y) \)
\( = 2y^4 + 3y^3 - 8y^2 - 12y \). Each term from the first bracket multiplies each term from the second.
(iii) To find the product of \( (m^2 - n)(5m^2n^2 - n^2) \), we distribute each term:
\( (m^2 - n)(5m^2n^2 - n^2) = m^2(5m^2n^2 - n^2) - n(5m^2n^2 - n^2) \)
\( = (m^2 \times 5m^2n^2) + (m^2 \times -n^2) - (n \times 5m^2n^2) - (n \times -n^2) \)
\( = 5m^4n^2 - m^2n^2 - 5m^2n^3 + n^3 \).
(iv) To find the product of \( 3(x - 5) \times 2(x - 1) \):
First, multiply the constant numbers: \( 3 \times 2 = 6 \).
Then, multiply the expressions in brackets: \( (x - 5)(x - 1) \).
\( (x - 5)(x - 1) = x(x - 1) - 5(x - 1) \)
\( = (x \times x) - (x \times 1) - (5 \times x) - (5 \times -1) \)
\( = x^2 - x - 5x + 5 \)
Combine like terms: \( = x^2 - 6x + 5 \).
Now, multiply this result by the constant \( 6 \):
\( = 6(x^2 - 6x + 5) \)
\( = (6 \times x^2) - (6 \times 6x) + (6 \times 5) \)
\( = 6x^2 - 36x + 30 \).
In simple words: To find the product of two sets of brackets, multiply each part of the first bracket by every part of the second bracket. Then, gather all the terms that are alike (like all the 'x' terms or all the 'x²' terms) and combine them by adding or subtracting. Make sure to be careful with negative signs.

🎯 Exam Tip: When multiplying binomials (two-term expressions), the FOIL method (First, Outer, Inner, Last) is a helpful mnemonic. For more complex polynomials, systematically multiply each term of one polynomial by every term of the other, then combine like terms.

Question 6. Find the missing term.
(i) \( 6xy \times \text{____} = -12x^3y \)
(ii) \( \text{____} \times (-15m^2n^3p) = 45m^3n^3p^2 \)
(iii) \( 2y(5x^2y - \text{____} + 3y^2) = 10x^2y^2 – 2xy + 6y^3 \)
Answer:
(i) To find the missing term, divide the result \( -12x^3y \) by \( 6xy \):
Missing term \( = \frac{-12x^3y}{6xy} \)
Divide the coefficients: \( \frac{-12}{6} = -2 \).
Divide the \( x \) terms: \( \frac{x^3}{x} = x^{3-1} = x^2 \).
Divide the \( y \) terms: \( \frac{y}{y} = y^{1-1} = y^0 = 1 \).
So, the missing term is \( -2x^2 \).
(ii) To find the missing term, divide the result \( 45m^3n^3p^2 \) by \( -15m^2n^3p \):
Missing term \( = \frac{45m^3n^3p^2}{-15m^2n^3p} \)
Divide the coefficients: \( \frac{45}{-15} = -3 \).
Divide the \( m \) terms: \( \frac{m^3}{m^2} = m^{3-2} = m \).
Divide the \( n \) terms: \( \frac{n^3}{n^3} = n^{3-3} = n^0 = 1 \).
Divide the \( p \) terms: \( \frac{p^2}{p} = p^{2-1} = p \).
So, the missing term is \( -3mp \).
(iii) The given equation is \( 2y(5x^2y - \text{____} + 3y^2) = 10x^2y^2 – 2xy + 6y^3 \).
First, expand the left side as if the missing term were \( A \):
\( 2y(5x^2y - A + 3y^2) = (2y \times 5x^2y) - (2y \times A) + (2y \times 3y^2) \)
\( = 10x^2y^2 - 2yA + 6y^3 \).
Now compare this with the right side of the given equation: \( 10x^2y^2 – 2xy + 6y^3 \).
We can see that \( -2yA \) must be equal to \( -2xy \).
So, \( -2yA = -2xy \).
Divide both sides by \( -2y \): \( A = \frac{-2xy}{-2y} = x \).
Thus, the missing term is \( x \). This problem utilizes the distributive property in reverse.
In simple words: To find a missing term in a multiplication, you can divide the total answer by the known term. For terms with letters and powers, divide the numbers, and then subtract the powers of the letters. If the missing term is inside brackets, expand what you can and then compare it to the given answer to find the missing part.

🎯 Exam Tip: When finding a missing term in multiplication or division, treat coefficients and each variable's exponents separately. Remember the rules of exponents, especially \( \frac{a^m}{a^n} = a^{m-n} \) and \( a^0=1 \).

Question 7. Match the following
a) \( 4y^2x - 3y \)
b) \( -2xy(5x^2 - 3) \)
c) \( 5x(x^2 - y^2 + xy) \)
d) \( (2x+3)(2x-3) \)
e) \( 5x(4xy-4) \)
(i) \( 20x^2y - 20x \)
(ii) \( 5x^3-5xy^2 + 5x^2y \)
(iii) \( 4x^2 - 9 \)
(iv) \( -12y^3 \)
(v) \( -10x^3y+6xy \)
(A) iv, v, ii, i, iii
(B) v, iv, iii, ii, i
(C) iv, v, ii, iii, i
(D) iv, v, iii, ii, i
Answer: (D) iv, v, iii, ii, i
a) \( 4y^2x - 3y \) matches (iv) \( -12y^3 \) (This looks like a typo in the original question's options for (a). \( 4y^2x - 3y \) is an expression, not a product. If the question intended it to be \( 4y^2 \times (-3y) \), then it would be \( -12y^3 \). Assuming the match is correct as per the provided answer, we are looking for a product that results in \( -12y^3 \)). The initial question's option 'a' seems like a result, not an item to expand. Let's re-evaluate based on the matched solution. The solution matches 'a' with 'iv'. Given 'a' is \( 4y^2x - 3y \) and 'iv' is \( -12y^3 \). This implies that 'a' itself might be a result of a multiplication, or the question source has a slight error in presenting 'a' as an item to be matched. Assuming that the match given in the provided answer (iv) is correct, and it is a result of a product, we'll proceed based on that. b) \( -2xy(5x^2 - 3) = -2xy \times 5x^2 - 2xy \times (-3) = -10x^3y + 6xy \). This matches (v). c) \( 5x(x^2 - y^2 + xy) = 5x \times x^2 - 5x \times y^2 + 5x \times xy = 5x^3 - 5xy^2 + 5x^2y \). This matches (ii). d) \( (2x+3)(2x-3) \). This is of the form \( (a+b)(a-b) = a^2 - b^2 \). Here \( a=2x \) and \( b=3 \). So, \( (2x)^2 - 3^2 = 4x^2 - 9 \). This matches (iii). e) \( 5x(4xy-4) = 5x \times 4xy - 5x \times 4 = 20x^2y - 20x \). This matches (i). So, the correct matching is: a-iv, b-v, c-ii, d-iii, e-i.
In simple words: For each item from 'a' to 'e', you need to do the multiplication or simplify the expression. Then, find the correct result from the Roman numeral list (i) to (v). Match them up. The option that shows all these correct pairs in order is the answer.

🎯 Exam Tip: For matching questions, expand each algebraic expression from one column first. Then, look for its simplified form in the other column. The difference of squares formula \( (a+b)(a-b) = a^2-b^2 \) is very useful for quick matching.

Question 8. A car moves at a uniform speed of (x + 30) km/hr. Find the distance covered by the car in (y + 2)hours. (Hint: distance = speed × time).
Answer:
Given:
Speed of the car \( = (x + 30) \) km/hr.
Time taken \( = (y + 2) \) hours.
Using the formula for distance: \( \text{Distance} = \text{Speed} \times \text{Time} \).
Substitute the given values into the formula:
\( \text{Distance} = (x + 30)(y + 2) \).
Now, expand this product by multiplying each term in the first bracket by each term in the second:
\( = x(y + 2) + 30(y + 2) \)
\( = (x \times y) + (x \times 2) + (30 \times y) + (30 \times 2) \)
\( = xy + 2x + 30y + 60 \).
So, the distance covered by the car is \( (xy + 2x + 30y + 60) \) km. This algebraic expression represents the total distance traveled.
In simple words: To find the total distance, you multiply the speed of the car by the time it traveled. Since both the speed and time are given as expressions with letters and numbers, you multiply these expressions like you would multiply two sets of brackets.

🎯 Exam Tip: Always state the formula you are using (e.g., Distance = Speed × Time). When multiplying two binomials like (a+b)(c+d), ensure you multiply each term in the first bracket by each term in the second bracket, then combine like terms.

Question 9. The product of \( 7p^3 \) and \( (2p^2)^2 \) is
(A) \( 14p^{12} \)
(B) \( 28p^7 \)
(C) \( 9p^7 \)
(D) \( 11p^{12} \)
Answer: (B) 28p⁷
In simple words: First, square the term in the bracket by squaring both the number and the power of 'p'. Then, multiply this result by the first term, remembering to add the powers of 'p'.

🎯 Exam Tip: When simplifying terms with exponents, remember that \( (a^m)^n = a^{m \times n} \) and \( a^m \times a^n = a^{m+n} \). Pay attention to the order of operations: simplify powers first, then multiply.

Question 10. The missing terms in the product \( -3m^3n \times 9(\text{____}) = m^4n^3 \) are
(A) \( mn^2, 27 \)
(B) \( m^2n, 27 \)
(C) \( m^2n^2, – 27 \)
(D) \( mn^2, – 27 \)
Answer: (D) mn², – 27
In simple words: We are looking for a term that, when multiplied by \( -3m^3n \times 9 \), gives \( m^4n^3 \). First, multiply \( -3 \) by \( 9 \) to get \( -27 \). So the expression becomes \( -27m^3n \times (\text{missing term}) = m^4n^3 \). To find the missing term, divide \( m^4n^3 \) by \( -27m^3n \). This means dividing \( 1 \) by \( -27 \) for the number, \( m^4 \) by \( m^3 \) for the 'm' part, and \( n^3 \) by \( n \) for the 'n' part.

🎯 Exam Tip: To find a missing term in an algebraic multiplication, divide the final product by the known terms. Remember to divide coefficients and subtract exponents for like variables.

Question 11. If the area of a square is \( 36x^4y^2 \) then, its side is
(A) \( 6x^4y^2 \)
(B) \( 8x^2y^2 \)
(C) \( 6x^2y \)
(D) \( -6x^2y \)
Answer: (C) 6x²y
In simple words: The area of a square is found by multiplying its side length by itself. So, to find the side length when you know the area, you need to find the square root of the area. Find the square root of the number and half the powers of the letters.

🎯 Exam Tip: The side of a square is found by taking the square root of its area. For a term like \( a^xb^y \), its square root is \( a^{x/2}b^{y/2} \). Always remember that side lengths are generally positive values.

Question 12. If the area of a rectangle is \( 48m^2n^3 \) and whose length is \( 8mn^2 \) then, its breadth is
(A) \( 6 mn \)
(B) \( 8m^2n \)
(C) \( 7m^2n^2 \)
(D) \( 6m^2n^2 \)
Answer: (A) 6 mn
In simple words: The area of a rectangle is found by multiplying its length by its breadth. So, if you know the area and the length, you can find the breadth by dividing the area by the length. Divide the numbers, then subtract the powers of the matching letters.

🎯 Exam Tip: Recall the formula for the area of a rectangle: Area = Length × Breadth. To find the breadth, divide the area by the length. Apply exponent rules \( \frac{a^m}{a^n} = a^{m-n} \) for each variable.

Question 13. If the area of a rectangular land is \( (a^2 – b^2) \) sq.units whose breadth is \( (a – b) \) then, its length is
(A) \( a - b \)
(B) \( a + b \)
(C) \( a^2 – b \)
(D) \( (a + b)^2 \)
Answer: (B) a + b
In simple words: We know that the area of a rectangle is length times breadth. To find the length, we divide the area by the breadth. Use the algebraic identity \( a^2 - b^2 = (a-b)(a+b) \) to easily simplify the division.

🎯 Exam Tip: This question tests your knowledge of algebraic identities, specifically the difference of squares: \( a^2 - b^2 = (a - b)(a + b) \). Recognizing this identity simplifies the division significantly.