Samacheer Kalvi Class 8 Maths Solutions Chapter 3 Algebra Exercise 3.5

 

Question 1. Subtract: \( -2(xy)² (y³ + 7x²y + 5) \) from \( 5y² (x²y³ – 2x²y + 10x²) \)
Answer: We need to subtract the first expression from the second one. This means we write the second expression first, then subtract the first.
\[ 5y² (x²y³ - 2x²y + 10x²) - [(-2) (xy)² (y³ + 7x²y + 5)] \] First, distribute \( 5y² \) into its bracket:
\[ = [5y² (x²y³) - 5y² (2x⁴y) + 5y² (10x²)] - [(-2) x²y² (y³ + 7x²y + 5)] \] Next, distribute \( -2x²y² \) into its bracket (remembering the negative sign outside the whole bracket):
\[ = (5x²y⁵ - 10x⁴y³ + 50x²y²) - [(-2x²y²) (y³) + (-2x²y²) (7x²y) + (-2x²y²) (5)] \] Multiply the terms inside each part:
\[ = 5x²y⁵ - 10x⁴y³ + 50x²y² - [-2x²y⁵ - 14x⁴y³ - 10x²y²] \] Now, remove the square bracket. Since there is a minus sign before it, change the sign of every term inside:
\[ = 5x²y⁵ - 10x⁴y³ + 50x²y² + 2x²y⁵ + 14x⁴y³ + 10x²y² \] Finally, combine the like terms (terms with the same variables and powers):
\[ = (5+2) x²y⁵ + (−10+14) x⁴y³ + (+50+10)x²y² \] \[ = 7x²y⁵ + 4x⁴y³ + 60x²y² \] This is the final simplified expression after subtraction.
In simple words: To subtract expressions, write the second one first, then minus the first. Distribute numbers into brackets, change signs if a minus is outside, and then combine terms that are exactly alike.

🎯 Exam Tip: Pay very close attention to negative signs, especially when distributing a negative number or when removing brackets that have a negative sign in front of them, as these are common sources of errors.

 

Question 2. Multiply \( (4x² + 9) \) and \( (3x – 2) \).
Answer: To multiply these two binomials, we use the distributive property. Each term in the first bracket multiplies each term in the second bracket.
\[ (4x² + 9)(3x – 2) = 4x²(3x – 2) + 9(3x – 2) \] Now, distribute \( 4x² \) and \( 9 \) into their respective brackets:
\[ = (4x²)(3x) – (4x²) (2) + 9(3x) – 9(2) \] Perform the multiplications:
\[ = (4 \times 3 \times x \times x²) – (4 \times 2 \times x²) + (9 \times 3 \times x) – 18 \] Simplify each term:
\[ = 12x³ – 8x² + 27x – 18 \] Since there are no like terms, this is the final simplified product.
In simple words: To multiply two brackets, make sure every part of the first bracket multiplies every part of the second bracket. Then, simplify by multiplying the numbers and adding the powers of the letters.

🎯 Exam Tip: Remember to combine like terms (terms with the same variable and exponent) after multiplication to simplify the expression fully. If there are no like terms, the expression is already in its simplest form.

 

Question 3. Find the simple interest on Rs. \( 5a²b² \) for \( 4ab \) years at \( 7b\% \) per annum.
Answer: The formula for simple interest (SI) is: \[ \text{Simple interest} = \frac{\text{Principal} \times \text{Years} \times \text{Rate}}{100} \] Given values are: Principal \( P = 5a²b² \) Years \( T = 4ab \) Rate \( R = 7b\% \) Substitute these values into the formula:
\[ \text{Simple interest} = \frac{5a²b² \times 4ab \times 7b}{100} \] Now, multiply the numerical coefficients and the variables separately:
\[ = \frac{(5 \times 4 \times 7) \times (a² \times b² \times a \times b \times b)}{100} \] Calculate the product of the numbers: \( 5 \times 4 \times 7 = 140 \)
Combine the powers of the variables: \( a^{2+1} = a³ \) and \( b^{2+1+1} = b⁴ \)
\[ = \frac{140 \times a³b⁴}{100} \] Simplify the fraction by dividing both numerator and denominator by 20:
\[ = \frac{140}{100} a³b⁴ \]
\[ = \frac{14}{10} a³b⁴ \]
\[ = \frac{7}{5} a³b⁴ \] So, the simple interest is \( \frac{7}{5} a³b⁴ \). This shows how interest can be calculated even with algebraic expressions.
In simple words: Simple interest is found by multiplying the main amount, the time in years, and the interest rate, then dividing by 100. Just put all the given numbers and letters into the formula and simplify.

🎯 Exam Tip: Always write down the formula first, then substitute the given values carefully to avoid errors. Simplify both the numerical and algebraic parts step by step.

 

Question 4. The cost of a notebook is Rs. \( 10ab \). If Babu has Rs. \( (5a²b + 20ab² + 40ab) \). Then how many notebooks can he buy?
Answer: To find out how many notebooks Babu can buy, we need to divide the total amount of money he has by the cost of one notebook. Total amount Babu has \( = 5a²b + 20ab² + 40ab \) Cost of one notebook \( = 10ab \) Number of notebooks \( = \frac{\text{Total amount}}{\text{Cost of 1 notebook}} \)
\[ = \frac{5a²b + 20ab² + 40ab}{10ab} \] We can split this single fraction into three separate fractions, dividing each term in the numerator by the denominator:
\[ = \frac{5a²b}{10ab} + \frac{20ab²}{10ab} + \frac{40ab}{10ab} \] Now, simplify each fraction. For each term, divide the numbers and subtract the powers of the common variables.
For the first term: \( \frac{5}{10} = \frac{1}{2} \), \( \frac{a²}{a} = a^{2-1} = a \), \( \frac{b}{b} = b^{1-1} = b⁰ = 1 \)
So, \( \frac{5a²b}{10ab} = \frac{1}{2}a \) For the second term: \( \frac{20}{10} = 2 \), \( \frac{a}{a} = a^{1-1} = a⁰ = 1 \), \( \frac{b²}{b} = b^{2-1} = b \)
So, \( \frac{20ab²}{10ab} = 2b \) For the third term: \( \frac{40}{10} = 4 \), \( \frac{ab}{ab} = 1 \)
So, \( \frac{40ab}{10ab} = 4 \) Combine the simplified terms:
\[ = \frac{1}{2}a + 2b + 4 \] So, Babu can buy \( \frac{1}{2}a + 2b + 4 \) notebooks. This shows a practical use of algebraic division in everyday scenarios.
In simple words: To find out how many notebooks can be bought, divide the total money by the cost of one notebook. Separate the expression into parts and simplify each part by dividing the numbers and letters.

🎯 Exam Tip: When dividing an expression with multiple terms, remember to divide *each* term in the numerator by the denominator. Treat the division of coefficients and variables separately.

 

Question 5. Factorise: \( (7y² – 19y – 6) \)
Answer: The expression \( 7y² – 19y – 6 \) is a quadratic trinomial of the form \( ax² + bx + c \), where \( a = 7 \), \( b = -19 \), and \( c = -6 \). To factorise it, we look for two numbers that have a product equal to \( a \times c \) and a sum equal to \( b \). Product \( a \times c = 7 \times (-6) = -42 \) Sum \( b = -19 \) Let's find pairs of factors for -42 and check their sums:

The numbers are 2 and -21. Now, we rewrite the middle term \( -19y \) using these two numbers as \( -21y + 2y \). So, the expression becomes:
\[ 7y² – 19y – 6 = 7y² – 21y + 2y – 6 \] Next, we group the terms and factor out the common factors from each pair:
\[ = 7y(y - 3) + 2(y - 3) \] Notice that \( (y - 3) \) is a common factor. Factor it out:
\[ = (y - 3)(7y + 2) \] This is the factored form of the quadratic expression. Factoring helps simplify complex algebraic expressions.
In simple words: To factor a quadratic expression like this, find two numbers that multiply to the first number times the last number, and add up to the middle number. Then, split the middle term using these numbers and group the terms to find common factors.

🎯 Exam Tip: Always double-check your factorization by multiplying the factored terms back out. If you get the original expression, your factorization is correct.

 

Question 6. A contractor uses the expression \( 4x² + 11x + 6 \) to determine the amount of wire to order when wiring a house. If the expression comes from multiplying the number of rooms times the number of outlets and he knows the number of rooms to be \( (x + 2) \), find the number of outlets in terms of 'x'. [Hint : factorise \( 4x² + 11x + 6 \)]
Answer: We are given that: Amount of wire needed \( = 4x² + 11x + 6 \) Number of rooms \( = x + 2 \) We also know that: Number of rooms \( \times \) Number of outlets \( = \) Amount of wire So, \( (x + 2) \times \) Number of outlets \( = 4x² + 11x + 6 \) To find the Number of outlets, we need to divide the total amount of wire by the number of rooms:
\[ \text{Number of outlets} = \frac{4x² + 11x + 6}{x + 2} \] The hint suggests we factorise the numerator \( 4x² + 11x + 6 \). This is a quadratic expression in the form \( ax² + bx + c \), where \( a = 4 \), \( b = 11 \), and \( c = 6 \). We need two numbers whose product is \( a \times c \) and whose sum is \( b \). Product \( a \times c = 4 \times 6 = 24 \) Sum \( b = 11 \) Let's find pairs of factors for 24 and check their sums:

The numbers are 3 and 8. Now, rewrite the middle term \( 11x \) as \( 8x + 3x \).
\[ 4x² + 11x + 6 = 4x² + 8x + 3x + 6 \] Group the terms and factor out common factors:
\[ = 4x(x + 2) + 3(x + 2) \] Factor out the common binomial factor \( (x + 2) \):
\[ = (x + 2)(4x + 3) \] Now, substitute this factored form back into the expression for the number of outlets:
\[ \text{Number of outlets} = \frac{(x + 2)(4x + 3)}{x + 2} \] We can cancel out the common factor \( (x + 2) \) from the numerator and denominator:
\[ \text{Number of outlets} = 4x + 3 \] So, the number of outlets in terms of 'x' is \( 4x + 3 \). This problem shows how factorization can simplify division of polynomials.
In simple words: To find the number of outlets, divide the total wire amount by the number of rooms. Factor the top part of the division into two smaller parts. One of these parts will match the bottom part, allowing you to cancel them out and find the answer.

🎯 Exam Tip: When a problem involves finding a missing factor, factorize the given polynomial first. This often reveals one of the factors, making the division straightforward.

 

Question 7. A mason uses the expression \( x² + 6x + 8 \) to represent the area of the floor of a room. If he decides that the length of the room will be represented by \( (x + 4) \), what will the width of the room be in terms of x?
Answer: We know that for a rectangular room, the area is calculated by multiplying its length and width. Given: Area of the room \( = x² + 6x + 8 \) Length of the room \( = x + 4 \) We need to find the width. So, we can write the formula as:
\[ \text{Length} \times \text{Breadth (Width)} = \text{Area} \] Substituting the given expressions:
\[ (x + 4) \times \text{Breadth} = x² + 6x + 8 \] To find the Breadth, we divide the Area by the Length:
\[ \text{Breadth} = \frac{x² + 6x + 8}{x + 4} \] To simplify this, we need to factorise the numerator, \( x² + 6x + 8 \). This is a quadratic expression of the form \( ax² + bx + c \), where \( a = 1 \), \( b = 6 \), and \( c = 8 \). We need to find two numbers that multiply to \( a \times c \) and add up to \( b \). Product \( a \times c = 1 \times 8 = 8 \) Sum \( b = 6 \) Let's find pairs of factors for 8 and check their sums:

The numbers are 2 and 4. Now, rewrite the middle term \( 6x \) as \( 2x + 4x \).
\[ x² + 6x + 8 = x² + 2x + 4x + 8 \] Group the terms and factor out common factors:
\[ = x(x + 2) + 4(x + 2) \] Factor out the common binomial factor \( (x + 2) \):
\[ = (x + 2)(x + 4) \] Substitute this factored form back into the expression for Breadth:
\[ \text{Breadth} = \frac{(x + 2)(x + 4)}{x + 4} \] Cancel out the common factor \( (x + 4) \) from the numerator and denominator:
\[ \text{Breadth} = x + 2 \] So, the width of the room in terms of x is \( x + 2 \). This demonstrates how polynomial division can be used in geometry.
In simple words: Area is length times width. To find the width, divide the area by the length. Factor the area expression to find its parts. One part will be the length, and the other part will be the width.

🎯 Exam Tip: In word problems involving geometric shapes and algebraic expressions, always relate the given information to the appropriate formula (e.g., Area = Length × Width) and then use factorization or division to find the unknown quantity.

 

Question 8. Find the missing term: \( y² + (\text{-})x + 56 = (y + 7)(y + \text{-}) \)
Answer: We need to fill in the missing terms in the given equation. Let's expand the right side of the equation using the distributive property (FOIL method). We know that \( (x + a)(x + b) = x² + (a + b)x + ab \). In our problem, \( (y + 7)(y + \text{-}) \), let the missing term be \( b \). So, \( (y + 7)(y + b) = y² + (7 + b)y + 7b \). Now, compare this with the left side of the given equation: \( y² + (\text{-})y + 56 \). By comparing the constant terms, we have \( 7b = 56 \). To find \( b \), divide 56 by 7:
\[ b = \frac{56}{7} \]
\[ b = 8 \] Now that we have \( b = 8 \), we can find the missing coefficient of the middle term (the coefficient of \( y \)). The middle term coefficient is \( (7 + b) \). Substitute \( b = 8 \):
\[ 7 + 8 = 15 \] So, the full expanded form is \( y² + 15y + 56 \). Therefore, the missing terms are 15 for the middle term coefficient and 8 for the second term in the second bracket. The complete equation is \( y² + 15y + 56 = (y + 7)(y + 8) \). This highlights the direct relationship between factors and coefficients.
In simple words: To find the missing numbers, use the rule that when you multiply two brackets like \( (y+7)(y+b) \), the last number in the answer is \( 7 \times b \), and the middle number is \( 7+b \). Use these facts to find the missing numbers.

🎯 Exam Tip: Always remember the general form of a quadratic expression \( x² + (a+b)x + ab \). By comparing the constant terms, you can usually find one of the missing factors, which then helps find the other terms.

 

Question 9. Factorise: \( 16p⁴ – 1 \)
Answer: The expression \( 16p⁴ – 1 \) is a difference of squares. We can rewrite it as \( (4p²)² – (1)² \). We use the algebraic identity \( a² – b² = (a + b)(a – b) \). Here, \( a = 4p² \) and \( b = 1 \). Applying the identity:
\[ 16p⁴ – 1 = (4p² + 1)(4p² – 1) \] Now, look at the second factor: \( (4p² – 1) \). This is also a difference of squares. We can rewrite it as \( (2p)² – (1)² \). Apply the identity \( a² – b² = (a + b)(a – b) \) again, where this time \( a = 2p \) and \( b = 1 \). So, \( (4p² – 1) = (2p + 1)(2p – 1) \). Substitute this back into the factored expression:
\[ 16p⁴ – 1 = (4p² + 1)(2p + 1)(2p – 1) \] This is the fully factorised form of the expression. Repeated application of the difference of squares identity is a common technique in algebra.
In simple words: If you have a number squared minus another number squared, you can factor it into \( (\text{first number} + \text{second number})(\text{first number} - \text{second number}) \). You might need to do this step more than once if parts of the answer are still squares.

🎯 Exam Tip: Always check if any of the factors obtained from a difference of squares can be further factorized. Often, expressions like \( (a^4 - b^4) \) require applying the identity twice.

 

Question 10. Factorise: \( 3x³ – 45x²y + 225xy² – 375y³ \)
Answer: First, look for a common factor in all terms. All the coefficients (3, -45, 225, -375) are divisible by 3. Factor out 3:
\[ 3x³ – 45x²y + 225xy² – 375y³ = 3(x³ – 15x²y + 75xy² – 125y³) \] Now, look at the expression inside the bracket: \( (x³ – 15x²y + 75xy² – 125y³) \). This looks similar to the expansion of \( (a - b)³ = a³ - 3a²b + 3ab² - b³ \). Let's compare terms: The first term is \( x³ \), so \( a = x \). The last term is \( -125y³ \), which is \( -(5y)³ \), so \( b = 5y \). Now, let's check the middle terms using \( a = x \) and \( b = 5y \): \( -3a²b = -3(x)²(5y) = -15x²y \) (This matches the second term) \( +3ab² = +3(x)(5y)² = +3x(25y²) = +75xy² \) (This matches the third term) Since all terms match, the expression inside the bracket is indeed \( (x - 5y)³ \). So, the fully factorised expression is:
\[ 3(x – 5y)³ \] This problem demonstrates how to recognize and factorize perfect cubes.
In simple words: First, take out any number that divides all parts of the expression. Then, check if the remaining part looks like \( (a-b) \text{ cubed} \), which means \( a \times a \times a \) minus \( 3 \times a \times a \times b \) plus \( 3 \times a \times b \times b \) minus \( b \times b \times b \). If it matches, write it in the cubed form.

🎯 Exam Tip: When factorizing polynomials with four terms, always first check for a common factor among all terms. Then, try to identify if it's a perfect cube expansion or if grouping can be applied to find common binomial factors.