Samacheer Kalvi Class 8 Maths Solutions Chapter 3 Algebra Exercise 3.4

Question 1. Factorise the following by taking out the common factor
(i) \( 18xy - 12yz \)
Answer: We need to find the common factors in both parts of the expression. First, let's write out the prime factors and variables for each term.
\( 18xy - 12yz = (2 \times 3 \times 3 \times x \times y) - (2 \times 2 \times 3 \times y \times z) \)
Now, we take out the common factors, which are \( 2, 3 \), and \( y \).
\( = 2 \times 3 \times y (3 \times x - 2 \times z) \)
\( = 6y(3x - 2z) \)
In simple words: We look for numbers and letters that are common in both parts of the expression. We pull them out to the front, and what's left stays inside the bracket.

🎯 Exam Tip: Always break down numbers into their prime factors and list all variables to clearly identify the greatest common factor (GCF) before factoring.

Question 1. Factorise the following by taking out the common factor
(ii) \( 9x^5y^3 + 6x^3y^2 - 18x^2y \)
Answer: Let's find the common factors for each term in the expression. We can expand each term into its prime factors and variables.
\( 9x^5y^3 + 6x^3y^2 - 18x^2y = (3 \times 3 \times x^5 \times y^3) + (2 \times 3 \times x^3 \times y^2) - (2 \times 3 \times 3 \times x^2 \times y) \)
Taking out the common factors \( 3, x^2 \), and \( y \) from all terms, we get:
\( = 3 \times x^2 \times y (3 \times x^3 \times y^2 + 2 \times y - 6) \)
\( = 3x^2y (3x^3y^2 + 2xy - 6) \)
In simple words: We find the biggest common parts (numbers and letters) in all three terms. We take these common parts outside a bracket, and the remaining parts stay inside.

🎯 Exam Tip: When factoring, remember to take the lowest power of each common variable present in all terms.

Question 1. Factorise the following by taking out the common factor
(iii) \( x(b - 2c) + y(b - 2c) \)
Answer: In this expression, we can clearly see that \( (b - 2c) \) is a common factor in both terms. This is called a binomial factor.
Taking out the binomial factor \( (b - 2c) \) from each term, we have:
\( = (b - 2c)(x + y) \)
In simple words: Both parts of the expression have the same bracket, \( (b - 2c) \). We can pull this whole bracket out as a common factor, leaving the other terms, \( x \) and \( y \), in a new bracket.

🎯 Exam Tip: Always look for common binomial factors first when an expression is already partially grouped, as it simplifies the process quickly.

Question 1. Factorise the following by taking out the common factor
(iv) \( (ax + ay) + (bx + by) \)
Answer: We need to factor this expression by grouping terms.
First, we take out 'a' from the first group \( (ax + ay) \) and 'b' from the second group \( (bx + by) \).
\( (ax + ay) + (bx + by) = a(x + y) + b(x + y) \)
Now, we can see that \( (x + y) \) is a common binomial factor in both terms.
Taking out the binomial factor \( (x + y) \) from each term, we get:
\( = (x + y)(a + b) \)
In simple words: We first find common letters in each pair of terms. Then, we notice a common bracket in both pairs, which we pull out.

🎯 Exam Tip: When factoring by grouping, ensure that the binomial factors become identical after the first step so they can be factored out as a common term.

Question 1. Factorise the following by taking out the common factor
(v) \( 2x^2(4x - 1) - 4x + 1 \)
Answer: We need to make the second part of the expression match the binomial factor in the first part.
We can take out \( -1 \) from the last two terms to create the same binomial factor \( (4x - 1) \).
\( 2x^2(4x - 1) - 4x + 1 = 2x^2(4x - 1) - 1(4x - 1) \)
Now, we take out the common binomial factor \( (4x - 1) \), which gives us:
\( = (4x - 1)(2x^2 - 1) \)
In simple words: We change \( -4x + 1 \) into \( -1(4x - 1) \) so it matches the other bracket. Then, we pull out the common bracket from both parts.

🎯 Exam Tip: When dealing with terms like \( -a + b \), remember that you can factor out \( -1 \) to get \( -1(a - b) \), which is often useful for matching binomial factors.

Question 1. Factorise the following by taking out the common factor
(vi) \( 3y(x - 2)^2 - 2(2 - x) \)
Answer: We want to make the second bracket similar to the first one. Notice that \( (2 - x) \) is the negative of \( (x - 2) \).
So, we can rewrite \( -2(2 - x) \) as \( -2(-1)(x - 2) \), which simplifies to \( +2(x - 2) \).
\( 3y(x - 2)^2 - 2(2 - x) = 3y(x - 2)(x - 2) - 2(-1)(x - 2) \)
\( = 3y(x - 2)(x - 2) + 2(x - 2) \)
Now, we can take out the common binomial factor \( (x - 2) \) from each term:
\( = (x - 2)[3y(x - 2) + 2] \)
In simple words: We change the second part so its bracket becomes the same as the first one by taking out a \( -1 \). Then, we pull out the common bracket.

🎯 Exam Tip: Remember that \( (a - b) = -1(b - a) \). This property is very useful when you need to switch the order of terms within a binomial to create common factors.

Question 1. Factorise the following by taking out the common factor
(vii) \( 6xy - 4y^2 + 12xy - 2yzx \)
Answer: First, let's combine like terms and then factor by grouping.
The terms \( 6xy \) and \( 12xy \) are like terms, so we can add them.
\( 6xy - 4y^2 + 12xy - 2yzx = 18xy - 4y^2 - 2xyz \)
Now, we look for common factors in these three terms. All terms have \( 2 \) and \( y \) as common factors.
Taking out \( 2y \) from each term:
\( = 2y(9x - 2y - xz) \)
In simple words: First, we add any parts that are the same. Then, we find numbers and letters that are common to all parts and bring them outside a bracket.

🎯 Exam Tip: Always simplify the expression by combining like terms before attempting to factorise, as this often reveals simpler common factors.

Question 1. Factorise the following by taking out the common factor
(viii) \( a^3 - 3a^2 + a - 3 \)
Answer: We can factor this expression by grouping terms.
First, group the terms and find common factors within each pair. We group \( a^3 - 3a^2 \) and \( a - 3 \).
\( a^3 - 3a^2 + a - 3 = a^2(a - 3) + 1(a - 3) \)
Now, we see that \( (a - 3) \) is a common binomial factor.
Taking out the common factor \( (a - 3) \), we get:
\( = (a - 3)(a^2 + 1) \)
In simple words: We split the expression into two pairs. We find common parts in each pair, which gives us the same bracket in both. Then, we pull out that common bracket.

🎯 Exam Tip: When factoring by grouping, always ensure that the binomial factors become identical after extracting common factors from each group.

Question 1. Factorise the following by taking out the common factor
(ix) \( 3y^3 - 48y \)
Answer: First, we look for a common numerical and variable factor in both terms.
We can see that \( 3 \) and \( y \) are common factors.
\( 3y^3 - 48y = 3 \times y \times y^2 - 3 \times 16 \times y \)
Taking out \( 3y \), we get:
\( = 3y(y^2 - 16) \)
Now, we notice that \( y^2 - 16 \) is a difference of squares, which can be factored as \( (y - 4)(y + 4) \) because \( 16 = 4^2 \).
Comparing \( y^2 - 4^2 \) with the identity \( a^2 - b^2 = (a + b)(a - b) \), where \( a = y \) and \( b = 4 \).
\( y^2 - 4^2 = (y + 4)(y - 4) \)
Therefore, the completely factored form is:
\( 3y(y^2 - 16) = 3y(y + 4)(y - 4) \)
In simple words: First, we take out the common number and letter. What's left inside is a special kind of expression called "difference of squares", which we can split into two brackets.

🎯 Exam Tip: Always look for common factors first. After that, check if the remaining expression can be factored further using identities like difference of squares \( (a^2 - b^2) \) or perfect square trinomials.

Question 1. Factorise the following by taking out the common factor
(x) \( ab^2 - bc^2 - ab + c^2 \)
Answer: We will factor this expression by grouping the terms.
Let's group the first two terms and the last two terms.
\( ab^2 - bc^2 - ab + c^2 \)
Grouping suitably, we can take out \( b \) from the first two terms and \( -1 \) from the last two terms.
\( = b(ab - c^2) - 1(ab - c^2) \)
Now, we can see that \( (ab - c^2) \) is a common binomial factor.
Taking out the common factor \( (ab - c^2) \), we get:
\( = (ab - c^2)(b - 1) \)
In simple words: We pair up the terms. In each pair, we take out a common letter or number. If the brackets match, we pull out the whole bracket as one common factor.

🎯 Exam Tip: When grouping, you might need to factor out \( -1 \) from one of the groups to make the binomial factors identical, as seen in this problem.

Question 2. Factorise the following expressions
(i) \( x^2 + 14x + 49 \)
Answer: We need to factor this trinomial. We can compare it with the perfect square identity \( a^2 + 2ab + b^2 = (a + b)^2 \).
Here, \( x^2 \) is \( a^2 \) so \( a = x \). And \( 49 \) is \( b^2 \) so \( b = 7 \).
Now, let's check the middle term \( 2ab \):
\( 2(x)(7) = 14x \). This matches the middle term in the given expression.
So, the expression is a perfect square trinomial.
Therefore, \( x^2 + 14x + 49 = (x + 7)^2 \)
In simple words: This is a special type of expression called a "perfect square". We find two numbers that multiply to make the last number and add to make the middle number. In this case, it's \( (x + 7) \) multiplied by itself.

🎯 Exam Tip: Recognize perfect square trinomials quickly by checking if the first and last terms are perfect squares and if the middle term is twice the product of their square roots.

Question 2. Factorise the following expressions
(ii) \( y^2 - 10y + 25 \)
Answer: We can factor this trinomial by comparing it with the perfect square identity \( a^2 - 2ab + b^2 = (a - b)^2 \).
Here, \( y^2 \) is \( a^2 \) so \( a = y \). And \( 25 \) is \( b^2 \) so \( b = 5 \).
Now, let's check the middle term \( -2ab \):
\( -2(y)(5) = -10y \). This matches the middle term in the given expression.
So, the expression is a perfect square trinomial.
Therefore, \( y^2 - 10y + 25 = (y - 5)^2 \)
In simple words: This expression is a "perfect square" where we subtract. We find two numbers that multiply to the last number and add to the middle number. Here, it is \( (y - 5) \) multiplied by itself.

🎯 Exam Tip: For trinomials of the form \( x^2 + Bx + C \), look for two numbers that multiply to \( C \) and add to \( B \). If the first and last terms are perfect squares, consider if it's a perfect square trinomial.

Question 2. Factorise the following expressions
(iii) \( c^2 - 4c - 12 \)
Answer: This is a quadratic expression of the form \( ax^2 + bx + c \), where \( a = 1, b = -4 \), and \( c = -12 \).
We need to find two numbers that multiply to \( ac = 1 \times (-12) = -12 \) and add up to \( b = -4 \).
Let's list pairs of factors for -12:
Product \( = -12 \) | Sum
\( 1 \times (-12) = -12 \) | \( 1 + (-12) = -11 \)
\( 2 \times (-6) = -12 \) | \( 2 + (-6) = -4 \)
The numbers are \( 2 \) and \( -6 \). So, the middle term \( -4c \) can be written as \( 2c - 6c \).
\( c^2 - 4c - 12 = c^2 + 2c - 6c - 12 \)
Now, factor by grouping:
\( = c(c + 2) - 6(c + 2) \)
Taking out the common binomial factor \( (c + 2) \), we get:
\( = (c + 2)(c - 6) \)
Therefore, \( c^2 - 4c - 12 = (c + 2)(c - 6) \)
In simple words: We need to find two numbers that multiply to make \( -12 \) and add up to \( -4 \). These numbers are \( 2 \) and \( -6 \). We then use these numbers to split the middle term and factor by grouping.

🎯 Exam Tip: For quadratic trinomials \( ax^2 + bx + c \) where \( a=1 \), find two numbers that multiply to \( c \) and add to \( b \). This method is often called splitting the middle term.

Question 2. Factorise the following expressions
(iv) \( m^2 + m - 72 \)
Answer: This is a quadratic expression of the form \( ax^2 + bx + c \), where \( a = 1, b = 1 \), and \( c = -72 \).
We need to find two numbers that multiply to \( ac = 1 \times (-72) = -72 \) and add up to \( b = 1 \).
Let's list pairs of factors for -72:
Product \( = -72 \) | Sum
\( 1 \times (-72) = -72 \) | \( 1 + (-72) = -71 \)
\( 2 \times (-36) = -72 \) | \( 2 + (-36) = -34 \)
\( 3 \times (-24) = -72 \) | \( 3 + (-24) = -21 \)
\( 4 \times (-18) = -72 \) | \( 4 + (-18) = -14 \)
\( 6 \times (-12) = -72 \) | \( 6 + (-12) = -6 \)
\( 8 \times (-9) = -72 \) | \( 8 + (-9) = -1 \)
\( 9 \times (-8) = -72 \) | \( 9 + (-8) = 1 \)
The numbers are \( 9 \) and \( -8 \). So, the middle term \( m \) can be written as \( 9m - 8m \).
\( m^2 + m - 72 = m^2 + 9m - 8m - 72 \)
Now, factor by grouping:
\( = m(m + 9) - 8(m + 9) \)
Taking out the common binomial factor \( (m + 9) \), we get:
\( = (m + 9)(m - 8) \)
Therefore, \( m^2 + m - 72 = (m + 9)(m - 8) \)
In simple words: We need to find two numbers that multiply to \( -72 \) and add up to \( 1 \). These numbers are \( 9 \) and \( -8 \). We use them to split the middle term and then group to factor the expression.

🎯 Exam Tip: Systematically listing factor pairs for the product \( ac \) and checking their sum against \( b \) helps to find the correct numbers efficiently for splitting the middle term.

Question 2. Factorise the following expressions
(v) \( 4x^2 - 8x + 3 \)
Answer: This is a quadratic expression of the form \( ax^2 + bx + c \), where \( a = 4, b = -8 \), and \( c = 3 \).
We need to find two numbers that multiply to \( ac = 4 \times 3 = 12 \) and add up to \( b = -8 \).
Let's list pairs of factors for 12:
Product \( = 12 \) | Sum
\( (-1) \times (-12) = 12 \) | \( (-1) + (-12) = -13 \)
\( (-2) \times (-6) = 12 \) | \( (-2) + (-6) = -8 \)
The numbers are \( -2 \) and \( -6 \). So, the middle term \( -8x \) can be written as \( -2x - 6x \).
\( 4x^2 - 8x + 3 = 4x^2 - 2x - 6x + 3 \)
Now, factor by grouping:
\( = 2x(2x - 1) - 3(2x - 1) \)
Taking out the common binomial factor \( (2x - 1) \), we get:
\( = (2x - 1)(2x - 3) \)
Therefore, \( 4x^2 - 8x + 3 = (2x - 1)(2x - 3) \)
In simple words: We need to find two numbers that multiply to \( (4 \times 3 = 12) \) and add up to \( -8 \). These numbers are \( -2 \) and \( -6 \). We split the middle term using these numbers and then factor by grouping.

🎯 Exam Tip: When \( a \ne 1 \) in \( ax^2 + bx + c \), always multiply \( a \) and \( c \) first to find the target product for splitting the middle term.

Question 3. Factorise the following expressions using \( (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \) identity
(i) \( 64x^3 + 144x^2 + 108x + 27 \)
Answer: We need to fit the given expression into the identity \( (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \).
First, identify \( a^3 \) and \( b^3 \).
\( 64x^3 = (4x)^3 \), so \( a = 4x \).
\( 27 = 3^3 \), so \( b = 3 \).
Now, let's check the middle terms using these values of \( a \) and \( b \).
\( 3a^2b = 3(4x)^2(3) = 3(16x^2)(3) = 144x^2 \). This matches the second term.
\( 3ab^2 = 3(4x)(3)^2 = 3(4x)(9) = 108x \). This matches the third term.
Since all terms match the identity, we can write the expression as:
\( 64x^3 + 144x^2 + 108x + 27 = (4x + 3)^3 \)
In simple words: We look for the cubic terms to find 'a' and 'b'. Then we check if the other terms fit the pattern of the \( (a + b)^3 \) formula. If they do, we can write it in the simpler cubed form.

🎯 Exam Tip: When using cubic identities, correctly identify \( a^3 \) and \( b^3 \) first, then verify the intermediate terms \( 3a^2b \) and \( 3ab^2 \) to confirm the identity applies.

Question 3. Factorise the following expressions using \( (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \) identity
(ii) \( 27p^3 + 54p^2q + 36pq^2 + 8q^3 \)
Answer: We need to fit the given expression into the identity \( (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \).
First, identify \( a^3 \) and \( b^3 \).
\( 27p^3 = (3p)^3 \), so \( a = 3p \).
\( 8q^3 = (2q)^3 \), so \( b = 2q \).
Now, let's check the middle terms using these values of \( a \) and \( b \).
\( 3a^2b = 3(3p)^2(2q) = 3(9p^2)(2q) = 54p^2q \). This matches the second term.
\( 3ab^2 = 3(3p)(2q)^2 = 3(3p)(4q^2) = 36pq^2 \). This matches the third term.
Since all terms match the identity, we can write the expression as:
\( 27p^3 + 54p^2q + 36pq^2 + 8q^3 = (3p + 2q)^3 \)
In simple words: We find the cube roots of the first and last terms to get 'a' and 'b'. Then we check if the terms in the middle fit the pattern of \( 3a^2b \) and \( 3ab^2 \). If they do, we can write the expression as \( (a+b)^3 \).

🎯 Exam Tip: When working with two variables like \( p \) and \( q \), be careful to assign them correctly to \( a \) and \( b \) and keep track of their powers during verification.

Question 4. Factorise the following expressions using \( (a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 \) identity
(i) \( y^3 - 18y^2 + 108y - 216 \)
Answer: We need to fit the given expression into the identity \( (a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 \).
First, identify \( a^3 \) and \( b^3 \).
\( y^3 \), so \( a = y \).
\( 216 = 6^3 \), so \( b = 6 \).
Now, let's check the middle terms using these values of \( a \) and \( b \).
\( -3a^2b = -3(y)^2(6) = -18y^2 \). This matches the second term.
\( 3ab^2 = 3(y)(6)^2 = 3(y)(36) = 108y \). This matches the third term.
Since all terms match the identity, we can write the expression as:
\( y^3 - 18y^2 + 108y - 216 = (y - 6)^3 \)
In simple words: We find 'a' and 'b' from the terms that are cubed. Then we check if the other terms match the specific pattern for \( (a - b)^3 \). If everything matches, we write it as a cubed difference.

🎯 Exam Tip: Pay close attention to the signs in the identity \( (a - b)^3 \); the terms alternate in sign, starting with positive, which helps confirm its application.

Question 4. Factorise the following expressions using \( (a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 \) identity
(ii) \( 8m^3 - 60m^2n + 150mn^2 - 125n^3 \)
Answer: We need to fit the given expression into the identity \( (a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 \).
First, identify \( a^3 \) and \( b^3 \).
\( 8m^3 = (2m)^3 \), so \( a = 2m \).
\( 125n^3 = (5n)^3 \), so \( b = 5n \).
Now, let's check the middle terms using these values of \( a \) and \( b \).
\( -3a^2b = -3(2m)^2(5n) = -3(4m^2)(5n) = -60m^2n \). This matches the second term.
\( 3ab^2 = 3(2m)(5n)^2 = 3(2m)(25n^2) = 150mn^2 \). This matches the third term.
Since all terms match the identity, we can write the expression as:
\( 8m^3 - 60m^2n + 150mn^2 - 125n^3 = (2m - 5n)^3 \)
In simple words: We find the cube roots of the first and last terms to get 'a' and 'b'. Then we check if the middle terms match the pattern of the cubic identity with alternating signs. If they do, we write it as \( (a-b)^3 \).

🎯 Exam Tip: When two variables are involved, make sure to correctly identify which parts correspond to \( a \) and \( b \) and pay attention to their coefficients and powers in each term.