Samacheer Kalvi Class 9 Maths Solutions Chapter 3 Algebra Exercise 3.5

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.5

Question 1. Factorise the following expressions:
(i) \( 2a^2 + 4a^2b + 8a^2c \)
(ii) \( ab - ac - mb + mc \)
Answer:
(i) To factorise \( 2a^2 + 4a^2b + 8a^2c \), we look for the common factor. The common factor in all terms is \( 2a^2 \).
So, we take \( 2a^2 \) out of each term:
\( 2a^2 + 4a^2b + 8a^2c = 2a^2(1 + 2b + 4c) \)

(ii) To factorise \( ab - ac - mb + mc \), we can group the terms in pairs.
First group: \( ab - ac \). Common factor is \( a \). So, \( a(b - c) \).
Second group: \( -mb + mc \). Common factor is \( -m \). So, \( -m(b - c) \).
Now we have \( a(b - c) - m(b - c) \). Here, \( (b - c) \) is a common factor.
So, we can write it as:
\( ab - ac - mb + mc = a(b - c) - m(b - c) = (b - c)(a - m) \)
In simple words: For the first part, find what is common in all numbers and letters, then pull it out. For the second part, group the terms, take out what's common in each group, and then you'll find another common part to factor out.

🎯 Exam Tip: Always look for the greatest common factor first in any expression. For four-term expressions, try grouping two terms at a time to find common binomial factors.

Question 2. Factorise the following expressions:
(i) \( x^2 + 4x + 4 \)
(ii) \( 3a^2 - 24ab + 48b^2 \)
(iii) \( x^5 - 16x \)
(iv) \( m^2 + \frac{1}{m^2} - 23 \)
(v) \( 6 - 216x^2 \)
(vi) \( a^2 + \frac{1}{a^2} - 18 \)
Answer:
(i) For \( x^2 + 4x + 4 \), we recognise this as a perfect square trinomial, which follows the pattern \( (a + b)^2 = a^2 + 2ab + b^2 \).
Here, \( a = x \) and \( b = 2 \).
\( x^2 + 4x + 4 = x^2 + 2(x)(2) + 2^2 \)
\( \implies (x + 2)^2 \)

(ii) For \( 3a^2 - 24ab + 48b^2 \), first, take out the common factor, which is 3.
\( 3a^2 - 24ab + 48b^2 = 3(a^2 - 8ab + 16b^2) \)
Now, the expression inside the bracket is a perfect square trinomial, following \( (a - b)^2 = a^2 - 2ab + b^2 \).
Here, \( a = a \) and \( b = 4b \).
\( a^2 - 8ab + 16b^2 = a^2 - 2(a)(4b) + (4b)^2 \)
\( \implies (a - 4b)^2 \)
So, the final factorisation is \( 3(a - 4b)^2 \).

(iii) For \( x^5 - 16x \), first, take out the common factor \( x \).
\( x^5 - 16x = x(x^4 - 16) \)
Now, we recognise \( x^4 - 16 \) as a difference of squares, \( a^2 - b^2 = (a - b)(a + b) \).
Here, \( a = x^2 \) and \( b = 4 \).
\( x(x^4 - 16) = x((x^2)^2 - 4^2) \)
\( \implies x(x^2 - 4)(x^2 + 4) \)
We can further factorise \( x^2 - 4 \) as another difference of squares, \( x^2 - 2^2 = (x - 2)(x + 2) \).
So, the final factorisation is \( x(x + 2)(x - 2)(x^2 + 4) \).

(iv) For \( m^2 + \frac{1}{m^2} - 23 \), we want to make it look like \( (a - b)^2 \) or \( (a + b)^2 \).
We know \( (m - \frac{1}{m})^2 = m^2 + \frac{1}{m^2} - 2 \).
So, we can rewrite \( -23 \) as \( -2 - 21 \) or \( -2 - 25 + 2 \). To get a perfect square, we aim for \( -2 \).
\( m^2 + \frac{1}{m^2} - 23 = m^2 + \frac{1}{m^2} - 2 - 21 \)
\( \implies (m - \frac{1}{m})^2 - 21 \)
Alternatively, the provided solution uses \( -23 = -25 + 2 \). Let's follow that:
\( m^2 + \frac{1}{m^2} - 23 = m^2 + \frac{1}{m^2} + 2 - 25 \) (We add and subtract 2 to form a perfect square).
\( \implies (m + \frac{1}{m})^2 - 5^2 \)
Now, this is in the form of \( a^2 - b^2 = (a - b)(a + b) \), where \( a = m + \frac{1}{m} \) and \( b = 5 \).
\( \implies (m + \frac{1}{m} - 5)(m + \frac{1}{m} + 5) \)

(v) For \( 6 - 216x^2 \), first, take out the common factor, which is 6.
\( 6 - 216x^2 = 6(1 - 36x^2) \)
Now, \( 1 - 36x^2 \) is a difference of squares, \( a^2 - b^2 = (a - b)(a + b) \).
Here, \( a = 1 \) and \( b = 6x \).
\( 6(1 - (6x)^2) \)
\( \implies 6(1 - 6x)(1 + 6x) \)

(vi) For \( a^2 + \frac{1}{a^2} - 18 \), we can rewrite \( -18 \) to form a perfect square.
We know \( (a - \frac{1}{a})^2 = a^2 + \frac{1}{a^2} - 2 \).
So, we can write \( -18 \) as \( -2 - 16 \).
\( a^2 + \frac{1}{a^2} - 18 = a^2 + \frac{1}{a^2} - 2 - 16 \)
\( \implies (a - \frac{1}{a})^2 - 16 \)
Now, this is a difference of squares, \( a^2 - b^2 = (a - b)(a + b) \).
Here, \( a = a - \frac{1}{a} \) and \( b = 4 \).
\( \implies (a - \frac{1}{a} - 4)(a - \frac{1}{a} + 4) \)
In simple words: Look for common factors first. Then, try to fit the expression into common algebraic formulas like a perfect square \( (a+b)^2 \) or a difference of squares \( (a^2-b^2) \). Sometimes you need to add and subtract a number to make it fit a formula.

🎯 Exam Tip: Remember the basic factorisation identities: \( a^2 \pm 2ab + b^2 = (a \pm b)^2 \) and \( a^2 - b^2 = (a - b)(a + b) \). These are very useful for many problems.

Question 3. Factorise the following expressions:
(i) \( 4x^2 + 9y^2 + 25z^2 + 12xy + 30yz + 20xz \)
(ii) \( 25x^2 + 4y^2 + 9z^2 - 20xy + 12yz - 30xz \)
Answer:
(i) We need to factorise \( 4x^2 + 9y^2 + 25z^2 + 12xy + 30yz + 20xz \).
This expression looks like the expansion of \( (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac \).
Let's identify \( a, b, c \):
\( a^2 = 4x^2 \implies a = 2x \)
\( b^2 = 9y^2 \implies b = 3y \)
\( c^2 = 25z^2 \implies c = 5z \)
Now check the cross-terms:
\( 2ab = 2(2x)(3y) = 12xy \) (Matches)
\( 2bc = 2(3y)(5z) = 30yz \) (Matches)
\( 2ac = 2(2x)(5z) = 20xz \) (Matches)
Since all terms match the identity, we can write:
\( 4x^2 + 9y^2 + 25z^2 + 12xy + 30yz + 20xz = (2x + 3y + 5z)^2 \)

(ii) We need to factorise \( 25x^2 + 4y^2 + 9z^2 - 20xy + 12yz - 30xz \).
This also follows the pattern of \( (a + b + c)^2 \), but some terms are negative. This means some of \( a, b, c \) must be negative.
Let's find the square roots first:
\( \sqrt{25x^2} = 5x \)
\( \sqrt{4y^2} = 2y \)
\( \sqrt{9z^2} = 3z \)
Now, let's look at the negative cross-terms: \( -20xy \) and \( -30xz \). Both involve \( x \). This suggests that \( 5x \) might be negative, or both \( 2y \) and \( 3z \) are negative (but if both are negative, \( 12yz \) would be positive, which it is). So, \( 5x \) must be negative.
Let \( a = -5x \), \( b = 2y \), \( c = 3z \).
Check the cross-terms with these values:
\( 2ab = 2(-5x)(2y) = -20xy \) (Matches)
\( 2bc = 2(2y)(3z) = 12yz \) (Matches)
\( 2ac = 2(-5x)(3z) = -30xz \) (Matches)
All terms match, so the factorisation is:
\( 25x^2 + 4y^2 + 9z^2 - 20xy + 12yz - 30xz = (-5x + 2y + 3z)^2 \)
We can also write this as \( (2y + 3z - 5x)^2 \).
In simple words: When you see three squared terms and three double product terms, think of the \( (a+b+c)^2 \) formula. Figure out which terms are \( a, b, \) and \( c \). If there are minus signs in the double product terms, some of \( a, b, \) or \( c \) must be negative to make the signs work out correctly.

🎯 Exam Tip: When dealing with \( (a+b+c)^2 \), if some 2ab/2bc/2ac terms are negative, carefully determine the signs of a, b, or c by checking which variable is common to the negative terms. Only one of them should be negative to make two cross terms negative.

Question 4. Factorise the following expressions:
(i) \( 8x^3 + 125y^3 \)
(ii) \( 27x^3 - 8y^3 \)
(iii) \( a^6 - 64 \)
Answer:
(i) For \( 8x^3 + 125y^3 \), we use the sum of cubes identity: \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \).
Here, \( a = 2x \) (since \( (2x)^3 = 8x^3 \)) and \( b = 5y \) (since \( (5y)^3 = 125y^3 \)).
So, \( 8x^3 + 125y^3 = (2x + 5y)((2x)^2 - (2x)(5y) + (5y)^2) \)
\( \implies (2x + 5y)(4x^2 - 10xy + 25y^2) \)

(ii) For \( 27x^3 - 8y^3 \), we use the difference of cubes identity: \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \).
Here, \( a = 3x \) (since \( (3x)^3 = 27x^3 \)) and \( b = 2y \) (since \( (2y)^3 = 8y^3 \)).
So, \( 27x^3 - 8y^3 = (3x - 2y)((3x)^2 + (3x)(2y) + (2y)^2) \)
\( \implies (3x - 2y)(9x^2 + 6xy + 4y^2) \)

(iii) For \( a^6 - 64 \), we can treat it as a difference of squares first, or a difference of cubes.
Let's treat it as a difference of squares: \( (a^3)^2 - 8^2 \).
Using \( x^2 - y^2 = (x - y)(x + y) \):
\( a^6 - 64 = (a^3 - 8)(a^3 + 8) \)
Now, we can factorise \( a^3 - 8 \) as a difference of cubes and \( a^3 + 8 \) as a sum of cubes.
For \( a^3 - 8 \): \( a^3 - 2^3 = (a - 2)(a^2 + 2a + 2^2) = (a - 2)(a^2 + 2a + 4) \)
For \( a^3 + 8 \): \( a^3 + 2^3 = (a + 2)(a^2 - 2a + 2^2) = (a + 2)(a^2 - 2a + 4) \)
Combining these, the final factorisation is:
\( a^6 - 64 = (a - 2)(a^2 + 2a + 4)(a + 2)(a^2 - 2a + 4) \)
Alternatively, we could treat it as a difference of cubes first: \( (a^2)^3 - 4^3 \).
Using \( x^3 - y^3 = (x - y)(x^2 + xy + y^2) \):
\( a^6 - 64 = (a^2 - 4)((a^2)^2 + a^2 \cdot 4 + 4^2) \)
\( \implies (a^2 - 4)(a^4 + 4a^2 + 16) \)
Then, \( a^2 - 4 \) is a difference of squares: \( (a - 2)(a + 2) \).
For \( a^4 + 4a^2 + 16 \), we use the identity \( (x^2 + xy + y^2)(x^2 - xy + y^2) = x^4 + x^2y^2 + y^4 \).
Or we can use the technique of adding and subtracting a term to make a perfect square: \( a^4 + 4a^2 + 16 = (a^4 + 8a^2 + 16) - 4a^2 \)
\( \implies (a^2 + 4)^2 - (2a)^2 \)
Now, this is a difference of squares: \( (a^2 + 4 - 2a)(a^2 + 4 + 2a) \).
So, \( a^6 - 64 = (a - 2)(a + 2)(a^2 - 2a + 4)(a^2 + 2a + 4) \). Both methods give the same result.
In simple words: When you see powers of 3 or 6, think of cube formulas like sum or difference of cubes. Also, if the power is even, you can try to use the difference of squares first, then factorise the cubic parts that result. There might be different ways to start, but the final answer should be the same.

🎯 Exam Tip: For \( a^6 - b^6 \), it's generally easier and less error-prone to factorise it as a difference of squares first, i.e., \( (a^3)^2 - (b^3)^2 \). This leads to \( (a^3 - b^3)(a^3 + b^3) \), which can then be factorised further using the sum and difference of cubes formulas.

Question 5. Factorise the following:
(i) \( x^3 + 8y^3 + 6xy - 1 \)
(ii) \( l^3 - 8m^3 - 27n^3 - 18lmn \)
Answer:
We will use the identity: \( a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ac) \).

(i) For \( x^3 + 8y^3 + 6xy - 1 \), we can rewrite this to match the identity.
Let \( a = x \), \( b = 2y \) (since \( (2y)^3 = 8y^3 \)), and \( c = -1 \) (since \( (-1)^3 = -1 \)).
Then \( 3abc = 3(x)(2y)(-1) = -6xy \).
So, the expression \( x^3 + 8y^3 - 1 - (-6xy) \) matches the identity.
\( x^3 + 8y^3 - 1 + 6xy = x^3 + (2y)^3 + (-1)^3 - 3(x)(2y)(-1) \)
Using the identity:
\( (x + 2y + (-1))(x^2 + (2y)^2 + (-1)^2 - (x)(2y) - (2y)(-1) - (-1)(x)) \)
\( \implies (x + 2y - 1)(x^2 + 4y^2 + 1 - 2xy + 2y + x) \)

(ii) For \( l^3 - 8m^3 - 27n^3 - 18lmn \), we rewrite the terms as cubes.
Let \( a = l \), \( b = -2m \) (since \( (-2m)^3 = -8m^3 \)), and \( c = -3n \) (since \( (-3n)^3 = -27n^3 \)).
Then \( 3abc = 3(l)(-2m)(-3n) = 18lmn \).
The expression is \( l^3 + (-2m)^3 + (-3n)^3 - (18lmn) \). This fits the identity structure.
Using the identity:
\( (l + (-2m) + (-3n))(l^2 + (-2m)^2 + (-3n)^2 - (l)(-2m) - (-2m)(-3n) - (-3n)(l)) \)
\( \implies (l - 2m - 3n)(l^2 + 4m^2 + 9n^2 + 2lm - 6mn + 3ln) \)
In simple words: When you see three terms raised to the power of 3 and a fourth term that is 3 times the product of their base numbers, use the special identity for \( a^3 + b^3 + c^3 - 3abc \). Pay close attention to the signs of \( a, b, \) and \( c \) to make sure the \( -3abc \) term matches.

🎯 Exam Tip: Carefully determine the signs of \( a, b, \) and \( c \) for the \( a^3 + b^3 + c^3 - 3abc \) identity, especially when negative terms are present. The sign of the \( 3abc \) term in the identity will dictate if you need to use negative values for \( a, b, \) or \( c \).