Samacheer Kalvi Class 10 Maths Solutions Chapter 3 Algebra Exercise 3.2

Question 1. Find the GCD of the given polynomials by Division Algorithm
(i) \( x^4 + 3x^3 – x – 3 \), \( x^3 + x^2 – 5x + 3 \)
Answer:
Let \( p(x) = x^4 + 3x^3 – x – 3 \)
Let \( g(x) = x^3 + x^2 – 5x + 3 \)
We perform polynomial long division of \( p(x) \) by \( g(x) \):

The remainder is \( 3x^2 + 6x – 9 \). We can factor out 3 from this remainder: \( 3(x^2 + 2x – 3) \). We will use \( x^2 + 2x – 3 \) as the new divisor for \( g(x) \). This makes calculations simpler.
Now dividing \( g(x) = x^3 + x^2 – 5x + 3 \) by the new remainder (leaving the constant 3):

Since the remainder is 0, the last divisor, \( x^2 + 2x – 3 \), is the Greatest Common Divisor (GCD). Understanding the common factors helps in simplifying fractions.
G.C.D. = \( x^2 + 2x – 3 \)
In simple words: We used long division for polynomials. We kept dividing until the remainder became zero. The last polynomial that we used to divide, which gave a zero remainder, is the GCD.

🎯 Exam Tip: Remember to simplify the remainder by taking out common constant factors before using it as the next divisor. This makes the division steps much easier.

Question 1. Find the GCD of the given polynomials by Division Algorithm
(ii) \( x^4 – 1 \), \( x^3 – 11x^2 + x − 11 \)
Answer:
Let \( p(x) = x^4 – 1 \)
Let \( g(x) = x^3 – 11x^2 + x − 11 \)
We perform polynomial long division of \( p(x) \) by \( g(x) \):

The remainder is \( 120x^2 + 120 \). We can factor out 120 from this remainder: \( 120(x^2 + 1) \). We will use \( x^2 + 1 \) as the new divisor for \( g(x) \). This helps simplify the next step in finding common factors.
Now dividing \( g(x) = x^3 – 11x^2 + x – 11 \) by the new remainder (leaving the constant 120):

Since the remainder is 0, the last divisor, \( x^2 + 1 \), is the Greatest Common Divisor (GCD). This polynomial division method is an important tool in algebra.
G.C.D. = \( x^2 + 1 \)
In simple words: We used the division algorithm to find the largest polynomial that divides both given polynomials. We kept dividing one polynomial by the remainder of the previous division until we got a remainder of zero. The polynomial that gave the zero remainder is our answer.

🎯 Exam Tip: When setting up long division, make sure to include terms with zero coefficients (like \( 0x^3 \) or \( 0x^2 \)) to keep the powers aligned correctly. This helps prevent errors.

Question 1. Find the GCD of the given polynomials by Division Algorithm
(iii) \( 3x^4 + 6x^3 – 12x^2 – 24x \), \( 4x^4 + 14x^3 + 8x^2 – 8x \)
Answer:
Let \( p(x) = 3x^4 + 6x^3 – 12x^2 – 24x \)
We can factor out \( 3x \) from \( p(x) \):
\( p(x) = 3x (x^3 + 2x^2 – 4x – 8) \)

Let \( g(x) = 4x^4 + 14x^3 + 8x^2 – 8x \)
We can factor out \( 2x \) from \( g(x) \):
\( g(x) = 2x (2x^3 + 7x^2 + 4x – 4) \)

The Greatest Common Divisor (GCD) of the common factors \( 3x \) and \( 2x \) is \( x \). This initial common factor must be included in the final GCD.
Now, we need to find the GCD of the remaining polynomials: \( (x^3 + 2x^2 – 4x – 8) \) and \( (2x^3 + 7x^2 + 4x – 4) \). Let's divide \( (2x^3 + 7x^2 + 4x – 4) \) by \( (x^3 + 2x^2 – 4x – 8) \):

The remainder is \( 3x^2 + 12x + 12 \). We can factor out 3: \( 3(x^2 + 4x + 4) \). We use \( x^2 + 4x + 4 \) as the new divisor. Factoring out constants makes the numbers smaller and easier to work with.
Now dividing \( p(x) \) (without the initial common factor \( 3x \)) by the new remainder (leaving the constant 3):
\( x^3 + 2x^2 – 4x – 8 \) by \( x^2 + 4x + 4 \)

Since the remainder is 0, the last divisor is \( x^2 + 4x + 4 \).
The G.C.D. for the entire original polynomials is the product of the initial common factor \( x \) and this last divisor. This ensures all common factors are accounted for.
G.C.D. = \( x(x^2 + 4x + 4) \)
In simple words: First, we took out any common terms from each polynomial. Then, we used polynomial long division on the remaining parts until we got no remainder. The last divisor, combined with the initial common term, gives us the biggest factor shared by both polynomials.

🎯 Exam Tip: Always look for common monomial factors (like \( x \), \( 2x \), \( 3x \)) in the given polynomials first. Factor them out to simplify the polynomials before starting the division algorithm.

Question 1. Find the GCD of the given polynomials by Division Algorithm
(iv) \( 3x^3 + 3x^2 + 3x + 3 \), \( 6x^3 + 12x^2 + 6x + 12 \)
Answer:
Let \( p(x) = 3x^3 + 3x^2 + 3x + 3 \)
We can factor out 3: \( p(x) = 3(x^3 + x^2 + x + 1) \)

Let \( g(x) = 6x^3 + 12x^2 + 6x + 12 \)
We can factor out 6: \( g(x) = 6(x^3 + 2x^2 + x + 2) \)

The Greatest Common Divisor (GCD) of the constants 3 and 6 is 3. This constant will be part of the final GCD.
Now, we find the GCD of \( (x^3 + x^2 + x + 1) \) and \( (x^3 + 2x^2 + x + 2) \). Let's divide \( (x^3 + 2x^2 + x + 2) \) by \( (x^3 + x^2 + x + 1) \):

The remainder is \( x^2 + 1 \). We use \( x^2 + 1 \) as the new divisor. This keeps the process moving towards finding the true common factor.
Now dividing \( (x^3 + x^2 + x + 1) \) by \( (x^2 + 1) \):

Since the remainder is 0, the last divisor is \( x^2 + 1 \).
The G.C.D. for the original polynomials is the product of the constant GCD (which was 3) and this last polynomial divisor. So, the G.C.D. is \( 3(x^2 + 1) \). This fully accounts for all common factors.
G.C.D. = \( 3(x^2 + 1) \) [3 is the G.C.D. of 3 and 6]
In simple words: First, we took out the biggest number that could divide all terms in each polynomial. Then, we divided the new, simpler polynomials until the remainder was zero. The polynomial that gave a zero remainder, multiplied by the number we took out initially, is our final Greatest Common Divisor.

🎯 Exam Tip: Always start by factoring out any common numerical coefficients from both polynomials. Find their GCD separately and include it in your final answer. This makes the polynomial division part less complex.

Question 2. Find the LCM of the given polynomials
(i) \( 4x^2y \), \( 8x^3y^2 \)
Answer:
First, we find the prime factors of the coefficients:
\( 4x^2y = 2 \times 2 \times x^2 \times y = 2^2 \times x^2 \times y \)
\( 8x^3y^2 = 2 \times 2 \times 2 \times x^3 \times y^2 = 2^3 \times x^3 \times y^2 \)
To find the Least Common Multiple (LCM), we take the highest power of each prime factor and variable present in any of the expressions. For numbers, this means finding the LCM of 4 and 8, which is 8. For variables, we choose the highest power. The LCM is the smallest expression that both given expressions can divide into evenly.
L.C.M. = \( 2^3 \times x^3 \times y^2 \)
L.C.M. = \( 8x^3y^2 \)
In simple words: To find the LCM, we look at the numbers and the letters separately. For the numbers, we find the smallest number that both 4 and 8 can divide into. For the letters, we take the highest power of each letter that appears in either expression. We then multiply these together.

🎯 Exam Tip: When finding the LCM of monomials, always identify the highest power for each variable and the LCM of the numerical coefficients. Combine these to get the final LCM.

Question 2. Find the LCM of the given polynomials
(ii) \( -9a^3b^2 \), \( 12a^2b^2c \)
Answer:
First, we factorize each expression into its prime components, including the sign:
\( -9a^3b^2 = -(3^2 \times a^3 \times b^2) \)
\( 12a^2b^2c = 2^2 \times 3 \times a^2 \times b^2 \times c \)
To find the Least Common Multiple (LCM), we take the highest power of each prime factor and variable. The LCM of the absolute values of the coefficients (9 and 12) is 36. Since one term is negative and the other is positive, the LCM is conventionally taken as negative if the "leading" term in a sequence is negative, or simply the smallest positive common multiple. Here, it is safe to use the negative form as shown in the source. This ensures that both original terms can divide into the LCM.
L.C.M. = \( -(2^2 \times 3^2 \times a^3 \times b^2 \times c) \)
L.C.M. = \( -36a^3b^2c \)
In simple words: We break down each term into its basic parts: numbers and letters. We find the smallest number that 9 and 12 can both go into, which is 36. For the letters, we pick the highest power of each letter. Since one of the original terms was negative, the LCM also becomes negative.

🎯 Exam Tip: When dealing with negative coefficients in LCM, factor out the negative sign and find the LCM of the absolute values. The final LCM will generally carry the negative sign if one of the original terms has it, representing the smallest common multiple that both terms can divide into.

Question 2. Find the LCM of the given polynomials
(iii) \( 16m \), \( -12m^2n^2 \), \( 8n^2 \)
Answer:
First, we factorize each expression into its prime components:
\( 16m = 2^4 \times m \)
\( -12m^2n^2 = -(2^2 \times 3 \times m^2 \times n^2) \)
\( 8n^2 = 2^3 \times n^2 \)
To find the Least Common Multiple (LCM), we take the highest power of each prime factor and variable. The LCM of 16, 12, and 8 is 48. We consider the highest power for each variable present. This makes sure the LCM can be divided by all three original terms.
L.C.M. = \( -(2^4 \times 3 \times m^2 \times n^2) \)
L.C.M. = \( -48m^2n^2 \)
In simple words: We find the prime factors of the numbers (16, 12, 8) and list all the letters with their highest powers (m, \( m^2 \), n, \( n^2 \)). We take the biggest common multiple of the numbers, which is 48. We then combine this with the highest powers of all the letters. Since one term was negative, the final LCM is also negative.

🎯 Exam Tip: For multiple terms, especially with negatives, determine the LCM of the absolute values of the numerical coefficients. Then, for each variable, take the highest power that appears in any of the terms. If any coefficient is negative, the overall LCM is conventionally taken as negative.

Question 2. Find the LCM of the given polynomials
(iv) \( p^2 – 3p + 2 \), \( p^2 – 4 \)
Answer:
First, we factorize each polynomial completely:
For \( p^2 – 3p + 2 \):
We look for two numbers that multiply to 2 and add to -3. These numbers are -2 and -1.
\( p^2 – 3p + 2 = p^2 – 2p – p + 2 \)
\( = p(p – 2) – 1(p – 2) \)
\( = (p – 2)(p – 1) \)

For \( p^2 – 4 \):
This is a difference of squares, \( a^2 – b^2 = (a + b)(a – b) \). Here \( a = p \) and \( b = 2 \).
\( p^2 – 4 = p^2 – 2^2 \)
\( = (p + 2)(p – 2) \)

To find the Least Common Multiple (LCM), we include all unique factors from both polynomials, using the highest power for any repeated factors. We list all factors present and ensure each appears with its maximum exponent from either expression.
L.C.M. = \( (p – 2)(p + 2)(p – 1) \)
In simple words: We first break down each expression into its simpler multiplied parts, or factors. For example, \( p^2 – 4 \) becomes \( (p – 2) \) and \( (p + 2) \). Then, we list every unique factor we found, making sure to include any factor multiple times if it appears more often in one expression than the other. This combined list of factors is the LCM.

🎯 Exam Tip: Always factorize each polynomial completely into its simplest irreducible factors first. Then, for the LCM, take every unique factor that appears, using the highest power it has in any of the factorizations.

Question 2. Find the LCM of the given polynomials
(v) \( 2x^2 – 5x – 3 \), \( 4x^2 – 36 \)
Answer:
First, we factorize each polynomial completely:
For \( 2x^2 – 5x – 3 \):
We look for two numbers that multiply to \( 2 \times -3 = -6 \) and add to -5. These numbers are -6 and 1.
\( 2x^2 – 5x – 3 = 2x^2 – 6x + x – 3 \)
\( = 2x(x – 3) + 1(x – 3) \)
\( = (x – 3)(2x + 1) \)

For \( 4x^2 – 36 \):
We can factor out a common term, 4, and then apply the difference of squares formula, \( a^2 – b^2 = (a + b)(a – b) \). This makes factoring much easier.
\( 4x^2 – 36 = 4(x^2 – 9) \)
\( = 4(x^2 – 3^2) \)
\( = 4(x + 3)(x – 3) \)

To find the Least Common Multiple (LCM), we include all unique factors from both polynomials, using the highest power for any repeated factors. We gather all distinct factors, including numerical constants, with their highest exponents.
L.C.M. = \( 4(x – 3)(x + 3)(2x + 1) \)
In simple words: We first break each polynomial down into its simpler parts that multiply together. For the second polynomial, we also take out the common number 4. Then, we list all the unique parts we found. If a part appears in both, we only list it once. We multiply all these unique parts, including the number 4, to get the LCM.

🎯 Exam Tip: Remember to factor out any common numerical coefficients (like the 4 in \( 4x^2 – 36 \)) before applying other factorization techniques like the difference of squares. This ensures the LCM is truly the "least" common multiple.

Question 2. Find the LCM of the given polynomials
(vi) \( (2x^2 – 3xy)^2 \), \( (4x – 6y)^3 \), \( (8x^3 – 27y^3) \)
Answer:
First, we factorize each expression completely:
For \( (2x^2 – 3xy)^2 \):
Factor out \( x \) from the term inside the parenthesis.
\( (2x^2 – 3xy)^2 = (x(2x – 3y))^2 \)
\( = x^2 (2x – 3y)^2 \)

For \( (4x – 6y)^3 \):
Factor out 2 from the term inside the parenthesis.
\( (4x – 6y)^3 = (2(2x – 3y))^3 \)
\( = 2^3 (2x – 3y)^3 \)
\( = 8 (2x – 3y)^3 \)

For \( (8x^3 – 27y^3) \):
This is a difference of cubes formula, \( a^3 – b^3 = (a – b)(a^2 + ab + b^2) \). Here \( a = 2x \) and \( b = 3y \). Recognizing these special forms helps in quick factorization.
\( (8x^3 – 27y^3) = (2x)^3 – (3y)^3 \)
\( = (2x – 3y)((2x)^2 + (2x)(3y) + (3y)^2) \)
\( = (2x – 3y)(4x^2 + 6xy + 9y^2) \)

To find the Least Common Multiple (LCM), we include all unique factors from all three polynomials, using the highest power for any repeated factors. We combine the highest powers of all numerical coefficients and each distinct algebraic factor.
L.C.M. = \( 8x^2 (2x – 3y)^3 (4x^2 + 6xy + 9y^2) \)
In simple words: We first break down each complex expression into its simplest multiplied parts. We use rules for factoring out common terms and special formulas like "difference of cubes". Then, we list every unique factor we found across all expressions. For any factor that appears multiple times, we choose the one with the highest power. Finally, we multiply all these chosen factors together to get the LCM.

🎯 Exam Tip: Be vigilant for special factorization forms like difference of squares or difference/sum of cubes. Properly identifying these can greatly simplify complex expressions and lead to the correct LCM more efficiently.