Samacheer Kalvi Class 10 Maths Solutions Chapter 3 Algebra Exercise 3.3

 

Question 1. Find the LCM and GCD for the following and verify that \( f(x) \times g(x) = \text{LCM} \times \text{GCD} \)
(i) \( 21x^2y, 35xy^2 \)
Answer:
For \( 21x^2y, 35xy^2 \):
First, we define the given polynomials:
\( p(x) = 21x^2y = 3 \times 7 \times x^2 \times y \)
\( g(x) = 35xy^2 = 5 \times 7 \times x \times y^2 \)

Next, we find the Greatest Common Divisor (GCD):
\( \text{G.C.D} = 7xy \)

Then, we find the Least Common Multiple (LCM):
\( \text{L.C.M} = 3 \times 5 \times 7 \times x^2 \times y^2 \)
\( = 105x^2y^2 \)

Now, let's verify the relationship \( p(x) \times g(x) = \text{LCM} \times \text{GCD} \):
Product of LCM and GCD:
\( \text{L.C.M} \times \text{G.C.D} = (105x^2y^2) \times (7xy) \)
\( = 735x^3y^3 \) ....(1)

Product of the polynomials:
\( p(x) \times g(x) = (21x^2y) \times (35xy^2) \)
\( = 735x^3y^3 \) ....(2)

From equations (1) and (2), we can see:
\( \text{L.C.M} \times \text{G.C.D.} = p(x) \times g(x) \)
In simple words: We found the common factors for GCD and all factors for LCM of the given expressions. Then, we checked if multiplying the two original expressions gives the same result as multiplying their LCM and GCD, and it does. This property holds for all polynomials.

🎯 Exam Tip: Remember to list all prime factors for coefficients and the highest/lowest powers for variables when finding LCM and GCD. The verification step is crucial to confirm your calculations.

 

Question 1. Find the LCM and GCD for the following and verify that \( f(x) \times g(x) = \text{LCM} \times \text{GCD} \)
(ii) \( (x^3 - 1)(x + 1), (x^3 + 1) \)
Answer:
Let's define the polynomials:
\( p(x) = (x^3 - 1)(x + 1) \)
First, factor \( x^3 - 1 \) using the difference of cubes formula \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \).
So, \( p(x) = (x - 1)(x^2 + x + 1)(x + 1) \)

Next, define the second polynomial:
\( g(x) = x^3 + 1 \)
Factor \( x^3 + 1 \) using the sum of cubes formula \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \).
So, \( g(x) = (x + 1)(x^2 - x + 1) \)

Now, find the Greatest Common Divisor (GCD):
The common factor is \( (x + 1) \).
\( \text{G.C.D} = (x + 1) \)

Next, find the Least Common Multiple (LCM):
\( \text{L.C.M} = (x + 1)(x - 1)(x^2 + x + 1)(x^2 - x + 1) \)

Let's verify the relationship \( p(x) \times g(x) = \text{LCM} \times \text{GCD} \):
Product of LCM and GCD:
\( \text{L.C.M} \times \text{G.C.D} = (x + 1)(x - 1)(x^2 + x + 1)(x^2 - x + 1) \times (x + 1) \)
\( = (x + 1)^2 (x - 1)(x^2 + x + 1)(x^2 - x + 1) \) ..........(1)

Product of the polynomials:
\( p(x) \times g(x) = (x - 1)(x^2 + x + 1)(x + 1) \times (x + 1)(x^2 - x + 1) \)
\( = (x + 1)^2 (x - 1)(x^2 + x + 1)(x^2 - x + 1) \) ..........(2)

From equations (1) and (2), we get:
\( \text{L.C.M} \times \text{G.C.D.} = p(x) \times g(x) \)
In simple words: We broke down each algebraic expression into its simplest factors. Then, we found the largest common factor (GCD) and the smallest common multiple (LCM). Finally, we showed that if you multiply the two original expressions, you get the same answer as when you multiply their GCD and LCM. This is a fundamental property of polynomials.

🎯 Exam Tip: When factoring cubic polynomials like \( x^3 \pm 1 \), remember the sum/difference of cubes formulas: \( a^3 + b^3 = (a+b)(a^2-ab+b^2) \) and \( a^3 - b^3 = (a-b)(a^2+ab+b^2) \).

 

Question 1. Find the LCM and GCD for the following and verify that \( f(x) \times g(x) = \text{LCM} \times \text{GCD} \)
(iii) \( (x^2y + xy^2), (x^2 + xy) \)
Answer:
Let's define the polynomials:
\( p(x) = x^2y + xy^2 \)
Factor out the common terms:
\( p(x) = xy(x + y) \)

Next, define the second polynomial:
\( g(x) = x^2 + xy \)
Factor out the common terms:
\( g(x) = x(x + y) \)

Now, find the Greatest Common Divisor (GCD):
The common factors are \( x \) and \( (x+y) \).
\( \text{G.C.D} = x(x + y) \)

Next, find the Least Common Multiple (LCM):
The LCM includes all factors, with the highest power for each.
\( \text{L.C.M} = xy(x + y) \)

Let's verify the relationship \( p(x) \times g(x) = \text{LCM} \times \text{GCD} \):
Product of LCM and GCD:
\( \text{L.C.M} \times \text{G.C.D} = [xy(x + y)] \times [x(x + y)] \)
\( = x^2y(x + y)^2 \) .....(1)

Product of the polynomials:
\( p(x) \times g(x) = [xy(x + y)] \times [x(x + y)] \)
\( = x^2y(x + y)^2 \) .....(2)

From equations (1) and (2), we get:
\( \text{L.C.M} \times \text{G.C.D.} = p(x) \times g(x) \)
In simple words: We first simplified both expressions by taking out common parts. Then we found their biggest shared part (GCD) and smallest shared multiple (LCM). We confirmed that when you multiply the original expressions, you get the same answer as when you multiply their GCD and LCM, proving the relationship works here too.

🎯 Exam Tip: Always look for common factors first when dealing with polynomial expressions. Factoring helps simplify the process of finding both LCM and GCD.

 

Question 2. Find the LCM of each pair of the following polynomials
(i) \( a^2 + 4a - 12, a^2 - 5a + 6 \) whose GCD is \( a - 2 \)
Answer:
We are given the two polynomials and their GCD.
First polynomial: \( p(x) = a^2 + 4a - 12 \)
To factor this, we look for two numbers that multiply to -12 and add to 4 (which are 6 and -2).
\( p(x) = a^2 + 6a - 2a - 12 \)
\( = a(a + 6) - 2(a + 6) \)
\( = (a + 6)(a - 2) \)

Second polynomial: \( g(x) = a^2 - 5a + 6 \)
To factor this, we look for two numbers that multiply to 6 and add to -5 (which are -3 and -2).
\( g(x) = a^2 - 3a - 2a + 6 \)
\( = a(a - 3) - 2(a - 3) \)
\( = (a - 3)(a - 2) \)

We are given that \( \text{G.C.D.} = (a - 2) \). This matches one of our factors, which is a good sign.

To find the LCM, we use the formula: \( \text{L.C.M.} = \frac{p(x) \times g(x)}{\text{G.C.D.}} \)
\( \text{L.C.M.} = \frac{(a+6)(a-2) \times (a-3)(a-2)}{(a-2)} \)
Cancel out one \( (a-2) \) from the numerator and denominator:
\( \text{L.C.M.} = (a+6)(a-3)(a-2) \)
In simple words: We factored both polynomial expressions into simpler terms. Since we already knew their Greatest Common Divisor (GCD), we used the special rule that says the LCM can be found by multiplying the two expressions and then dividing by their GCD. This gave us the final Least Common Multiple.

🎯 Exam Tip: When given GCD, it simplifies finding LCM. Always factor the polynomials completely before applying the formula \( \text{LCM} = \frac{p(x) \times g(x)}{\text{GCD}} \).

 

Question 2. Find the LCM of each pair of the following polynomials
(ii) \( x^4 - 27a^3x, (x - 3a)^2 \) whose GCD is \( (x - 3a) \)
Answer:
We are given the two polynomials and their GCD.
First polynomial: \( p(x) = x^4 - 27a^3x \)
Factor out the common term \( x \):
\( p(x) = x(x^3 - 27a^3) \)
Recognize \( 27a^3 \) as \( (3a)^3 \). Use the difference of cubes formula \( A^3 - B^3 = (A - B)(A^2 + AB + B^2) \):
\( p(x) = x[x^3 - (3a)^3] \)
\( = x(x - 3a)(x^2 + 3ax + 9a^2) \)

Second polynomial: \( g(x) = (x - 3a)^2 \)

We are given that \( \text{G.C.D.} = (x - 3a) \). This matches a factor in both polynomials.

To find the LCM, we use the formula: \( \text{L.C.M.} = \frac{p(x) \times g(x)}{\text{G.C.D.}} \)
\( \text{L.C.M.} = \frac{[x(x-3a)(x^2+3ax+9a^2)] \times (x-3a)^2}{(x-3a)} \)
Cancel out one \( (x-3a) \) from the numerator and denominator:
\( \text{L.C.M.} = x(x - 3a)^2 (x^2 + 3ax + 9a^2) \)
In simple words: We took the first polynomial and factored it completely, using the difference of cubes formula. The second polynomial was already in a factored form. Knowing the Greatest Common Divisor (GCD), we used a simple rule: multiply the two original polynomials and then divide by their GCD to find the Least Common Multiple (LCM). This method helps simplify complex expressions.

🎯 Exam Tip: Always look for common factors first, then apply algebraic identities like difference of cubes to factor polynomials. This makes GCD and LCM calculations easier.

 

Question 3. Find the GCD of each pair of the following polynomials
(i) \( 12(x^4 - x^3), 8(x^4 - 3x^3 + 2x^2) \) whose LCM is \( 24x^3(x - 1)(x - 2) \)
Answer:
We are given the two polynomials and their LCM.
First polynomial: \( p(x) = 12(x^4 - x^3) \)
Factor out the common term \( x^3 \):
\( p(x) = 12x^3(x - 1) \)

Second polynomial: \( g(x) = 8(x^4 - 3x^3 + 2x^2) \)
Factor out the common term \( x^2 \):
\( g(x) = 8x^2(x^2 - 3x + 2) \)
Now, factor the quadratic expression \( x^2 - 3x + 2 \). We need two numbers that multiply to 2 and add to -3 (which are -2 and -1).
\( g(x) = 8x^2(x - 2)(x - 1) \)

We are given the LCM: \( \text{L.C.M.} = 24x^3(x - 1)(x - 2) \)

To find the GCD, we use the formula: \( \text{G.C.D.} = \frac{p(x) \times g(x)}{\text{L.C.M.}} \)
\( \text{G.C.D.} = \frac{[12x^3(x-1)] \times [8x^2(x-2)(x-1)]}{24x^3(x-1)(x-2)} \)
Now, we simplify the expression by canceling common factors:
\( \text{G.C.D.} = \frac{12 \times 8 \times x^3 \times x^2 \times (x-1) \times (x-2) \times (x-1)}{24 \times x^3 \times (x-1) \times (x-2)} \)
\( \text{G.C.D.} = \frac{96x^5(x-1)^2(x-2)}{24x^3(x-1)(x-2)} \)
Divide the coefficients: \( \frac{96}{24} = 4 \)
Divide the \( x \) terms: \( \frac{x^5}{x^3} = x^{5-3} = x^2 \)
Divide the \( (x-1) \) terms: \( \frac{(x-1)^2}{(x-1)} = (x-1) \)
The \( (x-2) \) terms cancel out.
\( \text{G.C.D.} = 4x^2(x - 1) \)
In simple words: We first factored both given algebraic expressions into their simplest forms. We were provided with their Least Common Multiple (LCM). Using the rule that states the product of two polynomials is equal to the product of their LCM and GCD, we rearranged the formula to find the Greatest Common Divisor (GCD). After canceling out common terms, we got the final GCD.

🎯 Exam Tip: When finding GCD or LCM using the formula, always factor the polynomials completely first. This helps in correctly canceling terms and identifying the common/uncommon factors.

 

Question 3. Find the GCD of each pair of the following polynomials
(ii) \( (x^3 + y^3), (x^4 + x^2y^2 + y^4) \) whose LCM is \( (x^3 + y^3)(x^2 + xy + y^2) \)
Answer:
We are given the two polynomials and their LCM.
First polynomial: \( p(x) = x^3 + y^3 \)
Factor using the sum of cubes formula \( A^3 + B^3 = (A + B)(A^2 - AB + B^2) \):
\( p(x) = (x + y)(x^2 - xy + y^2) \)

Second polynomial: \( g(x) = x^4 + x^2y^2 + y^4 \)
This can be factored using a special identity: \( x^4 + x^2y^2 + y^4 = (x^2 + y^2)^2 - (xy)^2 \).
Now, use the difference of squares formula \( A^2 - B^2 = (A - B)(A + B) \):
\( g(x) = (x^2 + y^2 + xy)(x^2 + y^2 - xy) \)

We are given the LCM: \( \text{L.C.M.} = (x^3 + y^3)(x^2 + xy + y^2) \)
Substitute the factored form of \( x^3 + y^3 \):
\( \text{L.C.M.} = (x + y)(x^2 - xy + y^2)(x^2 + xy + y^2) \)

To find the GCD, we use the formula: \( \text{G.C.D.} = \frac{p(x) \times g(x)}{\text{L.C.M.}} \)
\( \text{G.C.D.} = \frac{[(x+y)(x^2-xy+y^2)] \times [(x^2+y^2+xy)(x^2+y^2-xy)]}{(x+y)(x^2-xy+y^2)(x^2+xy+y^2)} \)
Cancel out common factors from the numerator and denominator:
The terms \( (x+y) \), \( (x^2-xy+y^2) \), and \( (x^2+y^2+xy) \) cancel out.
\( \text{G.C.D.} = (x^2 + y^2 - xy) \)
In simple words: We first factored both given polynomials using algebraic identities like sum of cubes and a special factoring pattern for the fourth-degree expression. Since we already knew the Least Common Multiple (LCM), we used the relationship that the product of two polynomials is equal to the product of their LCM and GCD. By rearranging this rule and canceling out common parts, we found the Greatest Common Divisor (GCD).

🎯 Exam Tip: Recognizing special factoring patterns like \( x^4 + x^2y^2 + y^4 \) (which can be written as a difference of squares) is key to solving these types of problems efficiently.

 

Question 4. Given the L.C.M and G.C.D of the two polynomials \( p(x) \) and \( q(x) \) find the unknown polynomial in the following table

Answer:
We use the fundamental relationship: \( p(x) \times q(x) = \text{LCM} \times \text{GCD} \).
From this, we can find the unknown polynomial by: \( \text{Unknown polynomial} = \frac{\text{LCM} \times \text{GCD}}{\text{Known polynomial}} \)

(i) Find \( q(x) \):
Given:
\( \text{L.C.M.} = a^3 - 10a^2 + 11a + 70 \)
\( \text{G.C.D.} = a - 7 \)
\( p(x) = a^2 - 12a + 35 \)

First, factor the LCM polynomial. Since the GCD is \( (a-7) \), we know \( (a-7) \) is a factor of the LCM.
We can perform synthetic division or polynomial long division to factor \( a^3 - 10a^2 + 11a + 70 \) by \( (a-7) \).
Using synthetic division with root 7:
7 | 1 -10 11 70
| 7 -21 -70
----------------
1 -3 -10 0
So, \( \text{L.C.M.} = (a - 7)(a^2 - 3a - 10) \)
Further factor the quadratic: \( a^2 - 3a - 10 \). We need two numbers that multiply to -10 and add to -3 (which are -5 and 2).
\( \text{L.C.M.} = (a - 7)(a - 5)(a + 2) \)

Next, factor \( p(x) = a^2 - 12a + 35 \). We need two numbers that multiply to 35 and add to -12 (which are -5 and -7).
\( p(x) = (a - 5)(a - 7) \)

Now, substitute these into the formula for \( q(x) \):
\( q(x) = \frac{\text{LCM} \times \text{GCD}}{p(x)} \)
\( q(x) = \frac{[(a-7)(a-5)(a+2)] \times (a-7)}{[(a-5)(a-7)]} \)
Cancel out the common factors \( (a-5) \) and one \( (a-7) \):
\( q(x) = (a + 2)(a - 7) \)
\( q(x) = a^2 - 7a + 2a - 14 \)
\( q(x) = a^2 - 5a - 14 \)

(ii) Find \( p(x) \):
Given:
\( \text{L.C.M.} = (x^2 + y^2)(x^4 + x^2y^2 + y^4) \)
\( \text{G.C.D.} = x^2 - y^2 \)
\( q(x) = (x^4 - y^4)(x^2 + y^2 - xy) \)

First, let's factor the given terms.
Factor the LCM: \( x^4 + x^2y^2 + y^4 \) can be written as \( (x^2+y^2)^2 - (xy)^2 \).
\( (x^2+y^2)^2 - (xy)^2 = (x^2+y^2+xy)(x^2+y^2-xy) \)
So, \( \text{L.C.M.} = (x^2 + y^2)(x^2 + y^2 + xy)(x^2 + y^2 - xy) \)

Factor the GCD: \( x^2 - y^2 \)
\( \text{G.C.D.} = (x - y)(x + y) \)

Factor \( q(x) = (x^4 - y^4)(x^2 + y^2 - xy) \)
Factor \( x^4 - y^4 \) using difference of squares twice: \( (x^2)^2 - (y^2)^2 = (x^2 - y^2)(x^2 + y^2) = (x - y)(x + y)(x^2 + y^2) \)
So, \( q(x) = (x - y)(x + y)(x^2 + y^2)(x^2 + y^2 - xy) \)

Now, substitute these into the formula for \( p(x) \):
\( p(x) = \frac{\text{LCM} \times \text{GCD}}{q(x)} \)
\( p(x) = \frac{[(x^2+y^2)(x^2+y^2+xy)(x^2+y^2-xy)] \times [(x-y)(x+y)]}{[(x-y)(x+y)(x^2+y^2)(x^2+y^2-xy)]} \)
Cancel out the common factors: \( (x^2+y^2) \), \( (x^2+y^2-xy) \), \( (x-y) \), \( (x+y) \).
\( p(x) = (x^2 + y^2 + xy) \)
In simple words: For both parts of the question, we used the main rule that the product of two polynomials is equal to the product of their Least Common Multiple (LCM) and Greatest Common Divisor (GCD). We factored all the given expressions into their simplest forms. Then, we filled in the known values into the formula and canceled out matching terms from the top and bottom to find the missing polynomial. This strategy helps solve for an unknown polynomial when the others and their LCM/GCD are known.

🎯 Exam Tip: Always factor all given polynomials (LCM, GCD, and one of \( p(x) \) or \( q(x) \)) into their simplest forms before attempting to solve for the unknown. This makes cancellation much easier and reduces errors.