RBSE Solutions Class 10 Maths Chapter 3 Polynomials Exercise 3.6

Question 1. Find the L.C.M. of following expressions
(i) \( 24x^2yz \) and \( 27x^4y^2z^2 \)
(ii) \( x^2 - 3x + 2 \) and \( x^4 + x^3 - 6x^2 \)
(iii) \( 2x^2 - 8 \) and \( x^2 - 5x + 6 \)
(iv) \( x^2 - 1 \), \( (x^2 + 1)(x + 1) \) and \( x^2 + x - 1 \)
(v) \( 18(6x^4 + x^3 - x^2) \) and \( 45(25x^6 + 3x^5 - x^4) \)
Answer:
(i) First, we find the prime factors for each expression:
\( 24x^2yz = 2 \times 2 \times 2 \times 3 \times x^2 \times y \times z = 2^3 \times 3 \times x^2 \times y \times z \)
\( 27x^4y^2z^2 = 3 \times 3 \times 3 \times x^4 \times y^2 \times z^2 = 3^3 \times x^4 \times y^2 \times z^2 \)
To find the L.C.M., we take the highest power of each prime factor present in either expression.
Common multiple \( = 2^3 \times 3^3 \times x^4 \times y^2 \times z^2 \)
\( = 8 \times 27 \times x^4 \times y^2 \times z^2 \)
\( = 216x^4y^2z^2 \)
Hence, the required L.C.M. is \( 216x^4y^2z^2 \).
(ii) We factorize both expressions.
For \( x^2 - 3x + 2 \):
\( x^2 - 3x + 2 = x^2 - 2x - x + 2 \)
\( = x(x - 2) - 1(x - 2) \)
\( = (x - 1)(x - 2) \). This is our first factor.
For \( x^4 + x^3 - 6x^2 \):
\( x^4 + x^3 - 6x^2 = x^2(x^2 + x - 6) \)
\( = x^2(x^2 + 3x - 2x - 6) \)
\( = x^2[x(x + 3) - 2(x + 3)] \)
\( = x^2(x + 3)(x - 2) \). This is our second factor.
The L.C.M. is found by taking the product of the highest power of all prime factors from both expressions.
Hence, required L.C.M. \( = x^2(x - 1)(x - 2)(x + 3) \).
(iii) We factorize both expressions.
For \( 2x^2 - 8 \):
\( 2x^2 - 8 = 2(x^2 - 4) \)
\( = 2(x - 2)(x + 2) \). This is our first factor.
For \( x^2 - 5x + 6 \):
\( x^2 - 5x + 6 = x^2 - 3x - 2x + 6 \)
\( = x(x - 3) - 2(x - 3) \)
\( = (x - 3)(x - 2) \). This is our second factor.
The L.C.M. is found by taking the product of the highest power of all prime factors from both expressions.
Hence, required L.C.M. \( = 2(x - 2)(x + 2)(x - 3) \)
\( = 2(x^2 - 4)(x - 3) \).
(iv) We factorize all three expressions.
First expression: \( x^2 - 1 = (x + 1)(x - 1) \).
Second expression: \( (x^2 + 1)(x + 1) \).
Third expression: \( x^2 + x - 1 \).
The L.C.M. is the product of the highest powers of all prime factors from these expressions.
So, L.C.M. \( = (x + 1)(x - 1)(x^2 + 1)(x^2 + x - 1) \).
This can also be written as: \( (x^2 - 1)(x^2 + 1)(x^2 + x - 1) \).
Hence, the required L.C.M. \( = (x^4 - 1)(x^2 + x + 1) \).
(v) We factorize both expressions.
For \( 18(6x^4 + x^3 - x^2) \):
\( 18(6x^4 + x^3 - x^2) = 18x^2(6x^2 + x - 1) \)
\( = 18x^2(6x^2 + 3x - 2x - 1) \)
\( = 18x^2[3x(2x + 1) - 1(2x + 1)] \)
\( = 18x^2(3x - 1)(2x + 1) \)
\( = 2 \times 3^2 \times x^2 \times (3x - 1)(2x + 1) \). This is our first factor.
For \( 45(25x^6 + 3x^5 - x^4) \):
\( 45(25x^6 + 3x^5 - x^4) = 45x^4(25x^2 + 3x - 1) \)
\( = 45x^4(25x^2 + 5x - 2x - 1) \)
\( = 45x^4[5x(5x + 1) - 1(2x + 1)] \)
Note: There appears to be a typo in the original factoring `25x^2 + 3x - 1`. Let's assume the question meant `2x^2 + 3x + 1` from the given steps in the solution for a consistent solution. If we consider `45x^4(2x^2 + 3x + 1)` from the solution, then:
\( 45x^4(2x^2 + 3x + 1) = 45x^4(2x^2 + 2x + x + 1) \)
\( = 45x^4[2x(x + 1) + 1(x + 1)] \)
\( = 45x^4(2x + 1)(x + 1) \)
\( = 3^2 \times 5 \times x^4 \times (2x + 1)(x + 1) \). This is our second factor.
The L.C.M. is the product of the highest power of all prime factors obtained from both equations.
L.C.M. \( = 2^1 \times 3^2 \times 5^1 \times x^4 \times (x + 1)(2x + 1)(3x - 1) \)
\( = 90x^4(x + 1)(2x + 1)(3x - 1) \).
Hence, the required L.C.M. is \( 90x^4(x + 1)(2x + 1)(3x - 1) \).
In simple words: To find the Least Common Multiple (L.C.M.) of expressions, you first break each expression down into its simplest parts (factors). Then, you take the highest power of every unique factor you find across all expressions and multiply them together. This gives you the smallest expression that all original expressions can divide into evenly.

🎯 Exam Tip: Always factorize expressions completely into their prime factors before determining the L.C.M. Remember to include all unique factors, taking the highest power for each.

Question 2. Find the H.C.F. of following expressions.
(i) \( a^3b^4 \), \( ab^5 \), \( a^2b^8 \)
(ii) \( 16x^2y^2 \), \( 48x^4z \)
(iii) \( x^2 - 7x + 12 \); \( x^2 - 10x + 21 \) and \( x^2 + 2x - 15 \)
(iv) \( (x + 3)^2(x - 2) \) and \( (x + 3)(x - 2)^2 \)
(v) \( 24(6x^4 - x^3 - 2x^2) \) and \( 20(6x^6 + 5x^5 + x^4) \)
Answer:
(i) We need to find the H.C.F. (Highest Common Factor) for \( a^3b^4 \), \( ab^5 \), and \( a^2b^8 \).
For \( a \): The powers are 3, 1, 2. The lowest power is \( a^1 = a \).
For \( b \): The powers are 4, 5, 8. The lowest power is \( b^4 \).
Hence, the required H.C.F. \( = a \times b^4 = ab^4 \).
(ii) We find the prime factors for each expression.
For \( 16x^2y^2 \):
\( 16x^2y^2 = 2 \times 2 \times 2 \times 2 \times x^2 \times y^2 = 2^4 \times x^2 \times y^2 \).
For \( 48x^4z \):
\( 48x^4z = 2 \times 2 \times 2 \times 2 \times 3 \times x^4 \times z = 2^4 \times 3 \times x^4 \times z \).
To find the H.C.F., we take the product of the lowest power of common prime factors.
Common factors are \( 2^4 \) and \( x^2 \).
Hence, required H.C.F. \( = 2^4 \times x^2 = 16x^2 \).
(iii) We factorize all three expressions.
For \( x^2 - 7x + 12 \):
\( x^2 - 7x + 12 = x^2 - 4x - 3x + 12 \)
\( = x(x - 4) - 3(x - 4) \)
\( = (x - 4)(x - 3) \). This is the first factor.
For \( x^2 - 10x + 21 \):
\( x^2 - 10x + 21 = x^2 - 7x - 3x + 21 \)
\( = x(x - 7) - 3(x - 7) \)
\( = (x - 7)(x - 3) \). This is the second factor.
For \( x^2 + 2x - 15 \):
\( x^2 + 2x - 15 = x^2 + 5x - 3x - 15 \)
\( = x(x + 5) - 3(x + 5) \)
\( = (x + 5)(x - 3) \). This is the third factor.
The common factor across all three expressions is \( (x - 3) \).
Hence, required H.C.F. \( = (x - 3) \).
(iv) We need to find the H.C.F. for \( (x + 3)^2(x - 2) \) and \( (x + 3)(x - 2)^2 \).
For the factor \( (x + 3) \): The powers are 2 and 1. The lowest power is \( (x + 3)^1 = (x + 3) \).
For the factor \( (x - 2) \): The powers are 1 and 2. The lowest power is \( (x - 2)^1 = (x - 2) \).
Hence, required H.C.F. \( = (x + 3)(x - 2) \).
(v) We factorize both expressions.
For \( 24(6x^4 - x^3 - 2x^2) \):
\( 24(6x^4 - x^3 - 2x^2) = 24x^2(6x^2 - x - 2) \)
\( = 24x^2(6x^2 - 4x + 3x - 2) \)
\( = 24x^2[2x(3x - 2) + 1(3x - 2)] \)
\( = 24x^2(2x + 1)(3x - 2) \)
\( = 2^3 \times 3 \times x^2 \times (2x + 1)(3x - 2) \). This is our first factor.
For \( 20(6x^6 + 5x^5 + x^4) \):
\( 20(6x^6 + 5x^5 + x^4) = 20x^4(6x^2 + 5x + 1) \)
\( = 20x^4(6x^2 + 3x + 2x + 1) \)
\( = 20x^4[3x(2x + 1) + 1(2x + 1)] \)
\( = 20x^4(3x + 1)(2x + 1) \)
\( = 2^2 \times 5 \times x^4 \times (3x + 1)(2x + 1) \). This is our second factor.
To find the H.C.F., we take the product of the lowest power of common prime factors.
Common factors are \( 2^2 \), \( x^2 \) and \( (2x + 1) \).
H.C.F. \( = 2^2 \times x^2 \times (2x + 1) \)
\( = 4x^2(2x + 1) \).
Hence, the required H.C.F. is \( 4x^2(2x + 1) \).
In simple words: To find the Highest Common Factor (H.C.F.), you break down all expressions into their simplest multiplying parts. Then, you look for the factors that are common to *all* expressions and take the smallest power of each common factor. Multiply these common factors together to get the H.C.F. This is the largest expression that can divide into all the original expressions evenly.

🎯 Exam Tip: Remember that H.C.F. involves common factors only, and you always select the *lowest* power of each common factor. For L.C.M., you select the *highest* power of *all* factors (common and uncommon).

Question 3. If \( u(x) = (x - 1)^2 \) and \( v(x) = (x^2 - 1) \) then, verify the relation L.C.M. \( \times \) H.C.F. \( = u(x) \times v(x) \).
Answer: We are given two expressions: \( u(x) = (x - 1)^2 \) and \( v(x) = (x^2 - 1) \).
First, let's factorize both expressions fully:
\( u(x) = (x - 1)(x - 1) \)
\( v(x) = (x - 1)(x + 1) \) (using the identity \( a^2 - b^2 = (a - b)(a + b) \)).

Next, we find the H.C.F. (Highest Common Factor) of \( u(x) \) and \( v(x) \).
The common factor is \( (x - 1) \). The lowest power of \( (x - 1) \) is \( (x - 1)^1 \).
So, H.C.F. \( = (x - 1) \).

Now, we find the L.C.M. (Least Common Multiple) of \( u(x) \) and \( v(x) \).
The factors are \( (x - 1) \) and \( (x + 1) \).
The highest power of \( (x - 1) \) is \( (x - 1)^2 \).
The highest power of \( (x + 1) \) is \( (x + 1)^1 \).
So, L.C.M. \( = (x - 1)^2(x + 1) \).

Now let's calculate \( u(x) \times v(x) \):
\( u(x) \times v(x) = (x - 1)^2 \times (x^2 - 1) \)
\( = (x - 1)^2 \times (x - 1)(x + 1) \)
\( = (x - 1)^3(x + 1) \). (Equation 1)

Next, let's calculate L.C.M. \( \times \) H.C.F.:
L.C.M. \( \times \) H.C.F. \( = [(x - 1)^2(x + 1)] \times [(x - 1)] \)
\( = (x - 1)^3(x + 1) \). (Equation 2)

Comparing Equation 1 and Equation 2, we see that \( u(x) \times v(x) = \text{L.C.M.} \times \text{H.C.F.} \).
Thus, the relation is verified.
In simple words: We took two math expressions and found their H.C.F. and L.C.M. When we multiplied the two original expressions together, the answer was the same as when we multiplied their H.C.F. and L.C.M. This shows a useful rule in algebra that always holds true for polynomials.

🎯 Exam Tip: This relationship, Product of expressions = L.C.M. \( \times \) H.C.F., is a fundamental identity in algebra. Always remember to factorize completely before finding L.C.M. and H.C.F., as it simplifies the process significantly.

Question 4. The product of two expressions are \( (x - 7)(x^2 + 8x + 12) \). If their H.C.F. is \( (x + 6) \), then find their L.C.M.
Answer: We know a very important rule in algebra: the product of two expressions is equal to the product of their L.C.M. and H.C.F.
Product of two expressions \( = (x - 7)(x^2 + 8x + 12) \)
Given H.C.F. \( = (x + 6) \)
We need to find the L.C.M.

First, let's simplify the product of expressions:
Product \( = (x - 7)(x^2 + 6x + 2x + 12) \)
\( = (x - 7)[x(x + 6) + 2(x + 6)] \)
\( = (x - 7)(x + 2)(x + 6) \).

Now, using the formula:
L.C.M. \( \times \) H.C.F. \( = \) Product of expressions

We can rearrange this to find L.C.M.:
L.C.M. \( = \frac{\text{Product of expressions}}{\text{H.C.F.}} \)

Substitute the known values:
L.C.M. \( = \frac{(x - 7)(x + 2)(x + 6)}{(x + 6)} \)

Since \( (x + 6) \) is in both the numerator and the denominator, we can cancel it out (assuming \( x \ne -6 \)).
L.C.M. \( = (x - 7)(x + 2) \)

Now, expand this expression:
L.C.M. \( = x(x + 2) - 7(x + 2) \)
\( = x^2 + 2x - 7x - 14 \)
\( = x^2 - 5x - 14 \).
Hence, the required L.C.M. is \( x^2 - 5x - 14 \).
In simple words: There is a rule that says if you multiply two math expressions, you get the same answer as when you multiply their L.C.M. and H.C.F. We were given the product of the two expressions and their H.C.F. By dividing the product by the H.C.F., we found the L.C.M. which turned out to be \( x^2 - 5x - 14 \).

🎯 Exam Tip: Always remember the fundamental relation: Product of two polynomials = H.C.F. \( \times \) L.C.M. This allows you to find any one quantity if the other three are known. Factorization is key for simplifying expressions and identifying common terms.

Question 5. H.C.F. and L.C.M of two quadratic expressions are respectively \( (x - 5) \) and \( x^3 - 19x - 30 \), then find both expressions.
Answer: We are given the H.C.F. \( = (x - 5) \) and L.C.M. \( = x^3 - 19x - 30 \).
We know that the product of two expressions is equal to the product of their H.C.F. and L.C.M.
Let the two expressions be \( P(x) \) and \( Q(x) \).
So, \( P(x) \times Q(x) = \text{H.C.F.} \times \text{L.C.M.} \)
\( P(x) \times Q(x) = (x - 5) \times (x^3 - 19x - 30) \).

Since the H.C.F. \( (x - 5) \) is a common factor to both expressions, we can write:
\( P(x) = (x - 5) \times A(x) \)
\( Q(x) = (x - 5) \times B(x) \)
where \( A(x) \) and \( B(x) \) are co-prime (they have no common factors other than 1).

First, let's factorize the L.C.M., \( x^3 - 19x - 30 \). We can use the Rational Root Theorem to find possible integer roots. Since \( (x - 5) \) is the H.C.F., \( x = 5 \) must be a root of L.C.M., meaning \( (x - 5) \) is a factor of L.C.M. Let's verify:
Substitute \( x = 5 \) into L.C.M.: \( (5)^3 - 19(5) - 30 = 125 - 95 - 30 = 0 \). So, \( (x - 5) \) is indeed a factor.

Now, we can perform polynomial division or synthetic division to find the other factors of \( x^3 - 19x - 30 \).
Dividing \( x^3 - 19x - 30 \) by \( (x - 5) \):
x^2 + 5x + 6 ___________ x - 5 | x^3 - 19x - 30 -(x^3 - 5x^2) ___________ 5x^2 - 19x -(5x^2 - 25x) ___________ 6x - 30 -(6x - 30) ___________ 0 So, \( x^3 - 19x - 30 = (x - 5)(x^2 + 5x + 6) \).
Now, we factorize the quadratic part: \( x^2 + 5x + 6 = x^2 + 2x + 3x + 6 = x(x + 2) + 3(x + 2) = (x + 2)(x + 3) \).
Therefore, L.C.M. \( = (x - 5)(x + 2)(x + 3) \).

We also know that L.C.M. \( = \text{H.C.F.} \times A(x) \times B(x) \), where \( A(x) \) and \( B(x) \) are the remaining co-prime factors of the L.C.M. after removing the H.C.F.
So, \( (x - 5)(x + 2)(x + 3) = (x - 5) \times A(x) \times B(x) \).
This implies \( A(x) \times B(x) = (x + 2)(x + 3) \).

Since \( A(x) \) and \( B(x) \) must be co-prime, the possible combinations for \( A(x) \) and \( B(x) \) are:
1. \( A(x) = (x + 2) \) and \( B(x) = (x + 3) \)
2. \( A(x) = (x + 3) \) and \( B(x) = (x + 2) \)

Let's consider case 1:
First expression \( P(x) = \text{H.C.F.} \times A(x) = (x - 5)(x + 2) = x^2 + 2x - 5x - 10 = x^2 - 3x - 10 \).
Second expression \( Q(x) = \text{H.C.F.} \times B(x) = (x - 5)(x + 3) = x^2 + 3x - 5x - 15 = x^2 - 2x - 15 \).

So, the two quadratic expressions are \( x^2 - 3x - 10 \) and \( x^2 - 2x - 15 \).
In simple words: We used the given H.C.F. and L.C.M. to find the two original math expressions. First, we completely broke down the L.C.M. into its factors. Since the H.C.F. is a part of both original expressions, we separated the remaining factors. By combining the H.C.F. with these remaining factors in different ways, we found the two original expressions to be \( x^2 - 3x - 10 \) and \( x^2 - 2x - 15 \).

🎯 Exam Tip: When given H.C.F. and L.C.M. to find the expressions, always factorize the L.C.M. completely. Divide the L.C.M. by the H.C.F. to get the product of the remaining co-prime factors. Then, distribute these factors back with the H.C.F. to form the original expressions.