Exercise 5.4: HCF and LCM of Algebraic Expressions
1. Which one is the square of $(a – 5)$?
Explanation:
$(a – 5)^2 = a^2 – 2 \cdot a \cdot 5 + 5^2 = a^2 – 10a + 25$
Answer: b) $a^2 – 10a + 25$
2. Which one is the value of $(x + y)^2 + 2(x + y)(x – y) + (x – y)^2$?
Explanation: Let, $x + y = a$ and $x – y = b$
Therefore, Given expression $= a^2 + 2ab + b^2 = (a + b)^2$
$= \{(x + y) + (x – y)\}^2 = (2x)^2 = 4x^2$ [substituting values]
Answer: (c) $4x^2$
3. If $a + b = 4$ and $a – b = 2$, what is the value of $ab$?
Explanation:
$ab = \left(\frac{a + b}{2}\right)^2 – \left(\frac{a – b}{2}\right)^2 = \left(\frac{4}{2}\right)^2 – \left(\frac{2}{2}\right)^2 = (2)^2 – (1)^2 = 4 – 1 = 3$
Answer: (a) $3$
4. If a number is divisible without remainder by another number, what is called dividend in respect of divisor?
Explanation:
A multiple is a number that can be divided by another number without leaving a remainder.
Answer: (c) Multiple
5. What is the H.C.F. of $2a$ and $3b$?
Explanation:
H.C.F. of numerical coefficients $2$ and $3 = 1$
and there is no common factor between the algebraic variables $a$ and $b$.
Therefore, H.C.F. $= 1$
Answer: (a) $1$
6. What is the L.C.M. of $9x^2 – 25y^2$ and $15ax – 25ay$?
Explanation:
1st expression $= 9x^2 – 25y^2 = (3x)^2 – (5y)^2 = (3x + 5y)(3x – 5y)$
2nd expression $= 15ax – 25ay = 5a(3x – 5y)$
Required L.C.M. $= 5a(3x + 5y)(3x – 5y) = 5a(9x^2 – 25y^2)$
Answer: (d) $5a(9x^2 – 25y^2)$
7. What is the H.C.F of $x^3y^5$ and $a^2 – b^2$?
Explanation:
There is no common factor between the two expressions (other than 1). Therefore, the HCF will be $1$.
Answer: (d) $1$
8. If $x – \frac{1}{x} = 0$, then—
(i) $x = 1$
(ii) $x = -1$
(iii) $x = \pm 1$
Which of the following is correct?
Explanation: $x – \frac{1}{x} = 0$
or, $\frac{x^2 – 1}{x} = 0$
or, $x^2 – 1 = 0$
or, $x^2 = 1$
or, $x = \pm \sqrt{1} = \pm 1$
That means $x = 1$ or $x = -1$. Therefore, all three statements (i), (ii), and (iii) are correct.
Answer: d) i, ii, and iii
9. If $a + \frac{1}{a} = 4$, what is the value of $a^2 – 4a + 1$?
Explanation: $a + \frac{1}{a} = 4$
or, $\frac{a^2 + 1}{a} = 4$
or, $a^2 + 1 = 4a$
or, $a^2 – 4a + 1 = 0$
Answer: (d) $0$
10. What is the square of $(a + 5)$?
Explanation:
$(a + 5)^2 = a^2 + 2 \cdot a \cdot 5 + 5^2 = a^2 + 10a + 25$
Answer: (a) $a^2 + 10a + 25$
11. If $a + b = 8$ and $a – b = 4$, then $ab =$?
Explanation:
$ab = \left(\frac{a + b}{2}\right)^2 – \left(\frac{a – b}{2}\right)^2 = \left(\frac{8}{2}\right)^2 – \left(\frac{4}{2}\right)^2 = 4^2 – 2^2 = 16 – 4 = 12$
Answer: (c) $12$
Determine the H.C.F. (12–21):
12. $3a^3b^2c, 6ab^2c^2$
Solution:
1st expression $= 3a^3b^2c$
2nd expression $= 6ab^2c^2$
H.C.F. of numerical coefficients $3$ and $6 = 3$
H.C.F. of algebraic variables $= ab^2c$
Required H.C.F. $= 3ab^2c$
13. $5ab^2x^2, 10a^2by^2$
Solution:
1st expression $= 5ab^2x^2$
2nd expression $= 10a^2by^2$
H.C.F. of numerical coefficients $5$ and $10 = 5$
H.C.F. of algebraic variables $= ab$
Required H.C.F. $= 5ab$
14. $3a^2x^2, 6axy^2, 9ay^2$
Solution:
1st expression $= 3a^2x^2$
2nd expression $= 6axy^2$
3rd expression $= 9ay^2$
H.C.F. of numerical coefficients $3, 6, 9 = 3$
H.C.F. of algebraic variables $= a$
Required H.C.F. $= 3a$
15. $16a^3x^4y, 40a^2y^3x, 28ax^3$
Solution:
1st expression $= 16a^3x^4y$
2nd expression $= 40a^2y^3x$
3rd expression $= 28ax^3$
H.C.F. of numerical coefficients $16, 40, 28 = 4$
H.C.F. of algebraic variables $= ax$
Required H.C.F. $= 4ax$
16. $a^2 + ab, a^2 – b^2$
Solution:
1st expression $= a^2 + ab = a(a + b)$
2nd expression $= a^2 – b^2 = (a + b)(a – b)$
Required H.C.F. $= (a + b)$
17. $x^3y – xy^3, (x – y)^2$
Solution:
1st expression $= x^3y – xy^3 = xy(x^2 – y^2) = xy(x + y)(x – y)$
2nd expression $= (x – y)^2 = (x – y)(x – y)$
Required H.C.F. $= (x – y)$
18. $x^2 + 7x + 12, x^2 + 9x + 20$
Solution:
1st expression $= x^2 + 7x + 12 = x^2 + 4x + 3x + 12 = x(x + 4) + 3(x + 4) = (x + 4)(x + 3)$
2nd expression $= x^2 + 9x + 20 = x^2 + 5x + 4x + 20 = x(x + 5) + 4(x + 5) = (x + 5)(x + 4)$
Required H.C.F. $= (x + 4)$
19. $a^3 – ab^2, a^4 + 2a^3b + a^2b^2$
Solution:
1st expression $= a^3 – ab^2 = a(a^2 – b^2) = a(a + b)(a – b)$
2nd expression $= a^4 + 2a^3b + a^2b^2 = a^2(a^2 + 2ab + b^2) = a^2(a + b)^2 = a^2(a + b)(a + b)$
Required H.C.F. $= a(a + b)$
20. $a^2 – 16, 3a + 12, a^2 + 5a + 4$
Solution:
1st expression $= a^2 – 16 = a^2 – 4^2 = (a + 4)(a – 4)$
2nd expression $= 3a + 12 = 3(a + 4)$
3rd expression $= a^2 + 5a + 4 = a^2 + 4a + a + 4 = a(a + 4) + 1(a + 4) = (a + 4)(a + 1)$
Required H.C.F. $= (a + 4)$
21. $xy – y, x^3y – xy, x^2 – 2x + 1$
Solution:
1st expression $= xy – y = y(x – 1)$
2nd expression $= x^3y – xy = xy(x^2 – 1) = xy(x + 1)(x – 1)$
3rd expression $= x^2 – 2x + 1 = (x – 1)^2 = (x – 1)(x – 1)$
Required H.C.F. $= (x – 1)$
Determine the L.C.M. (22–31):
22. $6a^3b^2c, 9a^4bd^2$
Solution:
1st expression $= 6a^3b^2c$
2nd expression $= 9a^4bd^2$
L.C.M. of numerical coefficients $6$ and $9 = 18$
L.C.M. of algebraic variables $= a^4b^2cd^2$
Required L.C.M. $= 18a^4b^2cd^2$
23. $5x^2y^2, 10xz^3, 15y^3z^4$
Solution:
1st expression $= 5x^2y^2$
2nd expression $= 10xz^3$
3rd expression $= 15y^3z^4$
L.C.M. of numerical coefficients $5, 10, 15 = 30$
L.C.M. of algebraic variables $= x^2y^3z^4$
Required L.C.M. $= 30x^2y^3z^4$
24. $2p^2xy^2, 3pq^2, 6pqx^2$
Solution:
1st expression $= 2p^2xy^2$
2nd expression $= 3pq^2$
3rd expression $= 6pqx^2$
L.C.M. of numerical coefficients $2, 3, 6 = 6$
L.C.M. of algebraic variables $= p^2q^2x^2y^2$
Required L.C.M. $= 6p^2q^2x^2y^2$
25. $(b^2 – c^2), (b + c)^2$
Solution:
1st expression $= b^2 – c^2 = (b + c)(b – c)$
2nd expression $= (b + c)^2$
Required L.C.M. $= (b + c)^2(b – c)$
26. $x^2 + 2x, x^2 + 3x + 2$
Solution:
1st expression $= x^2 + 2x = x(x + 2)$
2nd expression $= x^2 + 3x + 2 = x^2 + 2x + x + 2 = x(x + 2) + 1(x + 2) = (x + 2)(x + 1)$
Required L.C.M. $= x(x + 1)(x + 2)$
27. $9x^2 – 25y^2, 15ax – 25ay$
Solution:
1st expression $= 9x^2 – 25y^2 = (3x)^2 – (5y)^2 = (3x + 5y)(3x – 5y)$
2nd expression $= 15ax – 25ay = 5a(3x – 5y)$
Required L.C.M. $= 5a(3x + 5y)(3x – 5y) = 5a(9x^2 – 25y^2)$
28. $x^2 – 3x – 10, x^2 – 10x + 25$
Solution:
1st expression $= x^2 – 3x – 10 = x^2 – 5x + 2x – 10 = x(x – 5) + 2(x – 5) = (x – 5)(x + 2)$
2nd expression $= x^2 – 10x + 25 = (x – 5)^2$
Required L.C.M. $= (x + 2)(x – 5)^2$
29. $a^2 – 7a + 12, a^2 + a – 20, a^2 + 2a – 15$
Solution:
1st expression $= a^2 – 7a + 12 = a^2 – 4a – 3a + 12 = a(a – 4) – 3(a – 4) = (a – 4)(a – 3)$
2nd expression $= a^2 + a – 20 = a^2 + 5a – 4a – 20 = a(a + 5) – 4(a + 5) = (a + 5)(a – 4)$
3rd expression $= a^2 + 2a – 15 = a^2 + 5a – 3a – 15 = a(a + 5) – 3(a + 5) = (a + 5)(a – 3)$
Required L.C.M. $= (a – 3)(a – 4)(a + 5)$
30. $x^2 – 8x + 15, x^2 – 25, x^2 + 2x – 15$.
Solution:
1st expression $= x^2 – 8x + 15 = x^2 – 5x – 3x + 15 = x(x – 5) – 3(x – 5) = (x – 5)(x – 3)$
2nd expression $= x^2 – 25 = x^2 – 5^2 = (x + 5)(x – 5)$
3rd expression $= x^2 + 2x – 15 = x^2 + 5x – 3x – 15 = x(x + 5) – 3(x + 5) = (x + 5)(x – 3)$
Required L.C.M. $= (x + 5)(x – 5)(x – 3)$ or $(x^2 – 25)(x – 3)$
31. $x + 5, x^2 + 5x, x^2 + 7x + 10$.
Solution:
1st expression $= x + 5$
2nd expression $= x^2 + 5x = x(x + 5)$
3rd expression $= x^2 + 7x + 10 = x^2 + 5x + 2x + 10 = x(x + 5) + 2(x + 5) = (x + 5)(x + 2)$
Required L.C.M. $= x(x + 5)(x + 2)$
32. If $a = 2x – 3$ and $b = 2x + 5$, then
a) Determine the value of $a + b$.
Solution:
Given that, $a = 2x – 3$ and $b = 2x + 5$
Therefore, $a + b = (2x – 3) + (2x + 5)$
$= 2x – 3 + 2x + 5$
$= 4x + 2$
Answer: $4x + 2$
b) Determine the value of $a^2$ using the formula.
Solution: $a^2 = (2x – 3)^2$
$= (2x)^2 – 2 \cdot 2x \cdot 3 + 3^2$
$= 4x^2 – 12x + 9$
Answer:$4x^2 – 12x + 9$
c) Determine the product of $a$ and $b$ using the formula. If $x = 2$, then $ab =$ ?
Solution:
Given that, $a = 2x – 3$, $b = 2x + 5$ and $x = 2$
We know that, $(x + a)(x + b) = x^2 + (a + b)x + ab$
Therefore,
$ab = (2x – 3)(2x + 5)$
$= (2x)^2 + (-3 + 5)(2x) + (-3)(5)$
$= 4x^2 + (2)(2x) – 15$
$= 4x^2 + 4x – 15$
Now, if $x = 2$,
$ab = 4(2)^2 + 4(2) – 15$
$= 4(4) + 8 – 15$
$= 16 + 8 – 15$
$= 24 – 15$
$= 9$
Answer: Product $4x^2 + 4x – 15$; Value $9$
33. $x^2 – 3x – 10, x^3 + 6x^2 + 8x$, and $x^4 – 5x^3 – 14x^2$ are three algebraic expressions.
a) Determine the square of $(3x – 2y + z)$.
Solution:
Square of $(3x – 2y + z)$
$= (3x – 2y + z)^2$
$= \{(3x – 2y) + z\}^2$
$= (3x – 2y)^2 + 2(3x – 2y)z + z^2$
$= \{(3x)^2 – 2 \cdot 3x \cdot 2y + (2y)^2\} + 6xz – 4yz + z^2$
$= 9x^2 – 12xy + 4y^2 + 6xz – 4yz + z^2$
$= 9x^2 + 4y^2 + z^2 – 12xy – 4yz + 6zx$
Answer:$9x^2 + 4y^2 + z^2 – 12xy – 4yz + 6zx$
b) Determine H.C.F. of the 1st and 2nd expressions.
Solution:
1st expression $= x^2 – 3x – 10 = x^2 – 5x + 2x – 10 = x(x – 5) + 2(x – 5) = (x – 5)(x + 2)$
2nd expression $= x^3 + 6x^2 + 8x = x(x^2 + 6x + 8) = x(x^2 + 4x + 2x + 8) = x\{x(x + 4) + 2(x + 4)\} = x(x + 4)(x + 2)$
Required H.C.F. $= (x + 2)$
c) Determine L.C.M. of the three expressions.
Solution: From ‘b’, we get,
1st expression $= (x – 5)(x + 2)$
2nd expression $= x(x + 4)(x + 2)$
And 3rd expression $= x^4 – 5x^3 – 14x^2$
$= x^2(x^2 – 5x – 14)$
$= x^2(x^2 – 7x + 2x – 14)$
$= x^2\{x(x – 7) + 2(x – 7)\}$
$= x^2(x – 7)(x + 2)$
Required L.C.M. $= x^2(x + 2)(x – 5)(x + 4)(x – 7)$
Sample Questions (Multiple Choice)
1. Which one is the Least Common Multiple (LCM) of $a, a^2, a(a + b)$?
Explanation:
1st expression $= a$
2nd expression $= a^2$
3rd expression $= a(a + b)$
Here, the LCM of $a$ and $a^2$ is $a^2$, and the remaining factor is $(a + b)$.
Therefore, Required LCM $= a^2(a + b)$
Answer: (d) $a^2(a + b)$
2. If $x = a + 7$ and $y = a – 7$, then—
(i) $x – y = 14$
(ii) $xy = a^2 – 49$
(iii) $y^2 = a^2 + 14a + 49$
Which of the following is correct?
Explanation:
(i) Verification:
$x – y = (a + 7) – (a – 7) = a + 7 – a + 7 = 14$
Therefore, (i) is correct.
(ii) Verification:
$xy = (a + 7)(a – 7) = (a)^2 – (7)^2 = a^2 – 49$
Therefore, (ii) is correct.
(iii) Verification:
$y^2 = (a – 7)^2 = a^2 – 2 \cdot a \cdot 7 + 7^2 = a^2 – 14a + 49$
In the question paper, $y^2 = a^2 + 14a + 49$ is given, which is incorrect (the sign should have been minus).
Therefore, (iii) is incorrect.
Thus, the correct statements are (i) and (ii).
Answer: (a) i and ii
$(x^3y – xy^3)$ and $(x – y)(x + 2y)$ are two algebraic expressions. Answer to the question no. 3 & 4 based on the above information:
3. Which one of the following is the factorization of first expression?
Solution:
First expression $= x^3y – xy^3 = xy(x^2 – y^2) = xy(x + y)(x – y)$
Answer: (d) $xy(x + y)(x – y)$
4. Which one of the following is the L.C.M. of the two algebraic expressions?
Solution:
First expression $= xy(x + y)(x – y)$
Second expression $= (x – y)(x + 2y)$
Required LCM $= xy(x + y)(x – y)(x + 2y) = xy(x^2 – y^2)(x + 2y)$
Answer: (c) $xy(x^2 – y^2)(x + 2y)$
Creative Questions:
5. $x^4 – 625$ and $x^2 + 3x – 10$ are two algebraic expressions.
a) Using a formula, find the product of $(m^2 – 3)$ and $(m^2 + 3)$.
Solution:
Required product $= (m^2 – 3)(m^2 + 3)$
$= (m^2)^2 – 3^2$
$= m^4 – 9$
Answer: $m^4 – 9$
b) Determine the H.C.F. of the two expressions.
Solution:
First expression $= x^4 – 625$
$= (x^2)^2 – 25^2$
$= (x^2 + 25)(x^2 – 25)$
$= (x^2 + 25)(x^2 – 5^2)$
$= (x^2 + 25)(x + 5)(x – 5)$
Second expression $= x^2 + 3x – 10$
$= x^2 + 5x – 2x – 10$
$= x(x + 5) – 2(x + 5)$
$= (x + 5)(x – 2)$
The common factor in both expressions is $(x + 5)$.
Required H.C.F. $= (x + 5)$
c) Determine the L.C.M. of the two expressions.
Solution: From ‘b’, we get,
First expression $= (x^2 + 25)(x + 5)(x – 5)$
Second expression $= (x + 5)(x – 2)$
Required LCM $= (x^2 + 25)(x + 5)(x – 5)(x – 2)$
$= (x^4 – 625)(x – 2)$
Short-Answer Questions
6. (a) Express $9x^2 – 24x + 16$ in the form of $(a – b)^2$.
Solution:
Given expression $= 9x^2 – 24x + 16$
$= (3x)^2 – 2 \cdot 3x \cdot 4 + 4^2$
$= (3x – 4)^2$
Answer: $(3x – 4)^2$
(b) If $a – b = 5$ and $a^2 + b^2 = 37$, find the value of $ab$.
Solution:
We know that, $a^2 + b^2 = (a – b)^2 + 2ab$
or, $37 = 5^2 + 2ab$
or, $37 = 25 + 2ab$
or, $2ab = 37 – 25$
or, $2ab = 12$
or, $ab = \frac{12}{2}$
or, $ab = 6$
Answer:$6$
(c) Using formula, find the square of $(ab – bc + ca)$.
Solution: Square of $(ab – bc + ca)$
$= (ab – bc + ca)^2$
$= \{(ab – bc) + ca\}^2$
$= (ab – bc)^2 + 2(ab – bc)(ca) + (ca)^2$
$= \{(ab)^2 – 2 \cdot ab \cdot bc + (bc)^2\} + 2a^2bc – 2abc^2 + c^2a^2$
$= a^2b^2 – 2ab^2c + b^2c^2 + 2a^2bc – 2abc^2 + c^2a^2$
$= a^2b^2 + b^2c^2 + c^2a^2 + 2a^2bc – 2ab^2c – 2abc^2$
Answer: $a^2b^2 + b^2c^2 + c^2a^2 + 2a^2bc – 2ab^2c – 2abc^2$
(d) If $x^2 + \frac{1}{x^2} = 7$, find the value of $\left(x – \frac{1}{x}\right)^2$.
Solution:
We know that, $x^2 + \frac{1}{x^2} = \left(x – \frac{1}{x}\right)^2 + 2 \cdot x \cdot \frac{1}{x}$
or, $7 = \left(x – \frac{1}{x}\right)^2 + 2$
or, $\left(x – \frac{1}{x}\right)^2 = 7 – 2$
or, $\left(x – \frac{1}{x}\right)^2 = 5$
Answer: $5$