Exercise 4.4: HCF and LCM of Algebraic Expressions
1. Which one of the following will be correct if we express $(3x – 5)(5 + 3x)$ in the form of the difference between two squares?
Solution: Given expression $= (3x – 5)(5 + 3x)$
$= (3x – 5)(3x + 5)$
$= (3x)^2 – (5)^2$
The form as the difference of two squares is $(3x)^2 – 5^2$.
Correct Answer: (c) $(3x)^2 – 5^2$
Answer the questions no. (2–4) in accordance with the information given below:
If $x^2 – \sqrt{3}x + 1 = 0$
2. Which one of the following is the value of $x + \frac{1}{x}$?
Solution: Given,
$x^2 – \sqrt{3}x + 1 = 0$
$\Rightarrow x^2 + 1 = \sqrt{3}x$
$\Rightarrow \frac{x^2 + 1}{x} = \frac{\sqrt{3}x}{x}$ [Dividing both sides by $x$]
$\Rightarrow \frac{x^2}{x} + \frac{1}{x} = \sqrt{3}$
$\Rightarrow x + \frac{1}{x} = \sqrt{3}$
Correct Answer: (d) $\sqrt{3}$
3. Which one of the following is the value of $x^2 + \frac{1}{x^2}$?
Solution: $x^2 + \frac{1}{x^2}$
$= \left(x + \frac{1}{x}\right)^2 – 2 \cdot x \cdot \frac{1}{x}$
$= (\sqrt{3})^2 – 2$
$= 3 – 2$
$= 1$
Correct Answer: (a) 1
4. Which one of the following is the value of $x^3 + \frac{1}{x^3}$?
Solution: $x^3 + \frac{1}{x^3}$
$= \left(x + \frac{1}{x}\right)^3 – 3 \cdot x \cdot \frac{1}{x}\left(x + \frac{1}{x}\right)$
$= (\sqrt{3})^3 – 3(\sqrt{3})$
$= 3\sqrt{3} – 3\sqrt{3}$
$= 0$
Correct Answer: (d) 0
5. Which one of the following expression is the factors of $x^2 – x – 30$?
Solution: $x^2 – x – 30$
$= x^2 – 6x + 5x – 30$
$= x(x – 6) + 5(x – 6)$
$= (x – 6)(x + 5)$ or $(x + 5)(x – 6)$
Correct Answer: (b) $(x + 5)(x – 6)$
6. In algebraic formulae:
i. $x^3 – y^3 = (x – y)(x^2 + xy + y^2)$ (Correct)
ii. $ab = \left(\frac{a + b}{2}\right)^2 – \left(\frac{a – b}{2}\right)^2$ (Correct)
iii. $x^3 + y^3 = (x + y)^3 + 3xy(x + y)$ (Incorrect, correct formula: $(x + y)^3 – 3xy(x + y)$)
Which one of the following is correct according to the above information?
Correct Answer: (a) i and ii
7. If $x + \frac{1}{x} = 2$, then
(1) What is the value of $\left(x – \frac{1}{x}\right)^2$?
Solution:
$\left(x – \frac{1}{x}\right)^2$
$= \left(x + \frac{1}{x}\right)^2 – 4 \cdot x \cdot \frac{1}{x}$
$= (2)^2 – 4$
$= 4 – 4$
$= 0$
Correct Answer: (a) 0
(2) What is the value of $x^3 + \frac{1}{x^3}$?
Solution: $x^3 + \frac{1}{x^3}$
$= \left(x + \frac{1}{x}\right)^3 – 3 \cdot x \cdot \frac{1}{x}\left(x + \frac{1}{x}\right)$
$= (2)^3 – 3(2)$
$= 8 – 6$
$= 2$
Correct Answer: (b) 2
(3) What is the value of $x^4 + \frac{1}{x^4}$?
Solution: $x^4 + \frac{1}{x^4}$
$= \left(x^2\right)^2 + \left(\frac{1}{x^2}\right)^2$
$= \left(x^2 + \frac{1}{x^2}\right)^2 – 2 \cdot x^2 \cdot \frac{1}{x^2}$
$= \left\{\left(x + \frac{1}{x}\right)^2 – 2\right\}^2 – 2$
$= \left\{(2)^2 – 2\right\}^2 – 2$
$= (4 – 2)^2 – 2$
$= (2)^2 – 2$
$= 4 – 2$
$= 2$
Correct Answer: (d) 2
Find the H.C.F. of the following (8–13):
8. $36a^2b^2c^4d^5$, $54a^5c^2d^4$, and $90a^4b^3c^2$
Solution:
H.C.F. of numerical coefficients $36, 54$, and $90 = 18$
Lowest powers of algebraic variables:
Lowest power of $a = a^2$
$b$ is not present in all expressions.
Lowest power of $c = c^2$
$d$ is not present in all expressions.
$\therefore$ Required H.C.F. $= 18a^2c^2$ (Answer)
9. $20x^3y^2a^3b^4$, $15x^4y^3a^4b^3$, and $35x^2y^4a^3b^2$
Solution:
H.C.F. of numerical coefficients $20, 15$, and $35 = 5$
Lowest powers of algebraic variables:
Lowest power of $x = x^2$
Lowest power of $y = y^2$
Lowest power of $a = a^3$
Lowest power of $b = b^2$
$\therefore$ Required H.C.F. $= 5a^3b^2x^2y^2$ (Answer)
10. $x^2 – 3x$, $x^2 – 9$, and $x^2 – 4x + 3$
Solution:
1st expression $= x^2 – 3x = x(x – 3)$
2nd expression $= x^2 – 9 = x^2 – 3^2 = (x + 3)(x – 3)$
3rd expression $= x^2 – 4x + 3$
$= x^2 – 3x – x + 3$
$= x(x – 3) – 1(x – 3)$
$= (x – 3)(x – 1)$
Common factor is $(x – 3)$
$\therefore$ Required H.C.F. $= x – 3$ (Answer)
11. $18(x + y)^3$, $24(x + y)^2$, and $32(x^2 – y^2)$
Solution:
H.C.F. of numerical coefficients $18, 24$, and $32 = 2$
1st expression $= 18(x + y)^3 = 18(x + y)(x + y)(x + y)$
2nd expression $= 24(x + y)^2 = 24(x + y)(x + y)$
3rd expression $= 32(x^2 – y^2) = 32(x + y)(x – y)$
Common factor is $(x + y)$
$\therefore$ Required H.C.F. $= 2(x + y)$ (Answer)
12. $a^2b(a^3 – b^3)$, $a^2b^2(a^4 + a^2b^2 + b^4)$, and $a^3b^2 + a^2b^3 + ab^4$
Solution:
1st expression $= a^2b(a^3 – b^3) = a \times a \times b \times (a – b)(a^2 + ab + b^2)$
2nd expression $= a \times a \times b \times b \times (a^4 + a^2b^2 + b^4)$
$= a \times a \times b \times b \times \{(a^2)^2 + 2a^2b^2 + (b^2)^2 – a^2b^2\}$
$= a \times a \times b \times b \times \{(a^2 + b^2)^2 – (ab)^2\}$
$= a \times a \times b \times b \times (a^2 + b^2 + ab)(a^2 + b^2 – ab)$
$= a \times a \times b \times b \times (a^2 + ab + b^2)(a^2 – ab + b^2)$
3rd expression $= a^3b^2 + a^2b^3 + ab^4$
$= ab^2(a^2 + ab + b^2)$
$\therefore$ Required H.C.F. $= ab(a^2 + ab + b^2)$ (Answer)
13. $a^3 – 3a^2 – 10a$, $a^3 + 6a^2 + 8a$, and $a^4 – 5a^3 – 14a^2$
Solution:
1st expression $= a^3 – 3a^2 – 10a$
$= a(a^2 – 3a – 10)$
$= a(a^2 – 5a + 2a – 10)$
$= a\{a(a – 5) + 2(a – 5)\}$
$= a(a – 5)(a + 2)$
2nd expression $= a^3 + 6a^2 + 8a$
$= a(a^2 + 6a + 8)$
$= a(a^2 + 4a + 2a + 8)$
$= a\{a(a + 4) + 2(a + 4)\}$
$= a(a + 4)(a + 2)$
3rd expression $= a^4 – 5a^3 – 14a^2$
$= a^2(a^2 – 5a – 14)$
$= a^2(a^2 – 7a + 2a – 14)$
$= a^2\{a(a – 7) + 2(a – 7)\}$
$= a^2(a – 7)(a + 2)$
Common factors: $a$ and $(a + 2)$
$\therefore$ Required H.C.F. $= a(a + 2)$ (Answer)
Find the L.C.M. of the following (14–19):
14. $a^5b^2c$, $ab^3c^2$, and $a^7b^4c^3$
Solution:
Highest powers of algebraic variables:
Highest power of $a = a^7$
Highest power of $b = b^4$
Highest power of $c = c^3$
$\therefore$ Required L.C.M. $= a^7b^4c^3$ (Answer)
15. $5a^2b^3c^2$, $10ab^2c^3$, and $15ab^3c$
Solution:
L.C.M. of numerical coefficients $5, 10$, and $15 = 30$
Highest powers of algebraic variables:
Highest power of $a = a^2$
Highest power of $b = b^3$
Highest power of $c = c^3$
$\therefore$ Required LCM $= 30a^2b^3c^3$ (Answer)
16. $3x^3y^2$, $4xy^3z$, $5x^4y^2z^2$, and $12xy^4z^2$
Solution:
L.C.M. of numerical coefficients $3, 4, 5$, and $12 = 60$
Highest powers of algebraic variables:
Highest power of $x = x^4$
Highest power of $y = y^4$
Highest power of $z = z^2$
$\therefore$ Required L.C.M. $= 60x^4y^4z^2$ (Answer)
17. $x^2 – 4$, $x^2 + 4x + 4$, and $x^3 – 8$
Solution:
1st expression $= x^2 – 4 = x^2 – 2^2 = (x + 2)(x – 2)$
2nd expression $= x^2 + 4x + 4 = x^2 + 2 \cdot x \cdot 2 + 2^2 = (x + 2)^2$
3rd expression $= x^3 – 8 = x^3 – 2^3 = (x – 2)(x^2 + 2x + 4)$
$\therefore$ Required L.C.M. $= (x + 2)^2(x – 2)(x^2 + 2x + 4)$ (Answer)
18. $6x^2 – x – 1$, $3x^2 + 7x + 2$, and $2x^2 + 3x – 2$
Solution:
1st expression $= 6x^2 – x – 1$
$= 6x^2 – 3x + 2x – 1$
$= 3x(2x – 1) + 1(2x – 1)$
$= (2x – 1)(3x + 1)$
2nd expression $= 3x^2 + 7x + 2$
$= 3x^2 + 6x + x + 2$
$= 3x(x + 2) + 1(x + 2)$
$= (x + 2)(3x + 1)$
3rd expression $= 2x^2 + 3x – 2$
$= 2x^2 + 4x – x – 2$
$= 2x(x + 2) – 1(x + 2)$
$= (x + 2)(2x – 1)$
$\therefore$ Required L.C.M. $= (2x – 1)(3x + 1)(x + 2)$ (Answer)
19. $a^3 + b^3$, $(a + b)^3$, $(a^2 – b^2)^2$, and $(a^2 – ab + b^2)^2$
Solution:
1st expression $= a^3 + b^3 = (a + b)(a^2 – ab + b^2)$
2nd expression $= (a + b)^3$
3rd expression $= (a^2 – b^2)^2 = \{(a + b)(a – b)\}^2 = (a + b)^2(a – b)^2$
4th expression $= (a^2 – ab + b^2)^2$
$\therefore$ Required L.C.M. $= (a + b)^3(a – b)^2(a^2 – ab + b^2)^2$ (Answer)
20.$3x – 5y + 3z$ and $3x + 5y – z$ are two algebraic expressions.
(a) Find out the square of the first expression.
Solution:
1st expression $= 3x – 5y + 3z$
$\therefore$ Square of the 1st expression $= (3x – 5y + 3z)^2$
$= (3x)^2 + (-5y)^2 + (3z)^2 + 2(3x)(-5y) + 2(-5y)(3z) + 2(3z)(3x)$
[We know, $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$]
$= 9x^2 + 25y^2 + 9z^2 – 30xy – 30yz + 18zx$ (Answer)
(b) Express the product of the two expressions in the form of difference of two squares.
Solution: Let, $A = 3x – 5y + 3z$ and $B = 3x + 5y – z$
We know, $AB = \left(\frac{A + B}{2}\right)^2 – \left(\frac{A – B}{2}\right)^2$
Here,
$A + B = (3x – 5y + 3z) + (3x + 5y – z) = 6x + 2z$
$\Rightarrow \frac{A + B}{2} = \frac{6x + 2z}{2} = 3x + z$
and,
$A – B = (3x – 5y + 3z) – (3x + 5y – z)$
$= 3x – 5y + 3z – 3x – 5y + z$
$= -10y + 4z = 4z – 10y$
$\Rightarrow \frac{A – B}{2} = \frac{4z – 10y}{2} = 2z – 5y$
$\therefore$ Expressed as the difference of two squares:
$= (3x + z)^2 – (2z – 5y)^2$ (or, $(3x + z)^2 – (5y – 2z)^2$) (Answer)
(c) If the second expression is ‘0’ (zero), prove that $27x^3 + 125y^3 + 45xyz = z^3$
Solution: According to the question, the value of the 2nd expression is zero,
i.e., $3x + 5y – z = 0$
$\Rightarrow 3x + 5y = z$
Now, LHS $= 27x^3 + 125y^3 + 45xyz$
$= (3x)^3 + (5y)^3 + 45xyz$
$= (3x + 5y)^3 – 3 \cdot 3x \cdot 5y (3x + 5y) + 45xyz$
$= z^3 – 45xy(z) + 45xyz$ [Substituting $3x + 5y = z$]
$= z^3 – 45xyz + 45xyz$
$= z^3 = \text{RHS}$ (Proved)
Sample Questions (Multiple Choice)
1. Which one of the following is the square of $-5 – y$?
Solution:
$(-5 – y)^2 = \{-(5 + y)\}^2 = (5 + y)^2 = 25 + 10y + y^2 = y^2 + 10y + 25$
Correct Answer: (a) $y^2 + 10y + 25$
2. What is the H.C.F. of $x^2 – 2x – 3$ and $x^2 + 2x – 3$?
Solution:
1st expression $= x^2 – 2x – 3 = x^2 – 3x + x – 3 = x(x – 3) + 1(x – 3) = (x – 3)(x + 1)$
2nd expression $= x^2 + 2x – 3 = x^2 + 3x – x – 3 = x(x + 3) – 1(x + 3) = (x + 3)(x – 1)$
Since there is no common algebraic factor between both expressions, their HCF is $1$.
Correct Answer: (c) 1
3. For the algebraic expressions $x^2 – 10x + 21$ and $(x – 7)(x – 2)$
Factorization:
1st expression $= x^2 – 10x + 21 = x^2 – 7x – 3x + 21 = x(x – 7) – 3(x – 7) = (x – 7)(x – 3)$
2nd expression $= (x – 7)(x – 2)$
Explanation:
i) H.C.F. $= x – 7$ (Correct)
ii) L.C.M. $= (x – 7)(x – 3)(x – 2)$ (Given L.C.M. in stem is $(x – 7)(x – 3)$, so it is incorrect)
iii) Product $= (x – 7)(x – 3) \times (x – 7)(x – 2) = (x – 7)^2(x – 3)(x – 2)$ (Correct)
Which of the following is correct?
Correct Answer: (b) i and iii
If $x + y = 5$ and $x – y = 3$, then Answer to the questions 4 & 5.
4. What is the value of $xy$?
Solution:
$xy = \left(\frac{x + y}{2}\right)^2 – \left(\frac{x – y}{2}\right)^2$
$= \left(\frac{5}{2}\right)^2 – \left(\frac{3}{2}\right)^2$
$= \frac{25}{4} – \frac{9}{4} = \frac{16}{4} = 4$
Correct Answer: (d) 4
5. What is the value of $x^2 – y^2$?
Solution:
$x^2 – y^2 = (x + y)(x – y) = 5 \times 3 = 15$
Correct Answer: (c) 15
Creative Question
6. Given three algebraic expressions are $P = 3x^2 – 5x + 2$, $Q = 3x^2 – 16x – 12$, and $R = 3x^2 – x – 2$
(a) Factorize $P$.
Solution: $P = 3x^2 – 5x + 2$
$= 3x^2 – 3x – 2x + 2$
$= 3x(x – 1) – 2(x – 1)$
$= (x – 1)(3x – 2)$ (Answer)
(b) If $P = 0$ and $x \neq 0$, find the value of $9x^2 + \frac{4}{x^2}$.
Solution: Given, $P = 0$
$\Rightarrow 3x^2 – 5x + 2 = 0$
$\Rightarrow 3x^2 + 2 = 5x$
Since $x \neq 0$, dividing both sides by $x$:
$\Rightarrow \frac{3x^2 + 2}{x} = \frac{5x}{x}$
$\Rightarrow 3x + \frac{2}{x} = 5$
Now,
$9x^2 + \frac{4}{x^2} = (3x)^2 + \left(\frac{2}{x}\right)^2$
$= \left(3x + \frac{2}{x}\right)^2 – 2 \cdot 3x \cdot \frac{2}{x}$
$= (5)^2 – 12$
$= 25 – 12$
$= 13$ (Answer)
(c) Find the L.C.M. of $Q$ and $R$.
Solution:
1st expression, $Q = 3x^2 – 16x – 12$
$= 3x^2 – 18x + 2x – 12$
$= 3x(x – 6) + 2(x – 6)$
$= (x – 6)(3x + 2)$
2nd expression, $R = 3x^2 – x – 2$
$= 3x^2 – 3x + 2x – 2$
$= 3x(x – 1) + 2(x – 1)$
$= (x – 1)(3x + 2)$
$\therefore$ Required LCM of $Q$ and $R = (3x + 2)(x – 6)(x – 1)$ (Answer)
7. If $a – \frac{1}{a} = 3$
(a) Find the value of $\left(a + \frac{1}{a}\right)^2$.
Solution: We know,
$\left(a + \frac{1}{a}\right)^2 = \left(a – \frac{1}{a}\right)^2 + 4 \cdot a \cdot \frac{1}{a}$
$= (3)^2 + 4$
$= 9 + 4$
$= 13$ (Answer)
(b) Prove that $a^4 = 119 – \frac{1}{a^4}$.
Solution: Given, $a – \frac{1}{a} = 3$
$\left(a – \frac{1}{a}\right)^2 = 3^2$ [Squaring both sides]
$\Rightarrow a^2 – 2 \cdot a \cdot \frac{1}{a} + \frac{1}{a^2} = 9$
$\Rightarrow a^2 + \frac{1}{a^2} – 2 = 9$
$\Rightarrow a^2 + \frac{1}{a^2} = 11$
Squaring both sides again:
$\left(a^2 + \frac{1}{a^2}\right)^2 = (11)^2$
$\Rightarrow (a^2)^2 + 2 \cdot a^2 \cdot \frac{1}{a^2} + \left(\frac{1}{a^2}\right)^2 = 121$
$\Rightarrow a^4 + 2 + \frac{1}{a^4} = 121$
$\Rightarrow a^4 + \frac{1}{a^4} = 121 – 2$
$\Rightarrow a^4 + \frac{1}{a^4} = 119$
$\Rightarrow a^4 = 119 – \frac{1}{a^4}$ (Proved)
(c) Find the value of $a^5 – \frac{1}{a^5}$.
Solution:
Given, $a – \frac{1}{a} = 3$
From (a), $\left(a + \frac{1}{a}\right)^2 = 13 \Rightarrow a + \frac{1}{a} = \sqrt{13}$
From (b), $a^2 + \frac{1}{a^2} = 11$
Now,
$a^3 – \frac{1}{a^3} = \left(a – \frac{1}{a}\right)^3 + 3 \cdot a \cdot \frac{1}{a}\left(a – \frac{1}{a}\right)$
$= (3)^3 + 3(3) = 27 + 9 = 36$
Multiplying $\left(a^2 + \frac{1}{a^2}\right)$ and $\left(a^3 – \frac{1}{a^3}\right)$:
$\left(a^2 + \frac{1}{a^2}\right)\left(a^3 – \frac{1}{a^3}\right) = 11 \times 36$
$\Rightarrow a^5 – a^2 \cdot \frac{1}{a^3} + \frac{1}{a^2} \cdot a^3 – \frac{1}{a^5} = 396$
$\Rightarrow a^5 – \frac{1}{a} + a – \frac{1}{a^5} = 396$
$\Rightarrow \left(a^5 – \frac{1}{a^5}\right) + \left(a – \frac{1}{a}\right) = 396$
$\Rightarrow \left(a^5 – \frac{1}{a^5}\right) + 3 = 396$
$\Rightarrow a^5 – \frac{1}{a^5} = 396 – 3$
$\Rightarrow a^5 – \frac{1}{a^5} = 393$ (Answer)
Short Answer Questions:
8.
(a) If $a + b = 10$ and $a – b = 6$, find the value of $a^2 + b^2$.
Solution: We know,
$2(a^2 + b^2) = (a + b)^2 + (a – b)^2$
$\Rightarrow 2(a^2 + b^2) = (10)^2 + (6)^2$ [Given $a + b = 10$ and $a – b = 6$]
$\Rightarrow 2(a^2 + b^2) = 100 + 36$
$\Rightarrow 2(a^2 + b^2) = 136$
$\Rightarrow a^2 + b^2 = \frac{136}{2}$
$\Rightarrow a^2 + b^2 = 68$
Answer:$68$
(b) Express $x^2 + 6x + 8$ as a difference of two squares.
Solution: $x^2 + 6x + 8$
$= x^2 + 4x + 2x + 8$
$= x(x + 4) + 2(x + 4)$
$= (x + 4)(x + 2)$
Let $A = x + 4$ and $B = x + 2$
We know, $AB = \left(\frac{A + B}{2}\right)^2 – \left(\frac{A – B}{2}\right)^2$
Here,
$\frac{A + B}{2} = \frac{(x + 4) + (x + 2)}{2} = \frac{2x + 6}{2} = x + 3$
$\frac{A – B}{2} = \frac{(x + 4) – (x + 2)}{2} = \frac{x + 4 – x – 2}{2} = \frac{2}{2} = 1$
$\therefore$ Form as the difference of two squares $= (x + 3)^2 – (1)^2$ (Answer)
(Alternative method)
$x^2 + 6x + 8 = x^2 + 2 \cdot x \cdot 3 + 3^2 – 1 = (x + 3)^2 – 1^2$
(c) If $a + \frac{1}{a} = 3$, find the value of $a^3 + \frac{1}{a^3}$.
Solution:
$a^3 + \frac{1}{a^3} = \left(a + \frac{1}{a}\right)^3 – 3 \cdot a \cdot \frac{1}{a}\left(a + \frac{1}{a}\right)$
$= (3)^3 – 3(3)$
$= 27 – 9$
$= 18$
Answer:$18$
(d) Factorize $2x^2 – 9x – 35$.
Solution: $2x^2 – 9x – 35$
$= 2x^2 – 14x + 5x – 35$ [$2 \times (-35) = -70 = -14 \times 5$]
$= 2x(x – 7) + 5(x – 7)$
$= (x – 7)(2x + 5)$
Answer:$(x – 7)(2x + 5)$
(e) Find the H.C.F. of $x^2 – 3x$ and $x^2 – 9$.
Solution:
1st expression $= x^2 – 3x = x(x – 3)$
2nd expression $= x^2 – 9 = x^2 – 3^2 = (x + 3)(x – 3)$
The common factor between both expressions is $(x – 3)$.
$\therefore$ Required HCF $= x – 3$ (Answer)