PRACTICE QUESTIONS ON HCF and LCM

a(x) = 8x(x ⁴ – x ² + 1) = 2 ³ x ³(x ⁴ - x ² + 1)
b(x) = 28((x ²) ³ + 1 ³)
= 2 ² 7 ¹ (x ² + 1)((x ²) ² - x ² + 1) [From a³ + b³ = (a + b)(a² — ab + b²)]
= 2 ² 7 ¹ (x ² + 1)(x ⁴ - x ² + 1)
HCF = 2 ² (x ⁴ - x ² + 1)
= 4(x ⁴ - x ² + 1)

a + b = 65, HCF (a, b) = 5
=>Both a and b are multiples of 5.
The possible pairs of numbers are: 5, 60; 10, 55; 15, 50; 20, 45; 25, 40; and 30, 35.
We have 6 solutions.

We know HCF(a, b) = 1
a and b both cannot be even, else their HCF would be 2.
Hence either both are odd or one is even, other is odd.
Case I: Both odd
Then a + b and a – b are even.
Therefore HCF (a + b, a – b) = 2
Case II: One is even, other is odd.
Then a + b and a – b are both odd.
Therefore HCF (a + b, a – b) = 1
HCF can be 1 or 2.

Let the numbers be x and y.
Then, x = 98a, y = 98b where a and b are coprime.
Sum = x + y = 98(a + b)
Clearly sum must be divisible by 98.
Of the given options only 1372 is divisible by 98.

a ² b ⁴ + 2a ² b ² = a ² b ² (b ² + 2)
a ⁷ b ⁷ - 4a ² b ⁹ = a ² b ² ( a ⁵ b ⁵ - 4b ⁷)
H.C.F. = a ² b ²

x ² + 2x – 8 = (x + 4)(x – 2)
x ³ - 4x ² + 4x = x (x – 2) ²
x ² + 4x = x(x + 4)
L.C.M. = x(x + 4)(x – 2) ²

HCF (a, b) = 1 => a and b are coprime.
HCF (a, c) = 1 => a and c are coprime.
a has no common factor other than 1 with b and c.
a has no common factor other than 1 with bc.
Hence HCF (a, bc) = 1

If n = 1, then m = 3 and k = 13
If n = 2, then m = 5 and k = 22
If n = 3, then m = 7 and k = 31
Therefore H.C.F. of m and k = 1.

HCF = x – 6
By Remainder Theorem x – 6 can be factor of x ² – kx – 6 only if x – 6 = 0 or x = 6
=> 6² – k x 6 – 6 = 0
=> 6k = 30 => k = 5

The first expression
36(3x⁴ + 5x³ – 2x²) = 36(x ²)(3x ² + 5x – 2)
= 36x ² (3x ² + 6x – x – 2)
= 36x ² (x + 2)(3x – 1)

The second expression
9(6x³ + 4x² – 2x) = 18x(3x ² + 2x – 1)
= 18x(3x – 1)(x + 1)

The third expression
54(27x⁴ – x) = 54x(27x ³ – 1)
= 54x(3x – 1)(9x ² + 3x + 1) [from the identity a³ – b³ = (a – b)(a² + ab + b²)]
HCF = 18x(3x – 1)

We know that H.C.F. of the numbers divides every number.
1. (x + y) and x ¹⁰ - y ¹⁰ are both divisible by x + y.
4. (x – y) and x ¹⁰ - y ¹⁰ are both divisible by x – y.
Hence, (c) is the correct answer.

The first expression x⁴ – 3x + 2 can be written as
x⁴ – x³ + x³ – x² + x² – x – 2x + 2
= x³ (x – 1) + x²(x – 1) + x(x – 1) – 2(x – 1)
= (x – 1)[x³ + x² + x – 2]

The second expression
x³ – 3x² + 3x – 1 = (x – 1)³

The third expression
x⁴ – 1 = (x – 1)(x + 1)(x² + 1)

HCF = x – 1

If (x + 2) is the H.C.F. of x ² + ax + b and x² + cx + d, then by remainder theorem
x² + ax + b = (–2)² + a(–2) + b = 0
i.e., 4 – 2a + b = 0 …(i)
also x² + cx + d = (–2) ² + c(–2) + d = 0
i.e., 4 – 2c + d = 0 …(ii)
=> (i) = (ii) which gives b + 2c = d + 2a

(x + K) is the H.C.F. of two expressions so each expression has (x + K) as its factor and so x + K = 0 or x = –K must make each expression zero.
Therefore aK ² + aK + b = 0 and K ² – cK + d = 0
This gives K = ad – b/ a(c – a)

The first expresstion x⁴ – x² – 6
= x⁴ – 3x² + 2x² – 6
= (x² – 3)(x² + 2)

The second expression x⁴ – 4x² + 3
= x⁴ – 3x² – x² + 3
= (x² – 3)(x² – 1)

Therefore H.C.F. = (x² – 3)

If x is 48, then three numbers will be 144, 48, 192. The H.C.F. of these numbers is 48. So if the H.C.F. is 12, then x cannot be 48.

The terms are 3/1, 6/5 and 3/50
H.C.F. of these terms = H.C.F. of 3, 6, 3/ L.C.M. of 1, 5, 50 = 3/50 = 0.06

q(x) = HCF x LCM/ p(x)
p(x) = 8x(x + 2)²
= 2x(x + 2) x 24x(x + 2)² (x – 2)/ 8x(x + 2)²
= 6x(x + 2)(x – 2)
= 6x(x² - 4)
= 6x ³ - 24x

f(x, y) = 4x ³ 3x ² y – 9xy ² 2y ³
g(x, y) = x ² + xy – 2y ²
Clearly (x – y) = 0 or x = y satisfy both equation, since for x = y makes both equations zero.
Hence, (x – y) is the H.C.F.
Similarly (x + 2y) = 0, i.e., x = –2y also makes both equations zero. Thus, (x + 2y)(x – y) is the H.C.F.