Algebra Chapter 9 Highest Common Factor
Elementary factor
If we cannot resolve a factor into any other factors, then such a factor is called an elementary factor. By factorization, we understand factorization as an elementary factor. The elementary factors of a3b3 are a and b because a and b cannot be resolved into any other factors.
Common factor
If two or more expressions are divisible by a factor then such a factor is called a common factor of those expressions. For example, if we consider two expressions a2b and ab2c then we see that a, b, and ab are their common factors.
The highest common factor
Two or more expressions may have more than one factor. Among the common factors of some quantities, the highest one (which is of the highest power) is called the Highest Common Factor or H.C.F. of those quantities. For example, the common factors of 2a2b3c2, 3a4b2c3 and 4a563c2 are a, b, c, a2, b2, c2, ab, be, ca, a2b, a2c, b2c, be2, abc, a2bc, ab2c, abc2, a2b2c2. Among them, a262c2 is of the highest power. Therefore, a2b2c2 will be the H.C.F.
Process for the determination of H.C.F.
1. At first, all the given expressions are resolved into factors.
2. The required H.C.F. is the product of the common elementary factors of the highest dimension that exactly divide each of the given expressions.
3. If numerical coefficients appear before the alphabetic expressions then after finding the H.C.F. of alphabetic expressions the arithmetical H.C.F. of the numerical coefficients is written before it.
Read And Learn More WBBSE Solutions For Class 8 Maths
Algebra Chapter 9 Highest Common Factor Some Examples
Example 1
To find the H.C.F. of 3xy2z2, 2yz2x2, and x2y2z.
Solution :
Given 3xy2z2, 2yz2x2, And x2y2z
x, y and z are the elementary factors of these three expressions.
The highest powers of them which exactly divide the given expressions are x, y, z.
∴ H.C.F. = xyz
WBBSE Class 8 Highest Common Factor Notes
Example 2
To find, the H.C.F. of 16a2b3x4y5, 40a3b2x3y4, 24a5b5x3y4.
Solution :
Given: 16a2b3x4y5, 40a3b2x3y4, 24a5b5x3y4
H.C.F. of 16, 40, and 24 = 8.
a, b, x, y are common elementary factors.
The highest powers of them which exactly divide the expressions are a2, b2, x3, and y4.
∴ The required H.C.F. = 8a2b2x3y4
Understanding Highest Common Factor (HCF)
Example 3
Find the H.C.F. of x4 – 1, x4 – x3 + x -1.
Solution :
Given x4 – 1, x4 – x3 + x -1
First expression
= (x4 – 1)
= (x2+1)(x2 – 1)
= (x2+1)(x+1)(x-1)
Second expression
= x4 – x3 + x – 1
= x3 (x – 1) + (x -1)
= (x – 1)(x3+1)
= (x-1)(x + 1)(x2-x + 1)
∴ The required H.C.F.
= (x +1)(x – 1)
= x2 – 1
x4 – 1, x4 – x3 + x -1 = x2 – 1
Example 4
Find the H.C.F. of x3 – 5x2 + 6x, x3 + 4x2 – 12x, x3 – 9x2 + 14x.
Solution :
Given: x3 – 5x2 + 6x, x3 + 4x2 – 12x, x3 – 9x2 + 14x
First expression
= x(x2 – 5x + 6)
= x(x2 – 2x – 3x + 6)
= x {x(x – 2) – 3(x – 2)}
= x (x – 2)(x – 3)
Second expression
= x (x2 + 4x – 12)
= x (x2 – 2x + 6x – 12)
= x {x (x – 2) + 6(x – 2)}
= x (x – 2) (x + 6)
Third expression
= x (x2 – 9x +14)
= x (x2 – 2x – 7x + 14)
= x {x (x – 2) – 7(x – 2)}
= x (x – 2) (x – 7)
.’. The required H.C.F. = x (x – 2)
x3 – 5x2 + 6x, x3 + 4x2 – 12x, x3 – 9x2 + 14x = x (x – 2)
Step-by-Step Guide to Finding HCF
Example 5
Find the H.C.F. of a2 – 1, a3 – 1 and a2 + a – 2.
Solution :
Given a2 – 1, a3 – 1 And a2 + a – 2
First expression
= a2 – 1
= (a + 1)(a – 1)
Second expression
= a3 – 1
= (a – 1)(a2 + a + 1)
Third expression
= a2 + a- 2
= a2-a + 2a-2
= a (a -1) + 2 (a – 1)
= (a – 1)(a + 2)
The required H.C.F. = a – 1
a2 – 1, a3 – 1 And a2 + a – 2 = a – 1
Example 6
Find the H.C.F. of x3 – 8, x2 + 3x – 10, x3 + 2x2 – 8x .
Solution :
Given x3 – 8, x2 + 3x – 10, x3 + 2x2 – 8x
First expression
= x3 – 8
= (x)3 – (2)3
= (x – 2){(x)2 + x.2 + (2)2}
= (x – 2)(x2 + 2x + 4)
Second expression
= x2 + 3x – 10
= x2 + 5x – 2x – 10
= x ( x + 5) – 2(x + 5)
= (x+ 5)(x – 2)
Third expression
= x3 + 2x2 – 8x
= x (x2 + 2x – 8)
= x {x2 + 4x – 2x – 8}
= x{x(x + 4) – 2(x + 4)}
= x (x + 4)(x – 2)
∴ The required H.C.F. = x – 2
x3 – 8, x2 + 3x – 10, x3 + 2x2 – 8x = x – 2
Practice Problems on Finding HCF
Example 7
Find the H.C.F. of 5(x3 – x), 4(x5 – x2), 7(x4 – 1).
Solution :
Given 5(x3 – x), 4(x5 – x2), 7(x4 – 1)
First expression
= 5(x2 – x)
= 5x (x2 – 1)
= 5x (x + 1)(x – 1)
Second expression
= 4(x5 – x2)
= 4x2 (x3 – 1)
= 4x2 (x – 1)(x2 + x + 1)
Third expression
= 7(x4 – 1)
= 7(x2 + 1)(x2 – 1)
= 7(x2 + 1)(x + 1) (x – 1)
∴ The required H.C.F. = x – 1
5(x3 – x), 4(x5 – x2), 7(x4 – 1). = x – 1
Example 8
If the H.C.F. of x2 + px + q and x2 + px + q’ be x + a then show that (p – p’) a = q – q.
Solution :
Given x2 + px + q And x2 + px + q’ be x + a
x + a is the H.C.F. of the two given expressions therefore, it is the factor of both expressions.
Therefore, the value of the expressions will be equal to 0 if its value is 0.
Now, see x + a= 0 if x = -a, therefore, both the expressions will be 0 if we put x = – a.
\(\text { Therefore, } \quad a^2-a p+q=0\) \(\text { and } \quad a^2-a p^{\prime}+q^{\prime}=0 \ldots \ldots \ldots \text { (2) }\) \( \text { (subtracting) }-a p+a p^{\prime}+q-q^{\prime}=0,\)or, – a(p – p’) = – (q – q’)
or, a(p – p’) = q – q’
or, ( p – p’) a = q – q’.
Conceptual Questions on Applications of HCF
Example 9
The H.C.F, and L.C.M. of the two expressions are a2 – a – 2 and a4 – 5a + 4. If one of the expressions is a3 + a2 + a – 4, find the other.
Solution :
Given
The H.C.F, and L.C.M. of the two expressions are a2 – a – 2 and a4 – 5a + 4
one of the expressions a3 + a2 + a – 4
The H.C.F, and L.C.M. of the two expressions are a2 – a – 2 and a4 – 5a + 4
The product of the two expressions
= Their H.C.F. x L.C.M.
Here, the product of the two expressions
= (a2 – a – 2)(a4 – 5a + 4)
and one expression = a3 + a2 + a – 4.
.’. The required other expression
= (a2 – a – 2)(a4 – 5a + 4) / (a3 + a2 + a – 4)
= (a2 – a – 2) (a -1)
= a3 – 2a2 – a + 2.
Examples of HCF Calculation Using Prime Factorization
Example 10
The H.C.F. and L.C.M. of two quadratic expressions are a + 1 and + 2a – a – 2, find the expressions.
Solution :
Given
The H.C.F. and L.C.M. of two quadratic expressions are a + 1 and + 2a – a – 2
Here, H.C.F. = a + 1
and L.C.M. = a3 + 2a2 – a – 2
= a2(a + 2) – 1 (a + 2)
= (a2 – 1)(a + 2) = (a + 1)(a – 1)(a + 2).
It is found that other than H.C.F. (a + 1), the two factors of L.C.M. are a – 1 and a + 2.
Therefore, the required quantities are (a + 1)(a – 1) and (a + 1) (a +2),
that means a2 – 1 and a2 + 3a + 2.