A polynomial f(x) is the product of two polynomials g(x) and h(x), then g(x) and h(x) are called the factors of f(x).
Example 1
f(x)=3x2–17x–20
=3x2–20x+3x–20
=x(3x–20)+1(3x–20)
=(x+1)(3x–20)
g(x)=x+1
h(x)=3x–20
are called the factors of f(x)
Example 2
Consider the polynomials
f(x)=2(x-2)2 (x-4) (2x-1)
g(x)=8(x-2)3 (2x-1) (3x+1)
Notice that
(x-2) is a common factor of f(x) and g(x)
and (2x-1) is a common factor of f(x) and g(x).
Also, 2 is a common factor of the polynomials and (x-2)2 is a common factor of f(x) and g(x).
Now, h(x)=2(x-2)2(2x-1) is a common factor of f(x)
and g(x) because
f(x)=2(x-2)2 (x-4) (2x-1)=h(x)(x-4)
g(x)=8(x-2)3 (2x-1) (3x+1)=h(x)[4(x-2)(3x+1)]
Are there any other factors not included in h(x)? No. This is because the remaining factors are of f(x)(x-4), and {4(x-2)(3x+1)} of g(x) has no common factor. So we say that
h(x) is the Highest Common Factor of f(x) and g(x).
In general, h(x) is the HCF(Highest Common Factor) of f(x) and g(x) if
- h(x) is a common factor of f(x) and g(x)
- every common factor of f(x) and g(x) is also a factor of h(x).
To find the HCF of two or more given polynomials, we can use the following steps.
Step 1: Express each polynomial as a product of powers of irreducible factors. The numerical factors must be expressed as the power of primes.
Step 2: If there is no common factor, the HCF is 1. If there are common irreducible factors, find the smallest (least) exponents of these irreducible factors in the factorized form of the given polynomials.
Step 3: Raise the common irreducible factors to the smallest exponents found in Step 2 and multiply to get the HCF.
Observe the following examples.
Example 3
Find the highest common factor of
f(x)=30(x2-3x+2)
g(x)=50(x2-2x+1)
We first write f(x) and g(x) as the product of powers of irreducible factors.
f(x)=30(x2-3x+2)
=2*3*5[x2-2x-x+2]
=2*3*5[x(x-2)-1(x-2)]
=2*3*5(x-1)(x-2)
g(x)=50(x2-2x+1)
=2*52*(x-1)2
On comparing x2-2x+1 with (a2-2ab+b2)=(a-b)2, common factors of f(x) and g(x) are
2, 5, (x-1)
h(x)=2*5*(x-1)
These are the irreducible factors. So HCF=h(x)=10(x-1)
Example 4
Find the HCF of the polynomials 8x(x+10)2(x-5) and
12x3(x+10)
Let f(x)=8x(x+10)2(x-5)
g(x)=12x3(x+10)
Reducing these polynomials to powers of irreducible factors
f(x)=23 * x * (x+10)2(x-5)
g(x)=22*3*x3*(x+10)
Common factors are 2, x, x+10
Least exponents are 2, 1, 1 respectively.
∴ h(x)=22 * x * (x+10)
∴ HCF=4x(x+10)
Try these questions
- Find the highest common factor of 42(x3-1) and 14(x2-1)
Answer:
Let f(x)=42(x3-1)
g(x)=14(x2-1)
Reducing f(x), g(x) to irreducible factors
f(x)=2*3*7*(x-1)(x2+x+1)
g(x)=2*7*(x-1)(x+1)
The common factors of f(x) and g(x) are 2*7*(x-1)
Least exponents are 1, 1, 1
∴ HCF=14(x-1)
- Find the highest common factor of the polynomials (x2+x+1)(x+13)2 and (x2+x+1)2 (x+13)
Answer:
Let f(x)=(x2+x+1) (x+13)2
g(x)=(x2+x+1)2(x+13)
Reducing f(x), g(x) to irreducible factors
f(x)=(x2+x+1) (x+13) (x+13)
g(x)=(x2+x+1) (x2+x+1) (x+13)
Common factors are (x2+x+1), (x+13)
Least exponents are 1, 1
∴HCF=(x2+x+1)(x+13)