Introduction For Prime Factors Binomials

Introduction For Prime Factors Binomials:

      Prime factors: A number written has no divisors and it divisible by itself and as a product of prime factors is said to in the prime   factor form.
     Example for prime factors:
             Prime factor form of 5 = 1*5 thus 1, 5  are prime factors.

     Binomials: An expression with two unlike terms is called a binomials.
     Example for binomials:
             x+y, m-5, mn+4m

Steps for Finding Prime Factors:

Step 1: Finding a common factor
      When terms of an algebraic equations A have a common factor B, we divide each term of A by B and get an expression C. Now, A is factored to be B × C.

Step 2: Grouping the terms
      When the terms of an algebraic equations it does not have a common factor, the terms may be grouped in an appropriate manner and a common factor is determined

Prime Factors Binomials:

Example 1:
Find prime factors for 25m^2 - 16n^2

Solution:
        We have - 25m^2- 16n^2 = (5m)2- (4n)2        where formula, a^2-b2 = (a+b)(a-b)
                                        = (5m+4n) (5m-4n)


Example 2:
 Find prime factors for x^2 – 7x + 12.

Solution :
        Since, a + b = – 7, ab = 12,and negative factors of 12 are – 1, – 2, – 3, – 4, – 6 and – 12, we find that a = – 4 and
b = – 3 (or a = – 3 and b = – 4). Hence,
        x^2 – 7x + 12 = x^2 + {(– 4) + (– 3)}x + (– 4) × (– 3)
                           = (x – 4) (x – 3)

Example  3:
 Find prime factors for x^2 + 3x – 10.

Solution :
Here, we have to find two numbers a and b such that a + b = 3 (the coefficient of x) and ab = – 10 (the constant term).
Now factors of –10 are ± 1, ± 2, ± 5 and ± 10. A little experimentation with these numbers tells us that we may take a and b as 5 and – 2. The sum of 5 and – 2 is 3, and product of 5 and – 2 = – 10. Hence,
              x^2 + 3x – 10 = x^2 + {5+ (– 2)}x + 5(– 2)
                                 = (x + 5) (x – 2)

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