An expression of the form axn + bxn-1 + cxn-2 + ….+kx+ l, where each variable has a constant accompanying it as its coefficient is called a polynomial of degree ‘n’ in variable x. Thus, a polynomial is an expression in which a combination of a constant and a variable is separated by an addition or a subtraction sign. Till now we have learnt about calculating zeroes of polynomials. Zeros of polynomials, when represented in the form of another linear polynomial are known as factors of polynomials. Let us learn about various ways of factoring polynomials.
Factor theorem: For a polynomial p(x) of degree greater than or equal to one,
- x-a is a factor of p(x), if p(a) = 0
- if p(a) = 0, x-a is a factor of p(x)
Where ‘a’ is a real number
Problems related to factoring polynomials:
Question 1: Check whether x+3 is a factor of x3 + 3x2 + 5x +15?
Solution: Let, q(x) = x + 3, for calculating zero of this polynomial
x + 3 = 0
=> x = -3
Now, p(x) = x3 + 3x2 + 5x +15, let us check the value of this polynomial for x = -3,
p(-3) = (-3)3 + 3 (-3)2 + 5(-3) + 15 = -27 + 27 – 15 + 15 = 0
As, p(-3) = 0, x+3 is a factor of x3 + 3x2 + 5x +15.
Question 2: Factorize x2 + 5x + 6.
Solution: Let us try factorizing this polynomial using splitting the middle term method
Factoring polynomials by splitting the middle term:
In this technique we need to find two terms ‘a’ and ‘b’ such that a + b =5 and ab = 6. On solving this we obtain, a = 3 and b = 2
Thus the above expression can be written as,
x2 + 3x + 2x + 6 = x(x + 3) + 2(x + 3) = (x + 3)(x + 2)
Thus, x+3 and x+2 are the factors of the polynomial x2 + 5x + 6.