NCERT Class 9 Mathematics

POLYNOMIALS 37

Step 5 : We multiply the divisor by the second
term of the quotient and subtract the product
from the dividend. That is, we multiply x + 1
by – 2 and subtract the product – 2x – 2
from the dividend – 2x – 1. This gives us 1
as the remainder.

This process continues till the degree of the new dividend is less than the degree of the
divisor. At this stage, this new dividend becomes the remainder and the sum of the
quotients gives us the whole quotient.

Step 6 : Thus, the quotient in full is 3x – 2 and the remainder is 1.

Let us look at what we have done in the process above as a whole:

3 x – 2

x + 1 3 x (^2) + x – 1
3 x^2 + 3x

• –
  
  
  • 2x – 1
  
  
  • 2x – 2
    ++
    1
  
  
  
  

Notice that 3x^2 + x – 1 = (x + 1) (3x – 2) + 1

i.e., Dividend = (Divisor × Quotient) + Remainder

In general, if p(x) and g(x) are two polynomials such that degree of p(x) ✄ degree of
g(x) and g(x) ✂ 0, then we can find polynomials q(x) and r(x) such that:
p(x) =g(x)q(x) + r(x),

where r(x) = 0 or degree of r(x) < degree of g(x). Here we say that p(x) divided by
g(x), gives q(x) as quotient and r(x) as remainder.

In the example above, the divisor was a linear polynomial. In such a situation, let us
see if there is any link between the remainder and certain values of the dividend.

In p(x) = 3x^2 + x – 1, if we replace x by –1, we have

p(–1) = 3(–1)^2 + (–1) –1 = 1

So, the remainder obtained on dividing p(x) = 3x^2 + x – 1 by x + 1 is the same as the
value of the polynomial p(x) at the zero of the polynomial x + 1, i.e., –1.

(x + 1)(–2) –2x – 1

= –2x – 2 –2x – 2

++

+ 1