NCERT Class 9 Maths Solutions Chapter - 2 Polynomials, Ex 2.3

Ex 2.3

Question 1.  Find the remainder when x³ + 3x² + 3x + 1 is divided by


Solution

Let p(x) = x³ + 3x² + 3x + 1.

(i)     x + 1
        Zero of x +1 is-1.
        i.e. p(-1) = (- 1)³ + 3 (- 1)² + 3 (-1) + 1 = 0
        So, the remainder is 0.

(ii)    

         Zero of   is 

(iii)     x
          Zero of x is 0.
          p(0) = (0)³ + 3(0)² + 3(0) + 1 = 1
          So, the remainder is 1.

(iv)     x + 
          Zero of x +  is:
          x +  = 0  x = - 
          p (- ) = (- )³ + 3(- )² + 3(- ) + 1 = - ³ + 3 ² - 3 + 1
          So, the remainder is - ³ + 3 ² - 3 + 1

(v)      5 + 2x
          Zero of 5+2x  is:
          5 + 2x = 0  2x = - 5
          i.e. x = - 

(i)     x + 1

        By long division

        

         So, remainder is 0.

(ii).    

         By long division

  So, remainder is  .

(iii)     x

          By long division

          

          So, remainder is 1.

(iv)     x + 
          By long division

      So, the remainder is 

(v)     5 + 2x 
         By long division

         

         So the remainder is -.

Concept Insight: The remainder of any polynomial p(x) when divided by another polynomial (ax+b) where a and b are real numbers  is p(-b/a).
Note that here -b/a  is the zero of polynomial ax+b.
This problem can also be solved using long division. For long division  first write the divisor and dividend in the standard form, i.e. arrange the terms in the descending order of their powers. The process of division is continued till the remainder is constant or the degree of new dividend is less than the degree of divisor. Do not forget to change the sign of terms while subtraction. For cross verification division algorithm
 Dividend = Quotient  Divisor + Remainder can be used.

Question 2.  Find the remainder when x³ - ax² + 6x - a is divided by x - a.

Solution

According to the remainder theorem, if p(x) is any polynomial of degree  1 and a is any real number, then when p(x) is divided by the linear polynomial x - a, then the remainder is p(a).

Here p(x) = x³ - ax² + 6x - a
        p(a) = (a)³ - a(a)² + 6a - a
               = 5a
So when x³ - ax² + 6x - a is divided by x - a, remainder comes to 5a.

OR
By long division

So when x³ - ax² + 6x - a is divided by x - a, remainder comes to 5a.

Concept Insight:  The remainder of any polynomial p(x) when divided by another polynomial (ax+b) where a and b are real numbers  is
p(-b/a).
Note that here -b/a  is the zero of polynomial ax+ b.
This question can also be solved using long division method however it is long and time consuming. Chances of making computational error are high in that method.

Question 3.  Check whether 7 + 3x is a factor of 3x³ + 7x.

Solution

Zero of 7 + 3x is:
7 + 3x = 0
Therefore, 

7+3x can be a factor of p(x) = 3x³ + 7x only if  
Here p(x) = 3x³ + 7x

 7 + 3x is not a factor of 3x³ + 7x.

OR

Let us divide (3x³ + 7x) by (7 + 3x). If remainder comes out to be 0 then  7 + 3x will be a factor of

3x³ + 7x.

By long division

As remainder is not zero so 7 + 3x is not a factor of 3x³ + 7x.

Concept Insight: Any linear polynomial 'ax+b' where a and b are real numbers  is a factor of the polynomial p(x) iff p(-b/a) = 0 i.e  -b/a is a zero of p(x) or both the polynomials has a common zero -b/a. This question can also be solved using long division method. Do not forget to change the sign of terms while subtraction in the long division.

Ex-2.4