Class 9 Maths Polynomials Find Remainder using Long Division

Find Remainder using Long Division

If you divide 7/3, you get 2 as quotient, 1 as remainder. Also 3 is divisor here & 7 is dividend.

So we can write 7 = (3 * 2)  +1  or   Dividend  = (Divisor * Quotient)  + Remainder.

Observe that Remainder is always less that Divisor.

Let’s try to divide two polynomials. E.g.  p(x)/ q(x)  where p(x) = 2x3 + x2 + x   &     q(x) =x.

 

 

In this case , x is common to each term of 2x3 + x2 + x. So we can write 2x3 + x2 + x as x(2x2 + x + 1). We say that x and 2x2 + x + 1 are factors of 2x3 + x2 + x, and 2x3 + x2 + x is a multiple of x as well as a multiple of 2x2 + x + 1.

Let us try to divided p(x)=  3x2 + x + 1  with q(x)= x.

(3x2 + x + 1) ÷ x = (3x2 ÷ x) + (x ÷ x) + (1 ÷ x)  =  3x + 1 +  1/x.

Since the last term (1) is not divisible by x, we will write this expression as

 3x2 + x + 1 = {(3x + 1) × x} + 1       à Dividend = (Divisor × Quotient) + Remainder

Notice that 1 is remainder here. In this case, 3x + 1 is the quotient and 1 is the remainder.  Also here, x is not a factor of 3x2 + x + 1, since the remainder is not zero.

 

Let us try to divided p(x)=  3x2 + x - 1  with q(x)= x+1.

 

 

3x2 + x - 1 = {(x + 1) ×( 3x-2)} + 1       à Dividend = (Divisor × Quotient) + Remainder

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