EXPLANATION

EXPLANATION

You know that a cubic polynomial has at most three zeroes. However, if you are given only one zero, can you find the other two? For this, let us consider the cubic polynomial x³ – 3x² – x + 3. If we tell you that one of its zeroes is 1, then you know that x – 1 is a factor of x³ – 3x² – x + 3. So, you can divide x³ – 3x² – x + 3 by x – 1, as you have learnt in Class IX, to get the quotient x² – 2x – 3.

Next, you could get the factors of x² – 2x – 3, by splitting the middle term, as (x + 1)(x – 3). This would give you x³ – 3x² – x + 3 = (x – 1)(x² – 2x – 3)= (x – 1)(x + 1)(x – 3)

So, all the three zeroes of the cubic polynomial are now known to you as 1, – 1, 3.

Let us discuss the method of dividing one polynomial by another in some detail.

Before noting the steps formally, consider an example.

Example 6 and 7

Example 6 and 7

EXAMPLE 6

Divide 2x² + 3x + 1 by x + 2.

SOLUTION

Note that we stop the division process when either the remainder is zero or its degree is less than the degree of the divisor. So, here the quotient is 2x – 1 and the remainder is 3. Also,

(2x – 1)(x + 2) + 3 = 2x² + 3x – 2 + 3 = 2x² + 3x + 1 i.e.,  2x² + 3x + 1 = (x + 2)(2x – 1) + 3

Therefore, Dividend = Divisor × Quotient + Remainder

Let us now extend this process to divide a polynomial by a quadratic polynomial.

EXAMPLE 7

Divide 3x³ + x² + 2x + 5 by 1 + 2x + x².

SOLUTION

We first arrange the terms of the dividend and the divisor in the decreasing order of their degrees. Recall that arranging the terms in this order is called writing the polynomials in standard form. In this example, the dividend is already in standard form, and the divisor, in standard form, is x² + 2x + 1.

Step 1 : To obtain the first term of the quotient, divide the highest degree term of the dividend (i.e., 3x³) by the highest degree term of the divisor (i.e., x²). This is 3x. Then carry out the division process. What remains is – 5x² – x + 5.

Step 2 : Now, to obtain the second term of the quotient, divide the highest degree term of the new dividend (i.e., –5x²) by the highest degree term of the divisor (i.e., x²). This gives –5. Again carry out the division process with – 5x² – x + 5.

Step 3 : What remains is 9x + 10. Now, the degree of 9x + 10 is less than the degree of the divisor x² + 2x + 1. So, we cannot continue the division any further.

So, the quotient is 3x – 5 and the remainder is 9x + 10. Also,

(x² + 2x + 1) × (3x – 5) + (9x + 10) = 3x³ + 6x² + 3x – 5x² – 10x – 5 + 9x + 10

= 3x³ + x² + 2x + 5

Here again, we see that

Dividend = Divisor × Quotient + Remainder

What we are applying here is an algorithm which is similar to Euclid’s division algorithm that you studied in Chapter 1.

This says that

If p(x) and g(x) are any two polynomials with 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).

This result is known as the Division Algorithm for polynomials. Let us now take some examples to illustrate its use

Example 8 and 9

Example 8 and 9

Divide 3x² – x³ – 3x + 5 by x – 1 – x², and verify the division algorithm.

SOLUTION

Note that the given polynomials are not in standard form. To carry out division, we first write both the dividend and divisor in decreasing orders of their degrees.

So, dividend = –x³ + 3x² – 3x + 5 and divisor = –x² + x – 1.

Division process is shown on the right side.

We stop here since degree (3) = 0 < 2 = degree (–x² + x – 1). So, quotient = x – 2, remainder = 3.

Now,

Divisor × Quotient + Remainder

= (–x² + x – 1) (x – 2) + 3

= –x³ + x² – x + 2x² – 2x + 2 + 3

= –x³ + 3x² – 3x + 5

= Dividend In this way, the division algorithm is verified.

EXAMPLE 9

Find all the zeroes of 2x⁴ – 3x³ – 3x² + 6x – 2, if you know that two of its zeroes are  and .

Solution

Since two zeroes are  and  is a factor of the given polynomial. Now, we divide the given polynomial by x² – 2.

First term of quotient is 

Second term of quotient is 

Third term of quotient is 

So, 2x⁴ – 3x³ – 3x² + 6x – 2 = (x² – 2)(2x² – 3x + 1).

Now, by splitting –3x, we factorise 2x² – 3x + 1 as (2x – 1)(x – 1). So, its zeroes are given by  and x = 1. Therefore, the zeroes of the given polynomial are