NCERT Class 9 Mathematics Textbook for Blind and Visually Impaired Students made Screen Readable by Professor T K bansal.

Let us consider two numbers 15 and 6. You know that when we divide 15 by 6, we get the quotient 2, and remainder 3. Do you remember how this fact is expressed? We write 15 as
15 = (6 × 2 ) + 3

We observe that the remainder 3 is less than the divisor 6. Similarly, if we divide 12 by 6, we get
12 = (6 × 2) + 0

What is the remainder here? Here the remainder is 0, and we say that 6 is a factor of 12 or 12 is a multiple of 6.

Now, the question is: can we divide one polynomial by another? To start with, let us try and do this when the divisor is a monomial. So, let us divide the polynomial \(2x^3\ +\ x^2\ +\ x\) by the monomial x.

We have

\[\frac{(2x^3\ +\ x^2\ +\ x)}{x}\ =\ \frac{2x^3}{x}\ +\ \frac{x^2}{x}\ +\ \frac{x}{x}\]

\[=\ 2x^2\ +\ x\ +\ 1\]

In fact, we may notice that x is common to each term of \(2x^3\ +\ x^2\ +\ x.\)

So we can write

\[2x^3\ +\ x^2\ +\ x\ =\ x\ ×\ (2x^2\ +\ x\ +\ 1).\]

We say that x and 2x^2 + x + 1 are factors of 2x^3 + x^2 + x, and 2x^3 + x^2 + x is a multiple of x as well as a multiple of 2x^2 + x + 1.

Consider another pair of polynomials 3x^2 + x + 1 and x. Here,

\[\frac{(3x^2\ +\ x\ +\ 1)}{x}\ =\ \frac{3x^2}{x}\ +\ \frac{x}{x}\ +\ \frac{1}{x}.\]

We see that we cannot divide 1 by x to get a polynomial term. So in this case we stop here, and note that 1 is the remainder. Therefore, we have

\[3x^2\ +\ x\ +\ 1\ =\ x\ ×\ (3x\ +\ 1)\ +\ 1\]

In this case, 3x + 1 is the quotient and 1 is the remainder. Do you think that x is a factor of 3x^2 + x + 1? Since the remainder is not zero, it is not a factor.

Now let us consider an example to see how we can divide a polynomial by any non−zero polynomial.

Example 6

Divide p(x) by g(x), where
\[p(x)\ =\ x\ +\ 3x^2\ −\ 1\]
and
\[g(x)\ =\ 1\ +\ x.\]

Solution :

We carry out the process of division by means of the following steps:

Step 1 : We write the dividend x + 3x^2 − 1 and the divisor 1 + x in the standard form, i.e., after arranging the terms in the descending order of their degrees. So, the dividend is 3x^2 + x −1 and divisor is x + 1.

Step 2 :We divide the first term of the dividend by the first term of the divisor, i.e., we divide 3x^2 by x, and get 3x. This gives us the first term of the quotient.

\[\frac{3x^2}{x}\ =\ 3x\]
= first term of the quotient

Step 3 :We multiply the divisor by the first term of the quotient, and subtract this product from the dividend, i.e., we multiply x + 1 by 3x and subtract the product 3x^2 + 3x from the dividend 3x^2 + x − 1. This gives us the remainder as − 2x − 1.

Step 4 : We treat the remainder − 2x − 1 as the new dividend. The divisor remains the same. We repeat Step 2 to get the next term of the quotient, i.e., we divide the first term − 2x of the (new) dividend by the first term x of the divisor and obtain − 2. Thus, − 2 is the second term in the quotient.

\[\frac{−2x}{x}\ =\ −2\]
= second term in the quotient
New Quotient = 3x − 2]

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 remainder is 0 or 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 complete 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:

{explain long division method}

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) greater than equal to 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.

Let us consider some more examples.

Example 7

Divide the polynomial

\[3x^4\ −\ 4x^3\ −\ 3x\ −\ 1\]
by x − 1.

Solution :

By long division, we have:
Here, the remainder is − 5. Now, the zero of x − 1 is 1. So, substituting x = 1 in p(x), we see that
p(1) = 3(1)^4 −4(1)^3 −3(1) −1
= 3 − 4 − 3 − 1
= − 5, which is the remainder.

Example 8

Find the remainder obtained on dividing p(x) = x^3 + 1 by x + 1.

Solution :

By long division, we find that the remainder is 0.
Here p(x) = x^3 + 1, and the root of x + 1 = 0 is x = −1. We see that
p(−1) = (−1)3 + 1
= −1 + 1
= 0,
which is equal to the remainder obtained by actual division.

Is it not a simple way to find the remainder obtained on dividing a polynomial by a linear polynomial?

We shall now generalise this fact in the form of the following theorem. We shall also show you why the theorem is true, by giving you a proof of the theorem.

Remainder Theorem (defination):

Let p(x) be any polynomial of degree greater than or equal to one and let a be any real number. If p(x) is divided by the linear polynomial x − a, then the remainder is p(a).

Proof :

Let p(x) be any polynomial with degree greater than or equal to 1. Suppose that when p(x) is divided by x − a, the quotient is q(x) and the remainder is r(x), i.e.,
p(x) = (x − a ) × q(x) + r(x)

Since the degree of x − a is 1 and the degree of r(x) is less than the degree of x − a, the degree of r(x) = 0. This means that r(x) is a constant, say r.
So, for every value of x, r(x) = r.
Therefore, p(x) = ( x − a ) × q(x) + r

In particular, if x = a, this equation gives us
p(a) = (a − a ) × q(a) + r = r,
which proves the theorem.

Let us use this result in another example.

Example 9

Find the remainder when x^4 + x^3 − 2x^2 + x + 1 is divided by x − 1.

Solution :

Here, p(x) = x^4 + x^3 − 2x^2 + x + 1, and the zero of x − 1 is 1.

So, p(1) = (1)^4 + (1)^3 − 2(1)^2 + 1 + 1 = 2

So, by the Remainder Theorem, 2 is the remainder when x^4 + x^3 − 2x^2 + x + 1 is divided by x − 1.

Example 10

Check whether the polynomial q(t) = 4t^3 + 4t^2 − t − 1 is a multiple of 2t + 1.

Solution :

As we know, q(t) will be a multiple of 2t + 1 only, if 2t + 1 divides q(t) leaving remainder zero. Now, taking 2t + 1 = 0, we have t = − 1/2.

Also, q (−1/2) = 4(−1/2) t^3 + 4(−1/2) ^2 − (−1/2) − 1
= −1/2+1+1/2−1
= 0

So the remainder obtained on dividing q(t) by 2t + 1 is 0.
So, 2t + 1 is a factor of the given polynomial q(t), that is q(t) is a multiple of 2t + 1.

EXERCISE 2.3

Q1. Find the remainder when
\[x^3\ +\ 3x^2\ +\ 3x\ +\ 1\]
is divided by
(i) x + 1
(ii) x − 1/2
(iii) x
(iv) x + π
(v) 5 + 2x

A1.
(i) 0
(ii) 27/8
(iii) 1
(iv) − pi^3 + 3 pi^2 − 3pi + 1
(v) – (27/8)

Q2. Find the remainder when
\[x^3\ −\ ax^2\ +\ 6x\ −\ a\]
is divided by x − a.

A2. 5a

Q3. Check whether 7 + 3x is a factor of 3x^3 + 7x.

A3. No, since remainder is not zero.

---