Please Wait... While Loading Full Video#### Class 9 chapter - 2 POLYNOMIALS

### Remainder Theorem

♦ Remainder Theorem

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 ` (2x^3 + x^2 +x ) ÷ x = (2 x^3)/x + x^2/x + x/x`

`= 2x^2 + x+ 1`

In fact, you may have noticed that x is common to each term of `2x^3 + x^2 + x`. So

we can write `2x^3 + x^2 + x` as `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, `(3x^2 + x + 1) ÷ x = (3x^2 ÷ x) + (x ÷ x) + (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.

`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 ` (2x^3 + x^2 +x ) ÷ x = (2 x^3)/x + x^2/x + x/x`

`= 2x^2 + x+ 1`

In fact, you may have noticed that x is common to each term of `2x^3 + x^2 + x`. So

we can write `2x^3 + x^2 + x` as `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, `(3x^2 + x + 1) ÷ x = (3x^2 ÷ x) + (x ÷ x) + (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.

Q 3230145912

Divide `p(x)` by `g(x)`, where `p(x) = x + 3x^2 – 1 `and `g(x) = 1 + x`.

Class 9 Chapter 2 Example 6

Class 9 Chapter 2 Example 6

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

`text (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`.

`text (Step 2 : )` We divide the first term of the dividend by the first term of the divisor, i.e., we divide `(3x^2)/x =3x` = first term of quotient of the quotient.

`text (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` .

`text (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 `(-2x)/x = -2` the first term `– 2x` of the (new) dividend by the first term `x` of the divisor and obtain = second term of quotient `– 2`. Thus, `– 2` is the second term in the quotient.

`text (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 whole quotient.

`text (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:

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

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

Let us consider some more examples.

Q 3260145915

Divide the polynomial `3x^4 – 4x^3 – 3x –1` by `x – 1`.

Class 9 Chapter 2 Example 7

Class 9 Chapter 2 Example 7

By long division, we have:

Here, the remainder is `– 5`. Now, the zero of `x – 1` is 1. So, putting `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.

Q 3270145916

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

Class 9 Chapter 2 Example 8

Class 9 Chapter 2 Example 8

By long division,

So, 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.

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)`.

`text (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.

`text (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.

Q 3210145919

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

Class 9 Chapter 2 Example 9

Class 9 Chapter 2 Example 9

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`.

Q 3210156010

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

Class 9 Chapter 2 Example 10

Class 9 Chapter 2 Example 10

As you 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)^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`.