**Polynomials**

**Exercise 2.3 Page: 36**

**1. Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:**

**(i)** **p(x) = x ^{3}-3x^{2}+5x–3 , g(x) = x^{2}–2**

**Solution:**

Given,

Dividend = p(x) = x^{3}-3x^{2}+5x–3

Divisor = g(x) = x^{2}– 2

Therefore, upon division we get,

Quotient = x–3

Remainder = 7x–9

**(ii) p(x) = x ^{4}-3x^{2}+4x+5 , g(x) = x^{2}+1-x**

**Solution:**

Given,

Dividend = p(x) = x^{4 }– 3x^{2 }+ 4x +5

Divisor = g(x) = x^{2} +1-x

Therefore, upon division we get,

Quotient = x^{2 }+ x–3

Remainder = 8

**(iii) p(x) =x ^{4}–5x+6, g(x) = 2–x^{2}**

**Solution:**

Given,

Dividend = p(x) =x^{4} – 5x + 6 = x^{4 }+0x^{2}–5x+6

Divisor = g(x) = 2–x^{2} = –x^{2}+2

Therefore, upon division we get,

Quotient = -x^{2}-2

Remainder = -5x + 10

**2. Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:**

**(i) t ^{2}-3, 2t^{4 }+3t^{3}-2t^{2}-9t-12**

**Solutions:**

Given,

First polynomial = t^{2}-3

Second polynomial = 2t^{4 }+3t^{3}-2t^{2 }-9t-12

As we can see, the remainder is left as 0. Therefore, we say that, t^{2}-3 is a factor of 2t^{2}+3t+4.

**(ii)x ^{2}+3x+1 , 3x^{4}+5x^{3}-7x^{2}+2x+2**

**Solutions:**

Given,

First polynomial = x^{2}+3x+1

Second polynomial = 3x^{4}+5x^{3}-7x^{2}+2x+2

As we can see, the remainder is left as 0. Therefore, we say that, x^{2} + 3x + 1 is a factor of 3x^{4}+5x^{3}-7x^{2}+2x+2.

**(iii) x ^{3}-3x+1, x^{5}-4x^{3}+x^{2}+3x+1**

**Solutions:**

Given,

First polynomial = x^{3}-3x+1

Second polynomial = x^{5}-4x^{3}+x^{2}+3x+1

As we can see, the remainder is not equal to 0. Therefore, we say that, x^{3}-3x+1 is not a factor of x^{5}-4x^{3}+x^{2}+3x+1 .

**3. Obtain all other zeroes of 3x ^{4}+6x^{3}-2x^{2}-10x-5, if two of its zeroes are √(5/3) and – √(5/3).**

**Solutions:**

Since this is a polynomial equation of degree 4, hence there will be total 4 roots.

**√(5/3) and – √(5/3) **are zeroes of polynomial f(x).

**∴ **(x –**√(5/3)**) (x+**√(5/3) **= x^{2}-(5/3) = 0

**(3x ^{2}−5)=0,** is a factor of given polynomial f(x).

Now, when we will divide f(x) by (3x^{2}−5) the quotient obtained will also be a factor of f(x) and the remainder will be 0.

Therefore, 3x^{4 }+6x^{3 }−2x^{2 }−10x–5 = (3x^{2 }–5)**(x ^{2}+2x+1)**

Now, on further factorizing (x^{2}+2x+1) we get,

**x ^{2}+2x+1** = x

^{2}+x+x+1 = 0

x(x+1)+1(x+1) = 0

**(x+1)(x+1) = 0**

So, its zeroes are given by: **x= −1 **and** x = −1.**

Therefore, all four zeroes of given polynomial equation are:

**√(5/3),- √(5/3) , −1 and −1.**

Hence, is the answer.

**4. On dividing x ^{3}-3x^{2}+x+2**

**by a polynomial g(x), the quotient and remainder were x–2 and –2x+4, respectively. Find g(x).**

**Solution:**

Given,

Dividend, p(x) = x^{3}-3x^{2}+x+2

Quotient = x-2

Remainder = –2x+4

We have to find the value of Divisor, g(x) =?

As we know,

Dividend = Divisor × Quotient + Remainder

∴ x^{3}-3x^{2}+x+2 = g(x)×(x-2) + (-2x+4)

x^{3}-3x^{2}+x+2-(-2x+4) = g(x)×(x-2)

Therefore, g(x) × (x-2) = x^{3}-3x^{2}+x+2

Now, for finding g(x) we will divide x^{3}-3x^{2}+x+2 with (x-2)

Therefore, **g(x) = (x ^{2}–x+1)**

**5. Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and**

**(i) deg p(x) = deg q(x)**

**(ii) deg q(x) = deg r(x)**

**(iii) deg r(x) = 0**

**Solutions:**

According to the division algorithm, dividend p(x) and divisor g(x) are two polynomials, where g(x)≠0. Then we can find the value of quotient q(x) and remainder r(x), with the help of below given formula;

Dividend = Divisor × Quotient + Remainder

∴ p(x) = g(x)×q(x)+r(x)

Where r(x) = 0 or degree of r(x)< degree of g(x).

Now let us proof the three given cases as per division algorithm by taking examples for each.

**(i) deg p(x) = deg q(x)**

Degree of dividend is equal to degree of quotient, only when the divisor is a constant term.

Let us take an example, 3x^{2}+3x+3 is a polynomial to be divided by 3.

So, (3x^{2}+3x+3)/3 = x^{2}+x+1 = q(x)

Thus, you can see, the degree of quotient is equal to the degree of dividend.

Hence, division algorithm is satisfied here.

**(ii) deg q(x) = deg r(x)**

Let us take an example , p(x)=x^{2}+x is a polynomial to be divided by g(x)=x.

So, (x^{2}+x)/x = x+1 = q(x)

Also, remainder, r(x) = 0

Thus, you can see, the degree of quotient is equal to the degree of remainder.

Hence, division algorithm is satisfied here.

**(iii) deg r(x) = 0**

The degree of remainder is 0 only when the remainder left after division algorithm is constant.

Let us take an example, p(x) = x^{2}+1 is a polynomial to be divided by g(x)=x.

So,( x^{2}+1)/x= x=q(x)

And r(x)=1

Clearly, the degree of remainder here is 0.

Hence, division algorithm is satisfied here.