**Divide the polynomial p(x) by the polynomialg(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

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

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

**Answer
1** :

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

**Solution:**

Here

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:**

Here

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:**

Here

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

**Check whether the first polynomial is a factorof the second polynomial by dividing the second polynomial by the firstpolynomial:**

_(i) t^{2} – 3, 2t^{4} + 3t^{3} – 2t^{2}– 9t – 12

_(ii) x^{2} + 3x + 1, 3x^{4} + 5x^{3} – 7x^{2} + 2x + 2

_(iii) x^{2} + 3x + 1, x^{5} – 4x^{3 }+ x^{2} + 3x + 1

**Answer
2** :

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

**Solutions:**

Here,

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:**

Here,

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:**

Here,

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 to0. Therefore, we say that, x^{3}-3x+1 is not a factor of x^{5}-4x^{3}+x^{2}+3x+1.

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

**Answer
3** :

**Solutions:**

Since this is a polynomial equation of degree4, 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 be0.

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=0

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 polynomialequation are:

**√(5/3),- √(5/3) , −1 and −1.**

Hence, is the answer.

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

**Answer
4** :

**Solution:**

Here,

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+2with (x-2)

Therefore, **g(x) = (x ^{2}–x+1)**

**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**

**Answer
5** :

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