**Q.1 If a polynomial x ^{3 }-3x^{2}+x+2 is divided by a polynomial g(x), the quotient and remainder obtained are (x-2) and (-2x+4), respectively. Find the equation of g(x).**

**Sol.**

Since, **Dividend = Divisor × Quotient + Remainder**

Therefore, x³ -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}+3x-2

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

Therefore, **g(x) = (x ^{2} – x +1)**

**Q.2 Find the quotient and remainder by dividing the polynomial f(x) by the polynomial g(x).**

(i) **f(x) = x ^{3 }+ 2x^{2 }– 9x + 5**,

**g(x) = x**

^{2}+5Therefore, Quotient is **(x+2)** and Remainder is **(-14x − 5)**

(ii) f(x) = **x ^{5}+2x^{4}-9x^{3}+5x^{2}-2x+1**, g(x) =

**x**

^{3 }+ x^{2 }– x+1Therefore, Quotient is **(x ^{2 }+ x − 9)** and Remainder is

**(14x**).

^{2}– 12x +10

(iii) f(x) = **2x ^{4 }+ 7x^{3 }+ 5x^{2 }+ 8x + 5**, g(x) =

**11− 2x**

^{3 }+ x^{2}Therefore, Quotient is **– (x + 4)** and Remainder is **(9x ^{2} + 19x +49).**

**Q3. Find all the zeroes of the polynomial equation 2x ^{4}-3x^{3}-3x^{2}+6x-2, if two of its zeroes are **

**Sol.**

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

Let, **f(x) = 2x ^{4 }– 3x^{3 }– 3x^{2 }+ 6x – 2**

∴**(x ^{2}− 2)** = g(x), is a factor of given polynomial f(x).

If we divide f(x) by g(x), the quotient will also be a factor of f(x) with remainder =0.

So, 2x^{4 }− 3x^{3 }− 3x^{2 }+ 6x – 2 = (x^{2 }– 2) **(2x ^{2} – 3x +1).**

Now, on further factorizing **(2x ^{2} – 3x +1)** we get,

2x^{2} – 3x +1 = 2x^{2} – 2x − x +1 = 0

2x (x − 1) – 1(x−1) = 0

(2x−1) (x−1) = 0

So, its zeroes are given by: x= ** 12** and x =

**1**

Therefore, all four zeroes of the given polynomial equation are:

**Q4. Find all zeroes of a polynomial equation x ^{4}-6x^{3}-26x^{2}+138x-35, if two of its zeroes are **

**Sol.**

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

Let, **f(x) = x ^{4 }– 6x^{3 }– 26x^{2}+ 138x – 35**

Since, ** (2+3–√)and(2−3–√)** are zeroes of given polynomial f(x).

**∴ [x−(2+3–√)][x−(2−3–√)]=0**

On multiplying the above equation we get,

**x ^{2 }−4x + 1**, this is a factor of a given polynomial f(x).

Now, if we will divide f(x) by g(x), the quotient will also be a factor of f(x) and the remainder will be 0.

So, x^{4 }− 6x^{3 }− 26x^{2 }+ 138x – 35 = (x^{2 }– 4x + 1) **(x ^{2} – 2x −35).**

Now, on further factorizing (x^{2} – 2x −35) we get,

**x ^{2} – (7−5)x − 35** = x

^{2}– 7x + 5x +35 = 0

x(x − 7) + 5 (x−7) = 0

** (x+5) (x−7) = 0**

So, its zeroes are given by: x= −5 and x = 7.

Therefore, all four zeroes of given polynomial equation are:

**Q5. Find all zeroes of a polynomial equation 3x ^{4 }+ 6x^{3 }– 2x^{2 }– 10x – 5, if two of its zeroes are **

**Sol.**

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

Let, **f(x)= 3x ^{4}+6x^{3}-2x^{2}-10x-5**

**∴ (x−53−−√)(x+53−−√)**

** (x2−53)=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 factor of f(x) and 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:

** 53−−√, −53−−√, −1 and −1**.