**EXERCISE 2.3**

**EXERCISE 2.3**

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

*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

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

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

*(i) p(x)*= x

^{3}-3x

^{2}+ 5x – 3,

*g (x)*= x

^{2}_ 2

*(ii) p(x)*=x

^{3}+

**3x**,

^{2}+ 4x*g(x)*= x

^{2}+ 1-x

*(iii) p(x)*=

**x**

^{4}– 5x+ 6,

*g(x)*= 2 –

*x*

**Sol. (i) We have,**

=> The quotient is x – 3 and the remainder is 7x – 9.

*(ii) *Here, the dividend is already in the standard form

and the divisor is also in the standard form.

We have,

=> The quotient is x^{2}+ x – 3 and the remainder is 8.

*(iii)* To carry out the division, we first write divisor in

the standard form.

So, divisor = – x^{2} + 2

We have,

=> The quotient is – x^{2} – 2 and the remainder is – 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, 2*t*^{4}+ 3*t*^{3}– 2*t*^{2} – 9*t* – 12

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

*(i)*

*t*

^{2}– 3, 2

*t*

^{4}+ 3

*t*

^{3}– 2

*t*

^{2}– 9

*t*– 12

*(ii)*x

^{2}+ 3x + 1, 3x

^{4}+ 5x

^{3}– 7x

^{2}+ 2x + 2

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

*(iii)*x

^{3}-3x + 1, x

^{5}– 4x

^{3}+ x

^{2}+ 3x + 1

Sol.*(i)* *t*^{2} – 3, 2*t*^{4}+ 3*t*^{3}– 2*t*^{2} – 9*t* – 12

We have,

*(i)*

*t*

^{2}– 3, 2

*t*

^{4}+ 3

*t*

^{3}– 2

*t*

^{2}– 9

*t*– 12

Since the remainder is zero, therefore,* t ^{2}-3* is a

factor of 2

*t*

^{4}+ 3

*t*

^{3}– 2

*t*

^{2}– 9

*t*– 12.

### *(ii) *Let us divide ** 3x**^{4} + 5x^{3} – 7x^{2} + 2x + 2 *by* **x**^{2} + 3x + 1

^{4}+ 5x

^{3}– 7x

^{2}+ 2x + 2

^{2}+ 3x + 1

We get

Since the remainder is zero, therefore, x^{2} + 3x + 1 is a factor of

3x^{4} + 5x^{3} – 7x^{2} + 2x + 2

(iii) Let us divide **x**^{5} – 4x^{3} + x^{2}+ 3x + 1 by **x**^{3}-3x + 1.

^{5}– 4x

^{3}+ x

^{2}+ 3x + 1

^{3}-3x + 1

we get

Here, remainder is 2_{} Therefore, x^{3}-3x + 1 is

not a factor of x^{5} – 4x^{3} + x^{2}+ 3x + 1.

**3.Obtain all the zeroes of 3x ^{4} + 6x^{3} – 2x^{2} – 10x – 5,**

=> (3x^{2}-5) is a factors of the given polynomial.

Applying the division algorithm to the given polynomial and 3x^{2}-5,

we have

### 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)*.

**Sol**. Since on dividing x^{3} – 3x^{2} + x + 2 by a polynomial *g(x)*, the quotient and remainder were (x – 2) and (- 2x +4)respectively, therefore,

=> Quotient × Divisor + Remainder = Dividend

= >(x-2) x *g(x)* + (-2x+4) = x^{3} – 3x^{2} + x + 2

= > ( – 2) x *g(x)* = x^{3} – 3x^{2} + x + 2+2x-4

### 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

**Sol.** There can be several examples for each of *(i), (ii) *and *(iii).*

However, one example for each case may be taken as under:

(i) *p(x) *= 2x^{2} – 2x + 14, *g(x)* = 2,

*g(x)* =x^{2}-x+7,*r(x)* = 0*(ii) p(x*)=X^{3}+x^{2}+x+1,*g(x) *= X^{2}– 1,

* q(x)* = x+1, *r(x)* = 2x + 2*(iii) p(r)* = x^{3} + 2x^{2} – x + 2,

*g(x) *=x^{2} -1 *q(x)*= x+ 2, *r(x)* = 4