**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**

**(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}**

Solution:

(i)p(x) = x^{3} – 3x^{2} + 5x – 3 ; g(x) = x^{2} – 2

Quotient = x − 3

Remainder = 7x − 9

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

Quotient = x^{2} + x − 3

Remainder = 8

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

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} – 2t^{2} – 9t – 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**

Solution:

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

t^{2} – 3 = t^{2 }+ 0.t – 3

Since the remainder is 0,

Hence, t^{2} – 3 is a factor of 2t^{4} + 3t^{3} – 2t^{2} – 9t – 12

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

Since the remainder is 0,

Hence 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

Since the remainder ≠ 0

Hence 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})**

Solution:

p(x) = 3x^{4 }+ 6x^{3} – 2x^{2} – 10x -5

Since the two zeros are √(^{5}/_{3}) and -√(^{5}/_{3})

Therefore, [x – √(^{5}/_{3})][x + √(^{5}/_{3})] = (x^{2} – ^{5}/_{3}) is a factor of 3x^{4 }+ 6x^{3} – 2x^{2} – 10x -5. Thus

we divide the given polynomial by x^{2} – ^{5}/_{3}

3x^{4 }+ 6x^{3} – 2x^{2} – 10x -5 = (x^{2} – ^{5}/_{3})(3x^{2} + 6x + 3)

= 3(x^{2} – ^{5}/_{3})(x^{2} + 2x + 1)

We factorise x^{2} + 2x + 1 = (x + 1 )^{2}

Therefore, its zero is given by x + 1 = 0 or x = −1

As it has the term (x+1)^{2} , therefore, there will be 2 zeroes at x = −1. Hence, the zeroes of the given polynomial are √(^{5}/_{3}) , – √(^{5}/_{3}), -1 and -1.

**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:

p(x) = x^{3} – 3x^{2} + x + 2 [dividend]

Quotient = (x -2)

Remainder = (-2x + 4)

Divisor = g(x)

Dividend = Divisor × Quotient + Remainder

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

x^{3} – 3x^{2} + x + 2 + 2x – 4 = g(x).(x – 2)

g(x) = [x^{3} – 3x^{2} + 3x – 2]/[x-2]

g(x) = x^{2} – x + 1

**5. Give examples of polynomial 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 **

Solution:

According to the division algorithm, if p(x) and g(x) are two polynomials with 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)

Degree of a polynomial is the highest power of the variable in the polynomial.

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

Degree of quotient will be equal to degree of dividend when divisor is constant ( i.e., when any polynomial is divided by a constant). Let us assume the division of 6x^{2} + 2x + 2 by 2.

Here, p(x) = 6x^{2} + 2x + 2

g(x) = 2 q(x) = 3x^{2} + x + 1 and r(x) = 0

Degree of p(x) and q(x) is the same i.e., 2.

Checking for division algorithm, p(x) = g(x) × q(x) + r(x)

6x^{2} + 2x + 2 = (2)* (3×2 + x + 1) + 0

Thus, the division algorithm is satisfied.

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

Let us assume the division of x^{3} + x by x^{2} ,

Here, p(x) = x^{3} + x g(x) = x^{2}

q(x) = x and r(x) = x

Clearly, the degree of q(x) and r(x) is the same i.e., 1.

Checking for division algorithm, p(x) = g(x) × q(x) + r(x)

x^{3} + x = (x^{2} ) × x + x

x^{3} + x = x^{3} + x

Thus, the division algorithm is satisfied.

(iii)deg r(x) = 0

Degree of remainder will be 0 when remainder comes to a constant.

Let us assume the division of x^{3} + 1 by x^{2} .

Here, p(x) = x^{3} + 1

g(x) = x_{2}

q(x) = x and r(x) = 1

Clearly, the degree of r(x) is 0.

Checking for division algorithm, p(x) = g(x) × q(x) + r(x) x 3 + 1 = (x^{2} ) × x + 1 x 3 + 1 = x^{3 }+ 1

Thus, the division algorithm is satisfied.

## 9 responses to “Polynomials – Exercise 2.3 – Class 10”

[…] Polynomials – Exercise 2.3 […]

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