**Question 1: ****Find the cubic polynomial whose zeros are 3, ½, -1. **

**Solution:**

Required polynomial = \((x-3)\left ( x-\frac{1}{2} \right )\left [ (x-(-1)) \right ]=(x-3)\left ( x-\frac{1}{2} \right )(x+1)\)

= \((x-3)(x+1)\left ( x-\frac{1}{2} \right )=(x^{2}-2x-3)\left ( x-\frac{1}{2} \right )\)

= \(\frac{2x^{3}-4x^{2}-6x-x^{2}+2x+3}{2}\)

p(x) = \(\frac{2x^{3}-4x^{2}-6x-x^{2}+2x+3}{2}\)

p(x) = \(\frac{1}{2}(2x^{3}-5x^{2}-4x+3)\)

Hence, required polynomial is \((2x^{3}-5x^{2}-4x+3)\)

**Question 2: ****On dividing 4x ^{3} – 8x^{2} + 8x + 1 by a polynomial g(x), the quotient and the remainder are (2x – 1) and (x + 3) respectively. Find g(x).**

**Solution:**

f(x) = 4x^{3} – 8x^{2} + 8x + 1

q(x) = (2x – 1)

r(x) = (x + 3)

By division algorithm, we have

Dividend = (Quotient x Divisor) + Remainder

f(x) = q(x) x g(x) + r(x)

(4x^{3} – 8x^{2} + 8x + 1) = (2x – 1) x g(x) + (x + 3)

g(x) = \(\frac{(4x^{3}-8x^{2}+8x+1)-(x+3)}{(2x-1)}\)

= \(\frac{4x^{3}-8x^{2}+7x-2}{2x-1}=2x^{2}-3x+2\)

**Question 3: ****Divide 2x ^{2} + x – 15 by x + 3 and verify the division algorithm.**

**Solution:**

The terms of dividend and divisor are in decreasing order

Clearly degree (of remainder) = 0 < degree (x + 3)

Therefore, Quotient = 2x – 5 and remainder = 0

\(\Rightarrow\)

= (2x – 5)(x + 3) + 0

= 2x^{2} + 6x – 5x – 15

= 2x^{2} + x – 15 = dividend

Thus, (Quotient x divisor) + remainder = dividend

**Question 4: ****Divide -5x ^{2} – 17x + 12 by -5x + 3 and verify the division algorithm.**

**Solution:**

First, we write the terms of dividend and divisor in decreasing order of their degree and then perform the division as shown below.

Clearly degree (of remainder) = 0 < degree (-5x + 3)

Therefore, Quotient = x + 4 and remainder = 0

\(\Rightarrow\)

= (x + 4)(-5x + 3) + 0

= -5x^{2} + 3x – 20x + 12 = 0

= -5x^{2} – 17x + 12 = dividend

Thus, (Quotient x divisor) + remainder = dividend

Hence, the division algorithm is verified.

**Question 5: ****Divide 3x ^{3} – 4x^{2 }+ 7x – 2 by x^{2} – x + 2 and verify the division algorithm.**

**Solution:**

First, we write the given polynomials in standard form in decreasing order of degree and then perform the division as shown below.

Clearly degree (of remainder) = 0 < degree (x^{2} – x + 2)

Therefore, Quotient = (3x – 1), Remainder = 0

\(\Rightarrow\)

= (3x – 1)(x^{2} – x + 2) + 0

= 3x^{3} – 3x^{2} + 6x – x^{2} + x – 2 = 0

= 3x^{3} – 4x^{2} + 7x -2 = dividend

Thus, (Quotient x divisor) + remainder = dividend

Hence, the division algorithm is verified.

**Question 6:**

**Divide -6x ^{3} + x^{2} + 19x + 6 by –3x^{2} + 5x + 2 and verify the division algorithm. **

**Solution:**

First, we write the given polynomials in standard form in decreasing order of degree and then perform the division as shown below.

Clearly degree (of remainder) = 0 < degree (-3x^{2} + 5x + 2)

Therefore, Quotient = (2x + 3), Remainder = 0

\(\Rightarrow\)

= (2x + 3)(-3x^{2} + 5x + 2) + 0

= -6x^{3} + 10x^{2} + 4x – 9x^{2} + 15x + 6

Therefore, -6x^{3} + x^{2} + 19x + 6 = dividend

Thus, (Quotient x divisor) + remainder = dividend

Hence, the division algorithm is verified.

**Question 7: ****It being given that 2 is a zero of the polynomial x ^{3} – 4x^{2} + x + 6, find its other zeros.**

**Solution:**

2 is the zero of the polynomial x^{3} – 4x^{2} + x + 6

Therefore, x – 2 is a factor of it

Dividing the given polynomial by x – 2

Remainder = 0, Quotient of q(x) = x^{2} – 2x – 3

Now, q(x) = x^{2} – 2x – 3

= x^{2} – 3x + x – 3 = x(x – 3) + (x – 3)

= (x – 3)(x + 1)

Other zeros of given cubic polynomial are zeros of q(x)

Therefore, q(x) = 0 \(\Rightarrow\)

\(\Rightarrow\)

Therefore, Either x = 3 or x = -1

Therefore, Other zeros of the given cubic polynomial are 3 and -1

**Question 8: ****It being given that -1 is a zero of the polynomial x ^{3} + 2x^{2} – 11x – 12, find its other zeros.**

**Solution:**

One zero of the polynomial x^{3} + 2x^{2} – 11x – 12 is -1

Therefore, x + 1 is a factor of x^{3} + 2x^{2} – 11x – 12

Dividing x^{3} + 2x^{2} – 11x – 12 by x + 1

Quotient q(x) = x^{2} + x – 12

= x^{2} + 4x – 3x – 12

= (x + 4)(x – 3)

Other zeros of given polynomial are the zeros of q(x)

Therefore, q(x) = 0

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

Therefore, -4, 3 are the zeros of q(x)

Therefore, The zeros of given polynomial are -4, -1 and 3

**Question 9: ****If 1 and -2 are two zeros of the polynomial (x ^{3} – 4x^{2} – 7x + 10), find its third zero.**

**Solution:**

1 and -2 are the two zeros of the polynomial

x^{3} – 4x^{2} – 7x + 10

Dividing the polynomial by x – 1

Quotient q(x) = x^{2} – 3x – 10

2 is a zero of given polynomial so it is a zero of x^{2} – 3x – 10

Dividing x^{2} – 3x – 10 by x + 2

Third zero of the given polynomial is given by x – 5 = 0 or x = 5

**Question 10: ****If 3 and -3 are two zeros of the polynomial (x ^{4} + x^{3} – 11x^{2} – 9x + 18), find all the zeros of the given polynomial.**

**Solution:**

3 and -3 are the two zeros of the polynomial x^{4} + x^{3} – 11x^{2} – 9x + 18

Therefore, (x – 3)(x + 3) = x^{2} – 9 is a factor of given polynomial

Therefore, Dividing the given polynomial by x^{2} – 9

Therefore, Quotient q(x) = x^{2} + x – 2

Other zeros of given polynomial are -2 and 1

So zeros of given polynomial are 1, -2,3 and -3

**Question 11: ****If 2 and -2 are two zeros of the polynomial (x ^{4} + x^{3} – 34x^{2} – 4x + 120), find all the zeros of the given polynomial.**

**Solution:**

2 and -2 are two polynomials

x^{4} + x^{3} – 34x^{2} – 4x + 120

Therefore, (x – 2)(x + 2) = x^{2} – 4 will divide the given polynomial completely.

Dividing x^{4} + x^{3} – 34x^{2} – 4x + 120 by x^{2} – 4

Therefore, Quotient q(x) = x^{2} + x – 30

= x^{2} + 6x – 5x – 30

= x(x + 6) -5(x + 6)

= (x + 6)(x – 5)

For finding zeros of q(x), q(x) = 0

(x + 6)(x – 5) = 0, x = -6 or 5

Other zeros of given polynomial are -6 and 5

So zeros of given polynomial are 2, -2, -6 and 5

**Question 12: ****Find all the zeros of (x ^{4} + x^{3} – 23x^{2} – 3x + 60), if it is given that two of its zeros are \(\sqrt{3}\)**

**Solution:**

\(\sqrt{3}\)

x^{4} + x^{3} – 23x^{2} – 3x + 60

Therefore, (x – \(\sqrt{3}\)^{2} – 3 will divide the given polynomial completely

Dividing x^{4} + x^{3} – 23x^{2} – 3x + 60 by x^{2} – 3

Quotient q(x) = x^{2} + x – 20 = x^{2} + 5x – 4x – 20

= x(x + 5) -4(x + 5)

= (x + 5)(x – 4)

Other zeros of the given polynomial are the zeros of q(x)

Therefore, q(x) = 0 \(\Rightarrow\)

Or x = -5 or 4

Hence the zeros of given polynomial are \(\sqrt{3}\)

**Question 13: ****Find all the zeros of (2x ^{4} – 3x^{3} – 5x^{2} + 9x – 3), it Being given that two of its zeros are \(\sqrt{3}\)**

**Solution:**

\(\sqrt{3}\)^{4} – 3x^{3} – 5x^{2} + 9x – 3

Therefore, (x – \(\sqrt{3}\)^{2} – 3 will divide the given polynomial completely

Dividing 2x^{4} – 3x^{3} – 5x^{2} + 9x – 3 by x^{2} – 3, we get:

Therefore, Quotient q(x) = 2x^{2} – 3x + 1 = 2x^{2} -2x –x + 1

= 2x(x – 1) – (x – 1) = (x – 1)(2x – 1)

Other zeros of given polynomial are given by

q(x) = 0 \(\Rightarrow\)

Therefore, x = 1, ½

Hence, zeros of given polynomial are \(\sqrt{3}\)

**Question 14: ****If two zeros of the polynomial (2x ^{4} – 11x^{3} + 7x^{2} + 13x – 7) are (3 + \(\sqrt{2}\)**

**Solution:**

Sum of 3 + \(\sqrt{2}\)

Product of (3 + \(\sqrt{2}\)

= (3 + \(\sqrt{2}\)

Polynomial whose zeros are 3 + \(\sqrt{2}\)

x^{2} – (sum of zeros)x + (product of zeros) = x^{2} – 6x + 7

Dividing p(x) by x^{2} – 6x + 7

Quotient = 2x^{2} + x – 1

Therefore, Other zeros of polynomial p(x) are also the zeros of q(x)

Therefore, q(x) = 2x^{2} + x – 1 = 2x^{2} + 2x – x – 1

= 2x(x + 1) – (x + 1) = (x + 1)(2x – 1)

q(x) = 0

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

Therefore, The zeros of given polynomial p(x) are ½, -1, (3 + \(\sqrt{2}\)

**Question 15: ****Obtain all zeros of the polynomial (x ^{4} + 4x^{3} – 2x^{2} – 20x – 15), if two of its zeros are \(\sqrt{5}\)**

**Solution:**

\(\sqrt{5}\)

x^{4} + 4x^{3} – 2x^{2} – 20x – 15

Therefore, (x – \(\sqrt{5}\)^{2} – 5 will divide the given polynomial completely.

Dividing x^{4} + 4x^{3} – 2x^{2} – 20x – 15 by x^{2} – 5, we get

Quotient = x^{2} + 4x + 3 = x^{2} + 3x + x + 3

= x(x + 3) + (x + 3) = (x + 3)(x + 1)

Other zeros of the given polynomial are the zeros of q(x)

Therefore, x = -3, -1

Thus, the zeros of the given polynomial are

\(\sqrt{5}\)

**Q.16: Find all the zeros of the polynomial \((2x^{4} – 11x^{3} + 7x^{2}+13x – 7)\), it being given that two of its zeros area \((3 + \sqrt{2}) \; and \; (3 – \sqrt{2})\).**

**Sol:**

The given polynomial is \(f(x) = 2x^{4} – 11x^{3} + 7x^{2}+13x – 7\)

Since \((3 + \sqrt{2}) \; and \; (3 – \sqrt{2})\)

It follows that each one of ( \(x + 3 + \sqrt{2}\)

Consequently , \(\left [ x – (3 + \sqrt{2} ) \right ] \left [ x – (3 – \sqrt{2} ) \right ] = \left [ (x- 3) – \sqrt{2} \right ] \left [ (x- 3) + \sqrt{2} \right ]\)

= \(\left [ (x – 3)^{2} – 2 ) \right ] = x^{2} – 6x + 7\)

On dividing f(x) by (\(x^{2} – 6x + 7\)

\(Therefore, f(x) = 0\)

\(\Rightarrow 2x^{4} – 11x^{3} + 7x^{2}+13x – 7 = 0 \\ \Rightarrow (x^{2} – 6x + 7) ( 2x^{2} + x – 1) = 0 \\ \Rightarrow (x + 3 + \sqrt{2}) (x+ 3 – \sqrt{2})(2x -1)( x + 1) =0\)

\(\Rightarrow x = -3 – \sqrt{2} \; or \; x = – 3 + \sqrt{2} \; or \; x = \frac{1}{2} \; or \; x = -1\)

Hence, all the zeros are \(\Rightarrow x = \left ( -3 – \sqrt{2} \right ) , \left ( -3 + \sqrt{2} \right ), \frac{1}{2} and -1\)