**Question 1:Â ****Find the cubic polynomial whose zeros are 3, Â½, -1. Â Â **

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

**Solution:**

f(x) = 4x^{3} â€“ 8x^{2} + 8x + 1

q(x) = (2x â€“ 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} â€“ 8x^{2} + 8x + 1) = (2x â€“ 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:Â ****Divide 2x ^{2} + x â€“ 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 â€“ 5 and remainder = 0

\(\Rightarrow\)

= (2x â€“ 5)(x + 3) + 0

= 2x^{2} + 6x â€“ 5x â€“ 15

= 2x^{2} + x â€“ 15 = dividend

Thus, (Quotient x divisor) + remainder = dividend

**Question 4:Â ****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 â€“ 20x + 12 = 0

= -5x^{2} â€“ 17x + 12 = dividend

Thus, (Quotient x divisor) + remainder = dividend

Hence, the division algorithm is verified.

**Question 5:Â ****Divide 3x ^{3} â€“ 4x^{2 }+ 7x â€“ 2 by x^{2} â€“ 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} â€“ x + 2)

Therefore, Quotient = (3x â€“ 1), Remainder = 0

\(\Rightarrow\)

= (3x â€“ 1)(x^{2} â€“ x + 2) + 0

= 3x^{3} â€“ 3x^{2} + 6x â€“ x^{2} + x â€“ 2 = 0

= 3x^{3} â€“ 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 â€“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 â€“ 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:Â ****It being given that 2 is a zero of the polynomial x ^{3} â€“ 4x^{2} + x + 6, find its other zeros.**

**Solution:**

2 is the zero of the polynomial x^{3} â€“ 4x^{2} + x + 6

Therefore, x â€“ 2 is a factor of it

Dividing the given polynomial by x â€“ 2

Remainder = 0, Quotient of q(x) = x^{2} â€“ 2x â€“ 3

Now, q(x) = x^{2} â€“ 2x â€“ 3

= x^{2} â€“ 3x + x â€“ 3 = x(x â€“ 3) + (x â€“ 3)

= (x â€“ 3)(x + 1)

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

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

\(\Rightarrow\)

Therefore, Either x Â = 3 or x = -1

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

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

**Solution:**

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

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

Dividing x^{3} + 2x^{2} â€“ 11x â€“ 12 by x + 1

Quotient q(x) = x^{2} + x â€“ 12

= x^{2} + 4x â€“ 3x â€“ 12

= (x + 4)(x â€“ 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:Â ****If 1 and -2 are two zeros of the polynomial (x ^{3} â€“ 4x^{2} â€“ 7x + 10), find its third zero.**

**Solution:**

1 and -2 are the two zeros of the polynomial

x^{3} â€“ 4x^{2} â€“ 7x + 10

Dividing the polynomial by x â€“ 1

Quotient q(x) = x^{2} â€“ 3x â€“ 10

2 is a zero of given polynomial so it is a zero of x^{2} â€“ 3x â€“ 10

Dividing x^{2} â€“ 3x â€“ 10 by x + 2

Third zero of the given polynomial is given by x â€“ 5 = 0 or x = 5

**Question 10:Â ****If 3 and -3 are two zeros of the polynomial (x ^{4} + x^{3} â€“ 11x^{2} â€“ 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} â€“ 11x^{2} â€“ 9x + 18

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

Therefore, Dividing the given polynomial by x^{2} â€“ 9

Therefore, Quotient q(x) = x^{2} + x â€“ 2

Other zeros of given polynomial are -2 and 1

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

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

**Solution:**

2 and -2 are two polynomials

x^{4} + x^{3} â€“ 34x^{2} â€“ 4x + 120

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

Dividing x^{4} + x^{3} â€“ 34x^{2} â€“ 4x + 120 by x^{2} â€“ 4

Therefore, Quotient q(x) = x^{2} + x â€“ 30

= x^{2} + 6x â€“ 5x â€“ 30

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

= (x + 6)(x â€“ 5)

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

(x + 6)(x â€“ 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:Â ****Find all the zeros of (x ^{4} + x^{3} â€“ 23x^{2} â€“ 3x + 60), if it is given that two of its zeros are \(\sqrt{3}\)**

**Solution:**

\(\sqrt{3}\)

x^{4} + x^{3} â€“ 23x^{2} â€“ 3x + 60

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

Dividing x^{4} + x^{3} â€“ 23x^{2} â€“ 3x + 60 by x^{2} â€“ 3

Quotient q(x) = x^{2} + x â€“ 20 = x^{2} + 5x â€“ 4x â€“ 20

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

= (x + 5)(x â€“ 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:Â ****Find all the zeros of (2x ^{4} â€“ 3x^{3} â€“ 5x^{2} + 9x â€“ 3), it Being given that two of its zeros are \(\sqrt{3}\)**

**Solution:**

\(\sqrt{3}\)^{4} â€“ 3x^{3} â€“ 5x^{2} + 9x â€“ 3

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

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

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

= 2x(x â€“ 1) â€“ (x â€“ 1) = (x â€“ 1)(2x â€“ 1)

Other zeros of given polynomial are given by

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

Therefore, x = 1, Â½

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

**Question 14:Â ****If two zeros of the polynomial (2x ^{4} â€“ 11x^{3} + 7x^{2} + 13x â€“ 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} â€“ (sum of zeros)x + (product of zeros) = x^{2} â€“ 6x + 7

Dividing p(x) by x^{2} â€“ 6x + 7

Quotient = 2x^{2} + x â€“ 1

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

Therefore, q(x) = 2x^{2} + x â€“ 1 = 2x^{2} + 2x â€“ x â€“ 1

= 2x(x + 1) â€“ (x + 1) = (x + 1)(2x â€“ 1)

q(x) = 0

\(\Rightarrow\)

\(\Rightarrow\)

\(\Rightarrow\)

Therefore, The zeros of given polynomial p(x) are Â½, -1, (3 + \(\sqrt{2}\)

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

**Solution:**

\(\sqrt{5}\)

x^{4} + 4x^{3} â€“ 2x^{2} â€“ 20x â€“ 15

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

Dividing x^{4} + 4x^{3} â€“ 2x^{2} â€“ 20x â€“ 15 by x^{2} â€“ 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 \(\Rightarrow x = \left ( -3 – \sqrt{2} \right ) , \left ( -3 + \sqrt{2} \right ), \frac{1}{2} and -1\)