**Q1. Which of the accompanying articulations are polynomials in a single variable and which are not? State the basis behind your responses.**

** (i) x**^{2}-5x+6

**(ii) y**^{2}

**(iii) 9 +t**

**(iv) y**

**(v) 5x**^{10}+ 10y^{3}+ 15t^{50}

**Solution: **

(i) x^{2}-5x+6 \( \rightarrow \) Polynomial in single variable x

(ii) y2 \( \rightarrow \) Polynomial in single variable y

(iii) 9 +t\( \rightarrow \) Not a polynomial since the power of the variable in the first term is not a whole number.

(iv) y \( \rightarrow \) Not a polynomial since the power of the variable in the second term is – 1, which is not a whole number.

(v) 5x^{10}+ 10y^{3}+ 15t^{50} \( \rightarrow \) Not a polynomial in one variable since there are three variables: x, y, t.

**Q2. Compose the coefficients of x**^{2} in every one of the accompanying

**(i) 2 + 3x**^{2} + x

**(ii) 4 –2x**^{2} +2 x^{3}

**(iii) 8x**^{2} +8x

**(iv)** \( \sqrt{11} \)** x – 5**

**Solution:**

- Coefficient of x
^{2} = 3
- Coefficient of x
^{2} = −2
- Coefficient of x
^{2} = 8
- Coefficient of x
^{2} = 0 , since there is no term of x^{2}

Q3. Give one example of a binomial of the degree 35 and a monomial of the degree 100.

**Solution:**

(i) Binomial of degree 35 can be 5x^{35} - 7.

(ii) A monomial of degree 100 can be \( \sqrt{5} \) y^{100}.

**Q4. Write the degree of each of the following polynomials.**

**(i) 5x**^{3}+2x^{2} + 2x

**(ii) 4 – 3y**^{2}

**(iii) 4t – **\( \sqrt{7} \)

**(iv) 5**

**Solution:**

(i) The given polynomial is 5x^{3}+2x^{2} + 2x. The highest power of the variable x is 3.

So, the degree of the polynomial is 3.

(ii) The given polynomial is 4 - 3y^{2}. The highest power of the variable y is 2.

So, the degree of the polynomial is 2.

(iii) The given polynomial is 4t – \( \sqrt{7} \). The highest power of variable t is 1.

So, the degree of the polynomial is 1.

(iv) Since, 5 = 5x° [\( \because \) x°=1]

So, the degree of the polynomial is 0.

**Q5. Find the zero of the polynomial in each of the following cases.**

**(i) p(x)=x+5**

**(ii) p(x) = x – 5**

**(iii) p(x) = 2x + 5**

**Solution:**

(i) We have, p(x) = x + 5. Since, p(x) = 0

\( \Rightarrow \) x + 5 = 0

\( \Rightarrow \) x = -5.

Thus, zero of x + 5 is -5.

(ii) We have, p(x) = x – 5.

Since, p(x) = 0 \( \Rightarrow \) x – 5 = 0 \( \Rightarrow \) x = -5

Thus, zero of x – 5 is 5.

(iii) We have, p(x) = 2x + 5. Since, p(x) = 0

\( \Rightarrow \) 2x + 5 =0

\( \Rightarrow \) 2x = -5

\( \Rightarrow \) x = −5/2

Thus, zero of 2x + 5 is −5/2

**Q6. Determine which of the following polynomials has (x +1) a factor.**

**(i) x**^{3}+x^{2}+x +1

**(ii) x**^{4} + x^{3} + x^{2} + x + 1

**(iii) x**^{4} + 3x^{3} + 3x^{2} + x + 1

**(iv) x**^{3} – x^{2} – (2 +\( \sqrt{2} \) )x + \( \sqrt{2} \)

**Solution:**

The zero of x + 1 is -1.

(i) Let p (x) =x^{3}+x^{2}+x +1

\( \therefore \) p (-1) = (-1)^{3} + (-1)^{2} + (-1) + 1.

= -1 + 1 – 1 + 1

\( \Rightarrow \) p (- 1) = 0

So, (x+ 1) is a factor of x^{3} + x^{2} + x + 1.

(ii) Let p (x) = x^{4} + x^{3} + x^{2} + x + 1

\( \therefore \) P(-1) = (-1)^{4} + (-1)^{3} + (-1)^{2} + (-1)+1

= 1 – 1 + 1 – 1 + 1

\( \Rightarrow \) P (-1) \( \neq \) 1

So, (x + 1) is not a factor of x^{4} + x^{3} + x^{2} + x + 1.

(iii) Let p (x) = x4 + 3x3 + 3x2 + x + 1.

∴ p (-1)= (-1)^{4} + 3 (-1)^{3} + 3 (-1)^{2} + (- 1) + 1

= 1 – 3 + 3 – 1 + 1 = 1

\( \Rightarrow \)p (-1) \( \neq \) 0

So, (x + 1) is not a factor of x^{4} + 3x^{3} + 3x^{2} + x + 1.

(iv) Let p (x) = x^{3} – x^{2} – (2 + \( \sqrt{2} \)) x + \( \sqrt{2} \)

\( \therefore \) p (- 1) =(- 1)^{3}- (-1)^{2} – (2 + \( \sqrt{2} \))(-1) + \( \sqrt{2} \)

= -1 – 1 + 2 + \( \sqrt{2} \) + \( \sqrt{2} \)

= 2\( \sqrt{2} \)

\( \Rightarrow \) p (-1) \( \neq \) 0

So, (x + 1) is not a factor of x^{3} – x^{2} – (2 + \( \sqrt{2} \)) x + ^{√2}.

**Q7. Factorise the following**

**(i) 16x**^{2} – 10x +1

**(ii) 2x**^{2} + 9x + 4

**(iii) 6x**^{2} -9x – 6

**(iv) 5x**^{2} – x – 6

Solution:

(i) We have,

16x^{2} – 10x + 1 = 16x^{2} – 8x- 2x + 1

= 8x (2x – 1 ) -1 (2x – 1)

= (8x -1) (2x -1)

Thus, 16x^{2} – 10x + 1 = (8x -1) (2x -1)

(ii) We have, 2x^{2} + 9x + 4 = 2x^{2} + x + 8x + 4

= x(2x + 1) + 4(2x + 1)

= (2x + 1)(x + 4)

Thus, 2x^{2} + 9x + 4 = (2x + 1)(x + 4)

(iii) We have, 6x^{2} -9x – 6 = 6x^{2} -12x +3x – 6

= 6x(x - 2) – 3(x -2)

= (6x-3)(x -2)

Thus, 6x^{2} -9x – 6 = (6x-3)(x -2)

(iv) We have, 5x^{2} – x – 6 = 5x^{2} +5x -6x – 6

= 5x(x +1) -6(x +1) = (5x – 6)(x + 1)

Thus, 5x^{2} – x – 6 = (5x – 6)(x + 1)

**Q8. Evaluate the following products without multiplying directly**

**(i) 103 \( \times \)107**

**(ii) 95 \( \times \) 96**

**(iii) 104 \( \times \) 96**

Solution:

(i) Using (x + a)(x + b) = x^{2} + (a + b)x + ab, we have

103 x 107 = (100 + 3) (100 + 7)

= ( 100)2 + (3 + 7) (100)+ (3 \( \times \) 7)

= 10000 + (10) \( \times \) 100 + 21

= 10000 + 1000 + 21=11021

(ii) We have, 95 \( \times \) 96 = (100 – 5) (100 – 4)

= ( 100)2 + [(- 5) + (- 4)] 100 + (- 5 \( \times \) – 4)

[Using (x + a)(x + b) = x^{2 }+ (a + b)x + ab]

= 10000 + (-9) + 20 = 9120

= 10000 + (-900) + 20 = 9120

(iii) We have 104 \( \times \) 96 = (100 + 4) (100 – 4)

= (100)^{2}- 42

[Using (a + b)(a -b) = a^{2}– b^{2}]

= 10000 – 16 = 9984

**Q9. Factorise**

**(i) x**^{3} – 2x^{2} – x + 2

**(ii) x**^{3} – 3x^{2} – 9x – 5

**(iii) x**^{3} + 13x^{2} + 32x + 20

**(iv) 2y**^{3} + y^{2} – 2y – 1

**Solution:**

(i) We have, x^{3} – 2x^{2} – x + 2

Rearranging the terms, we have x^{3} – x – 2x^{2} + 2

= x(x^{2} – 1) – 2(x^{2} -1) = (x^{2} – 1)(x – 2)

= [(x)^{2} – (1)2](x – 2)

= (x – 1)(x + 1)(x – 2)

[\( \because \) (a^{2} – b^{2}) = (a + b)(a-b)]

Thus, x^{3} – 2x^{2} – x + 2 = (x – 1)(x + 1)(x – 2)

(ii) We have, x^{3} – 3x^{2} – 9x – 5

= x^{3} + x^{2} – 4x^{2} – 4x – 5x – 5 ,

= x^{2} (x + 1) – 4x(x + 1) – 5(x + 1)

= (x + 1)(x^{2} – 4x – 5)

= (x + 1)(x^{2} – 5x + x – 5)

= (x + 1)[x(x – 5) + 1(x – 5)]

= (x + 1)(x – 5)(x + 1)

Thus, x^{3} – 3x^{2 }– 9x – 5 = (x + 1)(x – 5)(x +1)

(iii) We have, x^{3} + 13x^{2} + 32x + 20

= x^{3} + x^{2} + 12x^{2 }+ 12x + 20x + 20

= x^{2}(x + 1) + 12x(x +1) + 20(x + 1)

= (x + 1)(x^{2} + 12x + 20)

= (x + 1)(x^{2} + 2x + 10x + 20)

= (x + 1)[x(x + 2) + 10(x + 2)]

= (x + 1)(x + 2)(x + 10)

Thus, x^{3} + 13x^{2} + 32x + 20

= (x + 1)(x + 2)(x + 10)

(iv) We have, 2y3 + y2 – 2y – 1

= 2y^{3} – 2y^{2} + 3y^{2} – 3y + y – 1

= 2y^{2}(y – 1) + 3y(y – 1) + 1(y – 1)

= (y – 1)(2y^{2} + 3y + 1)

= (y – 1)(2y^{2} + 2y + y + 1)

= (y – 1)[2y(y + 1) + 1(y + 1)]

= (y – 1)(y + 1)(2y + 1)

Thus, 2y^{3} + y^{2} – 2y – 1

= (y – 1)(y + 1)(2y +1)

**Q10. Evaluate the following using suitable identities**

**(i) (97)**^{3}

**(ii) (105)**^{3}

**(iii) (992)**^{3}

Solution:

(i) We have, 97 = (100 - 3)

\( \therefore \) 97^{3} = (100 – 3)^{3}

= (100)^{3} – (3)^{3} – 3(100)(3)(100 -3)

[Using (a – b)^{3} = a^{3} – b^{3} – 3ab (a – b)]

= 1000000 – 27 – 900(100 – 3)

= 1000000 - 27 – 90000 + 270

= 1000270 – 90027 = 910243

(ii) We have, 105 =100 + 5

\( \therefore \) 105^{3} = (100 + 5)^{3}

= (100)3 + (5)^{3} + 3(100)(5)(100 + 5)

[Using (a + b)^{3} = a^{3} + b^{3} + 3ab (a + b)]

= 1000000 + 125 + 1500(100 + 5)

= 1000000 + 125 + 150000 + 7500 = 1157620

(iii) We have, 992 = 1000 – 8

\( \therefore \)(998)^{3} = (1000-8)^{3}

= (1000)^{3}– (8)^{3} – 3(1000)(8)(1000 – 8)

[Using (a – b)^{3} = a^{3} – b^{3} – 3ab (a – b)]

= 1000000000 – 512 – 24000(1000 – 8)

= 1000000000 – 512 – 24000000 +192000

= 976191488