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If (x ) is a polynomial in the highest power of (x)is called the degree of the polynomial (x).

- polynomial of degree 1 is called a linear polynomial . For example,4x-5 etc.
- A polynomial of degree 2 is called a quadratic polynomial. For example,x
^{2}+2x+3 etc. - A polynomial of degree 3 is called a cubic polynomial and so on. For example, x
^{3}+2x^{2}+x+5 etc.

**Relationship between Zeroes and Coefficients of a Polynomial**

**For Quadratic equation :**If α and β are zeroes of the quadratic equation ax^{2}+bx +c,a≠0, then α + β =-b/a α β=c/a**For cubic equation :**If α, β and γ are zeroes of the cubic equation ax^{3 }+bx^{2}+cx+d, a≠0 then α + β + γ =-b/a αβγ==d/aF.

**Zero of a polynomial :** if the value of p(x) at x=k is zero then K is called a zero of the polynomial p(x ).

**Graphical representation of zero of a polynomial**

- The zeroes of a polynomial p(x) are precisely the x-coordinates of the points, where the graph of y=p(x) intersects the x-axis.
- A quadratic polynomial can have at most 2 zeroes and a cubic polynomial can have at most 3 zeroes.
- For any quadratic polynomial ax
^{2}+bx +c,a≠0 its graph gas one of the two shapes either open upwards or open downwards or open downwards depending on whether a>0 or a<0. F

**Division Algorithm for Polynomials**

If p(x) and g(x) are any 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) or degree of r(x) <degree of g(x).This result is known as the Division Algorithm for polynomials.

**Question. Can x-2 be the remainder on division of a polynomial p(x) by x +3?****Sol.** No, as degree (x -2) = degree (x +3)

**Question. Find all the zeros of polynomial x ^{4}+x^{3}-9x^{2}-3x+18 if the two of its zeros are √3 and -√3.**

**Solution.**Other Zeroes 2 and -3

**Question. If α and β are the zeros of the quadratic polynomial x ^{2} -6x +a find the value of ‘a’ If 3α +2β=20**

**Solution.**a= -16

**Question. If the number of apples and mangoes are the zeroes of polynomial 3x ^{2}-8x-2k+1. and the number of apples are 7 times the number of mangoes, then find the number of zeroes and value of k.**

**Solution.**2 zeroes, K= -2/3

**Question. If α and β are the zeroes of a polynomial x ^{2}-x-30, then form a quadratic Polynomial whose zeroes are 2-α and 2- β**

**Solution.**x

^{2}-3x-28

**Question. On dividing 3x ^{3}-2x^{2}+5x-5 by the polynomial p(x), the quotient and remainder are x2-x+2 and -7 respectively. Find p(x).**

**Solution.**3x-1

**Question. Check whether x ^{2} +3x +1 is a factor of 3x^{4}+5x^{3} -7×2+2x+2?**

**Solution.**x

^{2}-2√3x+2

**Question. If α and β are the zeros of quadratic polynomial p(x) =x ^{2}-(k-6) x+ (2k+1) Find the value of k if α+β=αβ.**

**Solution.**K= -7

**Question. If one zero of the quadratic polynomial f(x) =4x ^{2}-8kx-9 is negative of the other, find the value of k.**

**Solution.**k= 0

**Question. On dividing 3x ^{3}+4x^{2}+5x-13by a polynomial g(x) ,the quotient and remainder are 3x+10and 16x-43 respectively. Find the polynomial g(x).**

**Solution.**g(x) =x

^{2}-2x+3

**Question. Find the value of k for which the quadratic polynomial 9x ^{2}-3kx+k has equal zeros.**

**Solution.**k= 0 or k=4

**Question. If the zeros of polynomial x ^{2}-5x+k are the reciprocal of the zeros then find the value of k.**

**Solution.**K=1

**Question. If the sum of squares of zeros of the polynomial 6x ^{2}+x+k is 25/36 find the value of k.**

**Solution.**K= -7

**Question**. Find a cubic polynomial with the sum of the zeroes, sum of the products of its zeroes taken two at a time, and the product of its zeroes as 2, – 7, -14 respectively.**Sol.** Let the cubic polynomial be p(x) = ax^{3} + bx^{2} + cx + d. Then

**Question. If the graph of a polynomial intersects the x-axis at only one point, it cannot be a quadratic polynomial.****Sol.** False, because every quadratic polynomial has at most two zeroes.**Question. Can a quadratic polynomial x ^{2} + kx + k have equal zeroes for some odd integer k >1?**

**Sol.**No, for equal zeroes, k = 0,4

⇒ k is even

Are the following statements ‘True’ or ‘False’? Justify your answer.

**Question**. If a and b are the zeroes of the quadratic polynomial f (x) = 2x^{2} – 5x + 7, find a polynomial whose zeroes are 2a + 3b and 3a + 2b.**Sol.** Since a and b are the ze roes of the quad ratic poly no mial f (x) = 2x^{2} – 5x + 7.

**Question. If the zeroes of a quadratic polynomial ax ^{2} + bx + c are both negative, then a, b and c all have the same sign.**

**Sol.**True, because – b/a

= sum of zeroes < 0, so that b/a > 0. Also the product of the zeroes = c/a > 0.

**Question**. Find the zeroes of the polynomial f (x) = x 3 – 5x^{2} – 2x + 24, if it is given that the product of its two zeroes is 12.**Sol.** Let a, b and g be the zer oes of polyn om ial f (x) such that ab =12.

**Question. What will the quotient and remainder be on division of ax ^{2} + bx + cby px^{3} + qx^{2} + rx +5, p ≠ 0?**

**Sol.**0, ax

^{2}+ bx + c

**1. Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:****(i) x ^{2} + 3x +1, 3x^{4} + 5x^{3} – 7x^{2} + 2x + 2 (ii) t^{2} – 3, 2t^{4} + 3t^{3} – 2t^{2} – 9t -12 **

**Sol.**

**Question. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.****(i) 6x ^{2} – 3 – 7x (ii) 4u^{2} + 8u (iii) 4s^{2} – 4s +1 **

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

p (x) = 6x

^{2}– 3 – 7x

⇒ p (x) = 6x

^{2}– 7x – 3 (In general form)

= 6x

^{2}– 9x + 2x – 3

= 3x (2x – 3) +1 (2x – 3) = (2x – 3) (3x +1)

The zeroes of polynomial p (x) is given by

p (x) = 0

⇒ (2x – 3) (3x +1) = 0 ⇒ x = 3/2 -1/3,

Thus, the zeroes of 6x

^{2}– 7x – 3 are α = 3/2 and β = – 1/3

Therefore, product of zeroes = Constant term/Coefficient of x^{2}

(ii) We have,

p (u) = 4u^{2} + 8u ⇒ p(u) = 4u (u + 2)

The zeroes of polynomial p (u) is given by

p (u) = 0 ⇒ 4u (u + 2) = 0

∴ u = 0, – 2

Thus, the zeroes of 4u^{2} + 8u are a = 0 and b = – 2

Now, sum of the zeroes = a + b = 0 – 2 = – 2

and (Coefficient of u )/ Coefficient of u 2=-8/4= -2

Therefore, sum of the zeroes = -(Coefficient of u) /(Coefficient of u2)

Again, product of the zeroes = αβ = 0 x (- 2) = 0 and Constant term/Coefficient of u2 =0/4=0

Therefore, product of zeroes = Constant term/Coefficient of u2

(iii) We have,

p (s) = 4s^{2} – 4s +1

⇒ p (s) = 4s^{2} – 2s – 2s +1 = 2s (2s -1) -1 (2s -1) = (2s -1) (2s -1)

The zeroes of polynomial p (s) is given by

p (s) = 0

⇒ (2s -1) (2s -1) = 0

⇒ s = 1/2-1/2,

Thus, the zeroes of 4s^{2} – 4s +1 are

and Constant term/Coefficient of s ^{2} = 1/4

∴ Product of zeroes = Constant term/Coefficient of s ^{2}

**Question. If the graph of a polynomial intersects the x-axis at exactly two points, it need not be a quadratic polynomial.****Sol.** True, x^{4} -1 is a polynomial intersecting the x-axis at exactly two points

**Question**.** Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.(i) – 1/4 , 1/4**

**(ii) √2,1/3**

**Sol.**

**Question. Verify that the numbers given alongside the cubic polynomial below are their zeroes. Also verify the relationship between the zeroes and the coefficients.**** x ^{3} – 4x 2 + 5x – 2; 2,1,1**

**Sol.**Let p(x) = x

^{3}– 4x

^{2}+ 5x – 2

On comparing with general polynomial p(x) = ax

^{3}+ bx

^{2}+ cx + d, we get a =1, b = – 4 , c = 5 and d = – 2.

Given zeroes 2, 1, 1.

∴ p(2) = (2)

^{3}– 4(2)

^{2}+ 5(2) – 2 = 8 -16 +10 – 2 = 0

and p (1) = (1)

^{3}– 4 (1)

^{2}+ 5(1) – 2 =1 – 4 + 5 – 2 = 0.

Hence, 2,1 and 1 are the zeroes of the given cubic polynomial.

Again, consider a = 2, b =1, g =1

∴ α + β + γ = 2 +1 +1 = 4

**Question. Determine the degree of the polynomial (x + 1) (x ^{2} – x – x^{4} + 1).**

**Solution.**Given polynomial in standard form is :

–x

^{5}– x

^{4}+ x

^{3}+ 1

So, its degree is 5

**Question. If the product of two zeros of the polynomial****p(x) = 2x ^{3} + 6x^{2} – 4x + 9 is 3, find the third zero of the polynomial.**

**Solution.**If α, β, and γ be the three zeros of p(x). Then,

αβγ = – 9/ 2

Since, αβ = 3,

we get γ = − × = − 9/ 2 × 1/3 = -3/2

Thus, the third zero of p(x) is − 3/ 2

**Question. Find the zeroes of following polynomials by factorisation method and verify relation between the zeroes and coeffcients of polynomials.****(A) 2x ^{2} + 2/ 7 x + 4 3 **

**(B) 2s**

^{2}− (1 + 2 √2)s + √2**(C) 7y**

^{2}– 11/3 y – 3/ 2**Solution.**(A) Let f(x) = 2x

^{2}+ 7/2 x + 3/4 = 8x

^{2}+ 14x + 3

(Multiplying the given equation by 4)

= 8x

^{2}+ (12x + 2x) + 3

= 8x

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

= 4x(2x + 3) + 1(2x + 3)

= (2x + 3)(4x + 1)

The zeroes of f(x) are given by f(x) = 0.

So, the value of 2x

^{2}+ 7/2 x + 3/4 is zero when

x = – 3/2 or x = –1/ 4

⇒ x = – 3/2 , – 1/4

**Question. If a and b are the zeros of the polynomial p(x) = 4x**

^{2}– 2x – 3, find the value of 1/α + 1/β .**Solution.**Here, α+ β = 2/ 4 or 1/ 2 and α.β = −3/ 4 .

So, 1/α + 1/β = α + β /αβ = 1/2/-3/4 =-2/3

**Question. If one of the zeros of polynomial p(x) =****(k − 1)x ^{2} – kx + 1 is −3, find the value of k.**

**Solution.**Since, (–3) is a zero of p(x), we have,

(k – 1)(–3)2–k(–3) + 1 = 0

⇒ 9k–9 + 3k + 1 = 0,

⇒ 12k = 8 ⇒ k = 2/ 3

**Question. If α and β be the roots of the equation x ^{2} − 1 = 0 , then show that: α + β = 1/α + 1/β**

**Solution.**Here, α + β = 0/ 1 = 0

[∵ Sum of roots = Coefficient of x / Coefficient of x

^{2}]

Also, 1/α + 1/β = α + β /αβ = 0/1 = 0

Thus, α + β = 1/α + 1/β

**Question. A teacher asked 10 of his students to write a polynomial in one variable on a paper and then to handover the paper. The following were the answer given by the students:****2x + 3, 3x ^{2} + 7x + 2, 4x^{3} + 3x^{2} + 2, x^{3} + √3x + 7, 7x + √7 , 5x^{3} – 7x + 2, 2x^{3} + 3 – 5 x , 5x – 1/2 , ax^{3} + bx^{2} + cx + d, x + 1 x .**

**Answer the following Questions:**

**(A) How many of the above ten are not polynomials?**

**(B) How many of the above ten are quadratic polynomials?**

**Solution.**(A) Three, namely: x

^{3}+ √3x + 7 , 2×2 + 3 − 5/x , x+1/x

(As they contain square roots of the variable and negative power of x)

(B) One, namely 3×2 + 7x + 2

**Question. If one of the zeroes of the quadratic polynomial f(x) = 4x ^{2} – 8kx – 9 is equal in magnitude but opposite in sign of the other, then find the value of k.**

**Solution**

**.**f(x) = 4x

^{2}– 8kx – 9

Let one of the zeroes of the polynomial be

and the other zeroes be – α

Sum of zeroes = (− b /a) = 8k/4

α + (- α) = 0

So, 8k/ 4 = 0 ⇒ k = 0

**Question. Can (x – 5) be the remainder on division of a polynomial p(x) by (x + 8)? ****Solution.** No. We know that we cannot divide the polynomials which have same degree.

As we can see that degree of (x – 5) = degree of (x + 8)

So, they are not divisible.

**Question. If the zeros of the polynomial x ^{3} – 2x^{2} + x + 1 are a – b, a and a + b, then find the values of a and b.**

**Solution.**As (a – b), a and (a + b) are zeros of x

^{3}– 3x

^{2}+ x + 1, we have:

a – b + a + a + b = 3

⇒ 3a = 3, or a = 1 …(i)

a (a – b) + a (a + b) + (a – b) (a + b) = 1

⇒ 3a

^{2}– b

^{2}= 1 …(ii)

and (a – b) a (a + b) = –1

⇒ a(a

^{2}– b

^{2}) = –1 …(iii)

From (i) and (ii), we have b = ± √2

Thus, a = 1, b = ± √2

**Question. Form a quadratic polynomial, the sum and product of whose zeroes are (–3) and 2 respectively. ****Solution.** A general form of a quadratic polynomial is

ax^{2} + bx + c

Here, α + β = − b/ a = – 3 and a.b = − b/ a = 2

where, a and b are the roots of given polynomial.

So, the required polynomial is x2 + 3x + 2.

**Question. Find the value of k for which the roots of the equation 3x ^{2} − 10x + k = 0 are reciprocal of each other. **

**Solution**. Given, equation is 3×2 – 10x + k = 0, where roots are reciprocals of each other.

Let the roots be α and 1/ α

∴ Product of roots = c/ a

⇒ α . 1/α = k /3 [∵ a = 3, b = – 10, c = k]

⇒ 1 = k /3

⇒ k = 3

**Question. What number should be added to the polynomial x ^{2} – 5x + 4 so that 3 is the zero of the polynomial?**

**Solution**. Let k be the number to be added to the given polynomial.. Then the polynomial becomes x

^{2}– 5x + (4 + k)

As 3 is the zero of the polynomial, we get:

(3)

^{2}– 5(3) + (4 + k) = 0

⇒ (4 + k) = 15 – 9

⇒ 4 + k = 6

⇒ k = 2

Thus, 2 is to be added to the polynomial

**Question. If the zeroes of a polynomial x ^{2} – 8x + k = 0, is the HCF of (6, 12), then find the value of k. **

**Solution**. HCF of (6, 12) = 6

So, 6 is one of the roots of the polynomial.

f(x) = x

^{2}– 8x + k = 0

f(6) = (6)

^{2}– 8(6) + k = 0

36 – 48 + k = 0

–12 + k = 0 ⇒ k = 12.

**Question. Find a quadratic polynomial whose zeroes are reciprocals of the zeroes of the polynomial f(x) = ax ^{2}+ bx + c, a ≠0 , c ≠ 0. **

**Solution**. Let α, β be the zeroes of f(x) = ax

^{2}+ bx + c. Thus

α + β = – b/ a and αβ = c/a

Now, 1/α + 1/β = α + β / αβ = -b/a/c/a = – b/c

1/α . 1/β = 1/αβ = 1/c/a = a/c

∴ Polynomial is: x

^{2}– (sum of roots) x

– product roots

x

^{2}– (-b/c)x + a/c = 0

⇒ cx

^{2}+ bx + a = 0

So, the required polynomial is cx

^{2}+ bx + a.

**Question. If the zeroes of the polynomial x ^{2} + px + q are double the value to the zeroes of 2x^{2} – 5x − 3, find the value of p and q. **

**Solution**. Let α and β are zeroes of the 2x

^{2}– 5x – 3

α + β = – b/a = 5/2

αβ = c/a =− 3/2

According to the question,

2α and 2β are zeroes of x

^{2}+ px + q

2α + 2β = –p ⇒ 2(α + β) = –p

2 x (5/2) = -p

= –p [from eqn. (i)]

p = –5

2α × 2β = q ⇒ 4αβ = q

4 x(- 3/2) q = –6

Hence, p = –5 and q = –6.

**Question. Find the value of k such that the polynomial x2 – (k + 6)x + 2(2k 1) has the sum of its zeros equal to half of their product. ****Solution. **Given polynomial is :

p(x) = x^{2} – (k + 6) x + 2(2k – 1)

In the given quadratic equation:

a = 1

b = – (k + 6)

c = 2(2k – 1)

Sum of zeroes = – b/ a

= k + 6 …(i)

Product of zeroes = c/a

= 2(2k – 1) …(ii)

According to the given condition:

Sum of the zeroes = 1 /2 × Product of zeroes

⇒ k + 6 = 1/ 2

×^{ 2} (2k – 1)

⇒ k + 6 = 2k – 1

⇒ 2k – k = 6 + 1

⇒ k = 7

Hence, the value of k is 7.

**Question. Find the quadratic polynomial sum and product of whose Zeroes are – 1 and – 20 espectively. also, find the zeroes of the polynomial so obtained.****Solution**. Let α and β the zeroes of the polynomial.

Given : sum of zeroes, α + β = – 1

product of zeroes, αβ = – 20

Equation of polynomial :

x^{2} – (sum of zeroes) x + product of zeroes = 0

∴ x^{2} – (– 1)x + (– 20) = 0

⇒ x^{2} + x – 20 = 0

On spliting the middle term,

x^{2} + 5x – 4x – 20 = 0

⇒ x(x + 5) – 4(x + 5) = 0

⇒ (x – 4) (x + 5) = 0

⇒ x = 4, – 5

Hence, the zeroes of the polynomial are 4 and –5.

**Question. Given that 2 is a zero of the cubic polynomial 6x ^{3} + √2 x^{2} + 10x – 4√ 2 , find its other two zeroes. **

**Solution.**Let p(x) = 6x

^{3}+ √2x

^{2}– 10x – 4 √2 As 2 is one of the zeroes of p(x).

⇒ g(x) = (x – √2) is one of the factors of p(x).

**Question. Given that x – 5 is a factor of the cubic polynomial x ^{3} – 3 √5 x + 13x – 3 √5 , find all the zeroes of the polynomial. **

**Solution.**Let p(x) = x

^{3}– 3 √5 x

^{2}+ 13x – 3 √5 As 5 is one of the zeroes of p(x).

⇒ (x – 5 ) is one of the factors of p(x).

x

^{2}– 2 5 x + 3

**Question. If one root of the equation 3x ^{2} – 8x + 2k + 1 = 0 is seven times the other, find the two roots and the value of k.**

**Solution.**Let α and 7α be the two roots of the equation:

3x

^{2}– 8x + (2k + 1) = 0

Then, α + 7α = 8α = 8 /3 …..(i)

and α.(7α) = 7α

^{2}= 2k + 1/3 …..(ii)

From (i) α = 1/ 3 . So, the two roots are 1/ 3 and 7/ 3

Using α = 1 3 in (ii), we have:

7( 1/3)2 =2k + 1/3

⇒ 2k + 1 = 7/3

⇒ 2k = 4/3

⇒ k = 2/3

**Question. Given that the zeroes of the cubic polynomial x ^{3} – 6x^{2} + 3x + 10 are of the form a, a + b, a + 2b for some real numbers a and b, find the values of a and b as well as the zeroes of the given polynomial. **

**Solution.**Let p(x) = x

^{3}– 6x

^{2}+ 3x + 10

**Question. Find a quadratic polynomial whose zeroes are 1 and -3. Verify the relation between the coefficients and Zeroes of polynomial. ****Solution.** Sum of zeroes, S = 1 + (–3) = –2 …(i)

Product of zeroes, P = 1 × (–3) = –3 …(ii)

Quadratic polynomial

p(x) = x2 – Sx + P

= x^{2} – (–2)x – 3 = x^{2} + 2x – 3

Here, a = 1, b = 2, c = –3

− b/ a = −2 /1 =- 2

Sum of zeroes = − b/ a = 2 [using eqn. (i)]

Also, c/a = -3/1 = -3

Product of zeroes = c/a = -3 [using eqn. (ii)]

**Question. For which values of a and b are the zeroes of q(x) = x ^{3} + 2x^{2} + a also the zeroes of the polynomial p(x) = x^{5} – x^{4} – 4x^{3} + 3x^{2} + 3x + b? **

**Solution.**Let p(x) = x

^{5}– x

^{4}– 4x

^{3}+ 3x

^{2}+ 3x + b and q(x) = x

^{3}+ 2x

^{2}+ a.

Since, the zeroes of the polynomial q(x) are also zeroes of p(x), we can say that q(x) is a factor of p(x).

Then, on dividing

**Question. Without actually calculating the zeroes, form a quadratic polynomial whose zeroes are reciprocals of the zeroes of the polynomial 5x ^{2} + 2x – 3.**

**Solution.**Let α and β be the zeroes of 5x

^{2}+ 2x – 3

**Question. Obtain other zeroes of the polynomial f(x) = 2x ^{4} + 3x^{3} – 5x^{2} – 9x – 3 if two of its zeroes are −√3 and − √3 **

**Solution.**Since √3 and − √3 are zeroes of f(x), (x − √3)(x + √3) i.e., (x

^{2}– √3) is a factor of f(x) to obtain other two zeroes, we shall determine the quotient, by dividing f(x) with (x

^{2}– √3)

**Question** Find the zeroes of the following polynomials and verify the relationship between the zeroes and the coefficients of the polynomials.**(i) 3x ^{2} + 4x – 4**

**(ii) 7y**

^{2}-(11/3) y – 2/3**(iii) 4x**

^{2}+5√2x -3**(iv) p**

^{2}-30**(v) √3x**

^{2}-11x +6√3**(vi) a(x**

^{2}+1) – x(a2 +1)**(vii) 6x**

^{2}+ x -2**(viii) y**

^{2}-1/2 y + 1/16**Answer**

(i) – 2, 2/3

(ii) -1/7 , 2/3

(iii) -3**√**2/2 , 2/4

(iv) **√**30 ,**– √**30

(v) 2/**√**3 , 3**√**3

(vi) a , 1/a

(vii) -2/3 , 1/2

(viii) 1/4 , 1/4

**10. If the sum of the zeroes of the quadratic polynomial f(x) = kx ^{2} +2x +3k is equal to their product, find the value of k.**

**Answer**

k = -2/3

**Question**. (i) Obtain all other zeroes of 2x^{4} +7x^{3} -19x^{2} -14x +30, if two of its zeroes are 2 and – 2.**(ii) Obtain all other zeroes of 2x ^{3} + x^{2} -6x -3, if two of its zeroes are – 3 and 3.**

**Answer**

(i) – 5, 3/2

(ii) -1/2

**Question**. Find a quadratic polynomial each with the given numbers as the sum and product of the zeroes respectively.**(iv) 21/8 , 5/16****Also find the zeroes of those polynomials by factorisation.**

**Answer**

(iv) 1/16 (16x^{2} – 42x +5); (1/8) , (5/2)

**Question**. Check whether g(x) is a factor of p(x) by dividing the first polynomial by the second polynomial:**(i) p(x) = 4x ^{3} +8x +8x^{2} +7, g(x) = 2x^{2} – x +1**

(ii) p(x) = x4 -5x +6, g(x) = 2 – x^{2}

(iii) p(x) =13x^{3} -19x^{2} +12x +14, g(x) = 2 -2x + x^{2}

**Answer**

(i) No

(ii) No

(iii) Yes

**Question**. If (x -2) is a factor of x 3 + ax 2 + bx +16 and b = 4a, find the values of a and b.

**Answer**

a = – 2,

b = – 8

**Question**. 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) = 0(iii) deg r(x) = 0**

**Answer**

(i) 2x^{2} -3= 2(x^{2} +1) -5

(ii) x^{3} +1= 0.(x^{4} ) +(x^{3} +1)

(iii) x^{2} +1=1(x^{2} -1) +2

**Question**. If a and b are the zeroes of the quadratic polynomial f(x) = 3x^{2} -5x -2, then evaluate**(i) α ^{2} +β^{2} (ii) α^{3} +β^{3} (iii) α^{2}/β+β^{2}/α**

**Answer**

(i) 37/9

(ii) 215/27

(iii) -215/18

**Question**. Find the cubic polynomial with the sum, sum of the products of its zeroes taken two at a time, and the products of its zeroes as –3, –8 and 2 respectively.

**Answer**

x^{3} +3x^{2} -8x -2

**Question**. A cubic polynomial always has degree three.

**Answer**

Yes, 4x^{3} +3x^{2} +2x +1 is a cubic polynomial

**Question**. The degree of a binomial cannot be more than two.

**Answer**

False, x^{3} +1 is a binomial of degree 3

**Question**. We can have a trinomial having degree 7.

**Answer**

Yes, x^{7} + x -1.

**Question**. There is only one term of degree one in a monomial.

**Answer**

False, 4x^{2} is a monomial of degree 2

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