Polynomials Chapter 2: NCERT Exemplars for Class 10

NCERT Exemplar Class 10 Math Chapter 2 Polynomials are provided here for students to prepare for the board exam. These solutions are prepared as per the NCERT guidelines and with respect to the updated syllabus of 2024-25. Simply Acad provides a platform to ace the examinations with accountability by providing all the study material at one place in pdf format for free. With the help of exemplar problems and solutions, students will be able to revise the complete chapter and score better marks in the exam.

This chapter brings a treasure with the concept of polynomials which has been discussed in detail. They will also study important topics like the relationship between coefficients and zeroes, the division algorithm for polynomials and the geometrical meaning of the zeros of a polynomial .As students need to be thorough with these topics from in depth, here is the detailed solutions for chapter 2 respectively.

 

Question 1: when one of the zeroes of the quadratic polynomial (k – 1) x2 + kx + 1 is –3, the value of k is

(A) 4/3

(B) −4/3

(C) 2/3

(D) −2/3

Answer 1: (A) 4/3

Explanation: Let α and β be the zeroes of the given polynomial and α = −3

We know that,

α+β=−b/a

−3+β=−k/(k−1) …..(1)

αβ=c/a

−3β=1/(k−1)

β=−1/3(k−1)

Put the value of β in (1), we get,

−3+−1/3(k−1)

= −kk−1kk−1+−13(k−1)

= 33k−13(k−1)

= 33k−1

= 9k−98

= 6kk

= 43

so the correct answer is option (A).

 

Question 2: The quadratic polynomial, whose zeroes are –3 and 4, is

(A) x2 – x + 12

(B) x2 + x + 12

(C) x2/2-x/2-6

(D) 2×2 + 2x – 24

Answer 2: (C) x2/2-x/2-6

Explanation: Let α and β be the zeroes of a quadratic polynomial.

Then the quadratic polynomial is k[x2−(α+β)x+αβ].

thus, a quadratic polynomial with zeros −3 and 4 is

k[x2−(−3+4)x+(−3)*4]

=k(x2−x−12)

=1/2(x2−x−12) (Putting k=1/2)

=x2/2−x/2−6

thus, the correct answer is an option (C)

 

Question 3: When the zeroes of the quadratic polynomial x2 + (a + 1) x + b are 2 and –3,

(A) a = –7, b = –1

(B) a = 5, b = –1

(C) a = 2, b = –6

(D) a = 0, b = –6

Answer 3: (D) a = 0, b = –6

Explanation: Let alpha and beta be the zeroes of polynomials Ax2+Bx+C.

Then we know that

α+β=−B/A …..(i)

and αβ=C/A …..(ii)

As Given that 2 and −3 are the zeroes of the quadratic polynomial x^2 + (a + 1) x + b,

⇒A=1, B=a+1, C=b

using (i)

2+(−3)=−(a+1)/1

−1=−a−1

a=0

using (ii)

2×(−3)=b/1

b=−6

Hence, the correct answer is option (D)

 

Question 4:The number of the polynomials having zeros as –2 and 5 is

(A) 1

(B) 2

(C) 3

(D) more than 3

Answer 4: (D) more than 3

Explanation: Quadratic polynomials having zeros α and β are given by k[x2−(α+β)x+αβ].

thus, the quadratic polynomial having zeros −2 and 5 is given by

k[x2−(−2+5)x+(−2)×5]

=k[x2−3x−10],

here, k is an arbitrary constant.

Thus, by putting k = 1, 2, 3, 4….., we can have infinitely many polynomials.

Therefore, the correct answer is option (D).

 

Question 5: It is Given that one of the zeroes of the cubic polynomial ax3 + bx2 + cx + d is zero, then the product of the other two zeroes is

(A) −c/a

(B) c/a

(C) 0

(D) −b/a

Answer 5:(B) c/a

Explanation: Let α, β, and γ be the zeroes of the cubic polynomial ax3 + bx2 + cx + d.

We know that

αβ+βγ+γα=c/a

Let α=0

⇒0×β+β×γ+γ×0=c/a

⇒βγ=c/a

Thus, the correct answer is option (B).

 

Question 6: When one of the zeroes of the cubic polynomial x3 + ax2 + bx + c is –1, the product of the other two zeroes is

(A) b – a + 1

(B) b – a – 1

(C) a – b + 1

(D) a – b –1

Answer 6:(A) b – a + 1

Explanation: Let α, β, γ be the zeroes of the cubic polynomial Ax3 + Bx2 + Cx + D.

We know that

α+β+γ=−B/A

⇒α+β+(−1)=−a (∵ γ=−1)

⇒α+β=−a+1 …..(1)

and

αβ+βγ+γα=C/A

⇒αβ+β(−1)+(−1)α=b/1

⇒αβ−β−α=b

⇒αβ=b+(α+β)

⇒αβ=b−a+1 [using (i)]

Hence, the correct answer is an option (A)

 

Question 7: Zeroes of the quadratic polynomial x2 + 99x + 127 are

(A) both positive

(B) both negative

(C) one positive and one negative

(D) both equal

Answer 7: (B) both negative

Explanation: To check the zeroes of a quadratic polynomial ax2+bx+c, we know that D=b2−4ac.

For x2 + 99x + 127 here we have, a=1, b=99, c=127

⇒D=(99)2−4×1×127

⇒D=9801−508

⇒D=9293>0

Thus, both the zeroes will be real and distinct.

Hence, a, b, and c are all positive. Therefore, both zeroes will be negative.

Thus, the correct answer is option (B)

 

Question 8: Zeroes of the quadratic polynomial x2+ kx + k, here, k ≠ 0,

(A) cannot both be positive

(B) cannot both be negative

(C) are always unequal

(D) are always equal

Answer 8 :(A) cannot both be positive.

Explanation: To check the nature of the zeroes of the quadratic polynomial ax2+bx+c, we know that D=b2−4ac.

For quadratic polynomials x2 + kx + k,

here we have a=1, b=k, and c=k.

⇒D=k2−4k

⇒D=k(k−4)

As Given that k≠0,

D=0 for k=4 ⇒roots will be same

D<0 for 0<k<4 ⇒no real roots

D>0 for k<0 or k>4 ⇒roots are real, that is roots can be positive or negative

Thus, in any case, zeroes of the given quadratic polynomial cannot be both positive.

we concluded, that the correct answer is option (A)

 

Question 9: When the zeroes of the quadratic polynomial ax2 + bx + c, c ≠ 0 are equal,

(A) c and a have opposite signs

(B) c and b have opposite signs

(C) c and a have the same sign

(D) c and b have the same sign

Answer 9: (C) c and a have the same sign

Explanation: As given that, zeroes of the quadratic polynomial ax2 + bx + c, c ≠ 0 are equal.

⇒D=0

we know that b2−4ac=0

b2 is always positive, that is, b2≥0 (hence, the square of any number is always greater than equal or zero)

(i) if a and c are of opposite signs,

ac<0

⇒4ac<0

⇒−4ac>0

⇒b2−4ac>0

(ii) When a and c are of the same sign,

ac>0

⇒4ac>0

⇒−4ac<0

⇒b2≥0

Thus, b2−4ac=0 is possible for some values of a, b, and c when a and c are of the same sign.

Hence, the correct answer is option (C).

 

Question 10: When one of the zeroes of a quadratic polynomial of the form x2 + ax + b is the negative of the other, it

(A) having no linear term, and a constant term is negative.

(B) having no linear term and a constant term is positive.

(C) can have a linear term, but a constant term is negative.

(D) can have a linear term, but a constant term is positive.

Answer 10: (A) has no linear term, and the constant term is negative.

Explanation:Let α & β be the zeros for quadratic polynomials x2 + ax + b.

it is Given that one of the zeros is negative of the other, i.e., β=−α, then

α+β=−a [hence, Sum of the zeros = −coefficient of x /coefficient of x2]

⇒α−α=−a

⇒a=0

αβ=b [Since, the product of zeros = constant term/ coefficient of x2]

⇒α×(−α)=b

⇒b=−α2

Thus, the quadratic polynomial becomes x2+0x−α2.

So, the correct answer is option (A).

Question 11:Which option is not the graph of a quadratic polynomial?

(A)

(B)

(C)

(D)

Answer 11: (D)

Explanation: For quadratic polynomial ax2 + bx + c, a ≠ 0, the curve for the quadratic polynomial crosses the X-axis on most two points.

Hence, in option (D), the curve crosses the X-axis on three points, thus it does not represent quadratic polynomial.

Thus, the correct answer is option (D).

 

Question 12. Answer the following and give an explanation:

(i) Could x2 – 1 be the quotient on the division for x6 + 2x3 + x – 1 with the polynomial in x having degree 5?

Answer 12: No, the x2 – 1 can’t be the quotient on the division for x6 + 2×3 + x – 1 with the polynomial in x having degree 5.

Clarification:

If a degree 6 polynomial is divided with the degree 5 polynomial,

The quotient will be having degree 1.

Let that (x2 – 1) divides the degree 6 polynomial and the quotient obtained is the degree 5 of the polynomial

As per our assumption,

degree for 6 polynomial = (x2 – 1)(degree for 5 polynomial) + r(x) [ Since, (a = bq + r)]

= (degree for 7 polynomial) + r(x) [ Since, (x2 term × x5 term = x7 term)]

= (degree for 7 polynomial)

From the given equation, it is clear that our predictions are contradicted.

x2 – 1 can’t be the quotient on the division with x6 + 2×3 + x – 1 for the polynomial in x for the degree 5

Hence, proved.

(ii) What will be the quotient and remainder be on the division for ax2 + bx + c by px3 + qx2 + rx + s, p ≠ 0?

Answer :

Degree for the polynomial px3 + qx2 + rx + s is 3

Degree for the polynomial ax2 + bx + c is 2

Where, degree for px3 + qx2 + rx + s is greater than degree of the ax2 + bx + c

Hence, the quotient will be zero,

Also, the remainder will be the dividend = ax3 + bx + c.

(iii) When on division for the polynomial p (x) by the polynomial g (x), the quotient is zero. What is the relation between degrees for p (x) and g (x)?

Answer :

We observe that,

p(x)= g(x) × q(x)+r(x)

As per the question,

q(x) =0

If q(x)=0, then r(x) is also = 0

Thus, now when we divide p(x) by g(x),

Now, p(x) should be equal to zero

Therefore, the relation between the degrees of p (x) and g (x) is the degree p(x)<degree g(x)

(iv) When on division for a non-zero polynomial p (x) by a polynomial g (x), the remainder is zero. What is the relation between degrees for p (x) and g (x)?

Answer :

To divide p(x) by g(x)

We observe,

The Degree for p(x) > degree for g(x)

And

Degree for p(x)= degree of g(x)

Therefore, we can see,

The relation between the degrees for p (x) and g (x) is degree for p(x) > degree of g(x)

(v) Can a quadratic polynomial x2 + kx + k have the same zeros of odd integers k > 1?

Answer :

A Quadratic Equation will have the same roots when it satisfies the condition:

b² – 4ac = 0

Given that the equation is x² + kx + k = 0

a = 1, b = k, x = k

Putting in the equation we get,

k² – 4 ( 1 ) ( k ) = 0

k² – 4k = 0

k ( k – 4 ) = 0

k = 0 , k = 4

But in the question, we can see that k is greater than 1.

Therefore, the value of k is 4 if the equation has common roots.

Thus, when the value of k = 4, then the equation ( x² + kx + k ) will have equal roots.

 

Question 13. The graphs for y = p(x) are given below Fig., for some polynomials p(x). Find out the number for zeros for p(x), in each case.

Answer 13:

Graphical method to get zeroes:-

Total no. of zeroes for any polynomial equation = total number of times any curve intersects the x-axis.

(i) For the given graph, the number of zeroes for p(x) is 0 as the graph is parallel to the x-axis as well as does not intersect it at any point.

(ii) For the given graph, the number of zeroes for p(x) is 1 as the graph cuts the x-axis at only one point.

(iii) For the given graph, the number of zeroes for p(x) is 3 as the graph intersects at the x-axis at any three points.

(iv) For the given graph, the number of zeroes for p(x) is 2 as the graph intersects the x-axis at two points.

(v) For the given graph, the number of zeroes for p(x) is 4 as the graph intersects the x-axis at four points.

(vi) For the given graph, the number of zeroes for p(x) is 3 as the graph intersects the x-axis at three points.

 

Question 14. Find out the zeroes of the following quadratic polynomials and check the relationship between the zeroes and the coefficients.

Answer 14:

(i) x2–2x –8

We have,

x2– 4x+2x–8 = x(x–4)+2(x–4) = (x-4)(x+2)

Hence, zeroes of polynomial equation x2–2x–8 are (4, -2)

Sum for zeroes = 4–2 = 2 = -(-2)/1 = -(Coefficient of x)/(Coefficient of x2)

Product for zeroes = 4×(-2) = -8 =-(8)/1 = (Constant term)/(Coefficient of x2)

(ii) 4s2–4s+1

We have,

4s2–2s–2s+1 = 2s(2s–1)–1(2s-1) = (2s–1)(2s–1)

Hence, zeroes of polynomial equation 4s2–4s+1 are (1/2, 1/2)

Sum for zeroes = (½)+(1/2) = 1 = -(-4)/4 = -(Coefficient of s)/(Coefficient of s2)

Product for zeroes = (1/2)×(1/2) = 1/4 = (Constant term)/(Coefficient of s2 )

(iii) 6×2–3–7x

We have,

6×2–7x–3 = 6×2 – 9x + 2x – 3 = 3x(2x – 3) +1(2x – 3) = (3x+1)(2x-3)

Hence, zeroes of polynomial equation 6×2–3–7x are (-1/3, 3/2)

Sum for zeroes = -(1/3)+(3/2) = (7/6) = -(Coefficient of x)/(Coefficient of x2)

Product for zeroes = -(1/3)×(3/2) = -(3/6) = (Constant term) /(Coefficient of x2 )

(iv) 4u2+8u

We have,

4u(u+2)

Hence, zeroes of polynomial equation 4u2 + 8u are (0, -2).

Sum for zeroes = 0+(-2) = -2 = -(8/4) = = -(Coefficient of u)/(Coefficient of u2)

Product for zeroes = 0×-2 = 0 = 0/4 = (Constant term)/(Coefficient of u2 )

(v) t2–15

We have,

t2 = 15 or t = ±√15

Hence, zeroes of polynomial equation t2 –15 are (√15, -√15)

Sum for zeroes =√15+(-√15) = 0= -(0/1)= -(Coefficient of t) / (Coefficient of t2)

Product for zeroes = √15×(-√15) = -15 = -15/1 = (Constant term) / (Coefficient of t2 )

(vi) 3×2–x–4

We have,

3×2–4x+3x–4 = x(3x-4)+1(3x-4) = (3x – 4)(x + 1)

Hence, zeroes of polynomial equation 3×2 – x – 4 are (4/3, -1)

Sum for zeroes = (4/3)+(-1) = (1/3)= -(-1/3) = -(Coefficient of x) / (Coefficient of x2)

Product for zeroes=(4/3)×(-1) = (-4/3) = (Constant term) /(Coefficient of x2 )

 

Question 15: When the zeroes for the polynomial x3 – 3×2 + x + 1 are a – b, a, a + b, then find out the value of a and b.

Answer 15:

Assume the given polynomial be:

For p(x) = x3 – 3×2 + x + 1

Given that,

The zeroes for the p(x) are a – b, a, and a + b.

Then, compare the given polynomial equation with general expression.

px3 + qx2 + rx + s = x3 – 3×2 + x + 1

Here, p = 1, q = -3, r = 1 and s = 1

For sum for zeroes:

Sum for zeroes will be = a – b + a + a + b

-q/p = 3a

Put the values q and p.

-(-3)/1 = 3a

a = 1

Therefore, the zeroes are 1 – b, 1, 1 + b.

For the product for zeroes:

Product for zeroes = 1(1 – b)(1 + b)

-s/p = 1 – ?2

=> -1/1 = 1 – ?2

And, b2 = 1 + 1 =2

Hence,, b = √2

We conclude, 1 – √2, 1, 1 + √2 are the zeroes of equation ?3 − 3?2 + ? + 1.

 

Question 16. Divide the polynomial p(x) with the polynomial g(x) and find out the quotient and the remainder in each of the following:

(i) p(x) = x3-3×2+5x–3 , g(x) = x2–2

Answer 16 (i):

Given that,

Dividend = p(x) = x3-3×2+5x–3

Divisor = g(x) = x2– 2

x-3

_______________

x2-2 ) x3– 3×2+ 5x – 3

_

x3 + 0x2 -2x

_____________

-3×2 + 7x – 3

_

-3×2 + 0x + 6

___________

7x- 9

 

Hence, upon division we get,

Quotient = x–3

Remainder = 7x–9

(ii) p(x) = x4-3×2+4x+5 , g(x) = x2+1-x

Answer (ii):

Given that,

Dividend = p(x) = x4 – 3×2 + 4x +5

Divisor = g(x) = x2 +1-x

 

x2 + x-3

____________________

x2-x+1 ) x4 – 0x3 – 3×2 + 4x + 5

_

x4 – x3 + x2

_____________________

x3 – 4×2 + 4x + 5

_

x3 – x2 + x

________________

-3×2 + 3x + 5

_

 

-3×2 + 3x – 3

_____________

8

Hence, upon division we observe,

Quotient = x2 + x–3

thus, Remainder = 8

(iii) p(x) =x4–5x+6, g(x) = 2–x2

Answer (iii):

its Given that,

Dividend = p(x) =x4 – 5x + 6 = x4 +0x2–5x+6

Divisor = g(x) = 2–x2 = –x2+2

– x2 -2

__________________

– x2 + 2) x4 + 0x3 + 0x2 -5x + 6

_

x4 + 0x3 – 2×2

_____________

2×2 – 5x + 6

_

2×2 + 0x – 4

________

– 5x + 10

 

Hence, upon division we get,

Quotient = -x2-2

Remainder = -5x + 10

 

Question 17. Check whether the first polynomial is the factor for the second polynomial dividing the second polynomial with the first polynomial:

(i) t2-3, 2t4 +3t3-2t2-9t-12

Answer 17 (i):

Given that,

First polynomial = t2-3

Second polynomial = 2t4 +3t3-2t2 -9t-12

 

3t2 + 3t + 4

____________________

t2 – 3 ) 2t4 + 3t3 -2t2 – 9t – 12

_

2t4 + 0t3 – 6t2

__________________

3t3 + 4t2 – 9t – 12

_

3t3 + 0t2 – 9t

______________

4t2 + 0t – 12

_

4t2 + 0t – 12

_____________

0

 

As we can observe, the remainder is left as 0. Hence, we can say that t2-3 is a factor of 2t4 +3t3-2t2 -9t-12.

(ii) x2+3x+1 , 3×4+5×3-7×2+2x+2

Answer (ii):

Given that,

First polynomial = x2+3x+1

Second polynomial = 3×4+5×3-7×2+2x+2

3×2 – 4x + 2

____________________

x2 + 3x + 1) 3×4 + 5×3 – 7×2 + 2x+ 2

_

3×4 – 9×3 + 3×2

_____________________

-4×3 – 10×2 + 2x + 2

_

-4×3 – 12×2 – x

________________

2×2 + 6x + 2

_

2×2 + 6x – 2

_____________

0

 

As we can observe, the remainder is left as 0. Hence, we can say that x2 + 3x + 1 is a factor of 3×4+5×3-7×2+2x+2.

(iii) x3-3x+1, x5-4×3+x2+3x+1

Answer (iii):

Given that,

First polynomial = x3-3x+1

Second polynomial = x5-4×3+x2+3x+1

 

x2 – 1

____________________

x3 – 3x + 1) x5 – 0x3 + x2 + 3x + 1

_

x5 + 0x4 – 3×3 + x2

__________________

– x3 + 0x2 + 3x + 1

_

-x3 + 0x2 + 3x – 1

_______

2

 

As we can observe, the remainder is not equal to 0.

Hence, we say that x3-3x+1 is not a factor of x5-4×3+x2+3x+1 .

 

Question 18. Obtain all other zeroes of 3×4+6×3-2×2-10x-5, when two of its zeroes are √(5/3) and – √(5/3).

Answer 18:

For a polynomial equation of degree 4, so, there will be a total of 4 roots.

√(5/3) as well as – √(5/3) are zeroes for polynomials f(x).

Therefore, (x –√(5/3)) (x+√(5/3) = x2-(5/3)2 = 0

(x2– 5/3) is a factor for the given polynomial f(x).

Then, if we divide f(x) with (x2– 5/3), the quotient obtained will also be a factor for f(x), and the remainder will have 0.

 

x2 + 2x + 1

____________________

3×2 – 5) 3×4 + 6×3 – 2×2 – 10x – 5

_

3×4 – 5×2

__________________

6×3 + 3×2 – 10x – 5

_

– 6×3 – 10x

__________________

+ 3×2 – 5

_

+ 3×2 – 5

___________

0

___________

 

Hence, 3×4 +6×3 −2×2 −10x–5 = 3 (x2– 5/3)(x2+2x+1)

Then, on further factoring (x2+2x+1) we observe,

x2+2x+1 = x2+x+x+1 = 0

x(x+1)+1(x+1) = 0

(x+1)(x+1) = 0

Thus, its zeroes are given by: x= −1.

As a result, all four zeroes for given polynomial equations are given:

√(5/3),- √(5/3), and −1.

We interpret it as the answer.

 

Question 19: α and β are zeroes of the quadratic polynomial x2 – 6x + y. Find out the value of ‘y’ if 3α + 2β = 20.

Answer 19:

Assume,

f(x) = x² – 6x + y

For the given,

3α + 2β = 20———————(i)

For f(x),

α + β = 6———————(ii)

Also,

αβ = y———————(iii)

Multiply equation (ii) with 2. If you subtract the whole equation from equation (i),

=> α = 20 – 12 = 8

Then, substitute the value in equation (ii),

=> β = 6 – 8 = -2

Put the values of α and β in the equation (iii) to get the value for y, such as;

y = αβ = (8)(-2) = -16

 

Question 20. When dividing x3-3×2+x+2 with a polynomial g(x), the quotient and remainder was x–2 and –2x+4, respectively. Find out g(x).

Answer 20:

Given that,

Dividend, p(x) = x3-3×2+x+2

Quotient = x-2

Remainder = –2x+4

We have to find the value for divisor, g(x) =?

As we are aware,

Dividend = Divisor × Quotient + Remainder

∴ x3-3×2+x+2 = g(x)×(x-2) + (-2x+4)

x3-3×2+x+2-(-2x+4) = g(x)×(x-2)

Hence, g(x) × (x-2) = x3-3×2+3x-2

Then, for finding g(x) we will divide x3-3×2+3x-2 with (x-2)

 

x2 – x + 1

______________

x-2) x3 – 3×2 + 3x – 2

_

x3 – 2×2

____________

-x2 + 3x – 2

_

-x2 + 2x

_________

x – 2

x – 2

_________

0

Thus, g(x) = (x2–x+1)

 

Question 21. Give examples for polynomials p(x), g(x), q(x) and r(x), the satisfy the division algorithm and

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

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

(iii) deg r(x) = 0

Answer 21:

As per the division algorithm,

dividend p(x), as well as divisor g(x), are two polynomials, here g(x)≠0.

Now,we can find out the value for quotient q(x) and remainder r(x), with the help of the below given formula;

Dividend = Divisor × Quotient + Remainder

therefore, p(x) = g(x)×q(x)+r(x)

here r(x) = 0 or degree of r(x)< degree of g(x).

Then, let us prove the three given cases as per division algorithm by taking examples for each.

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

The degree of dividend is equal to the degree for quotient only if the divisor is a constant term.

Taking an example, for p(x) = 3×2+3x+3 is a polynomial to be divided with g(x) = 3.

Hence, (3×2+3x+3)/3 = x2+x+1 = q(x)

Therefore, you can see the degree of quotient q(x) = 2, which is also equal to the degree of dividend p(x)

r(x) = 0

Checking for the division algorithm in this case:

p(x) = g(x) × q(x) + r(x)

= 3 (x2+x+1) + 0

= (3×2+3x+3)

So, the division algorithm is satisfied here.

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

• Taking an example, for p(x) = x3 + x is a polynomial to be divided with g(x) = x2

Hence, x3 + x = x2×(x) + (x)

As a result, quotient q(x) = x

And, remainder r(x) = x

Therefore, you can see that the degree of quotient q(x) = 1, which is also equal to the degree of remainder r(x).

Checking for division algorithm in this case,

p(x) = g(x) × q(x) + r(x)

= (x2 × x ) + x

= x3 + x

So, the division algorithm is satisfied here.

(iii) deg r(x) = 0

The degree of the remainder would be 0 only if the remainder left after the division algorithm is constant.

Taking an example, p(x) = x3 + 1 is a polynomial to be divided by g(x) = x2

Thus, x3 + 1 = x2 × (x) + 1

Therefore, quotient q(x) = x

Also remainder r(x) = 1

We observe, the degree of remainder here is 0.

Checking for division algorithm in this case,

p(x) = g (x) × q (x) + r (x)

= (x2 × x) + 1

= x3 + 1

So, the division algorithm is satisfied here.

 

Question 22: For which the value of k, is the polynomial f(x) = 3×4 – 9×3 + x2 + 15x + k completely divisible with 3×2 – 5?

Answer 22:

Given that,

for (x) = 3×4 – 9×3 + x2 + 15x + k

And,

g(x) = 3×2 – 5

For dividing f(x) with g(x),

 

x2 – 3x + 2

____________________

3×2 – 5 ) 3×4 – 9×3 + x2 + 15x + k

_

3×4 + 0x3.– 5×2

__________________

-9×3 + 6×2 + 15x + k

_

– 9×3 + 0x2 + 15x

__________________

6×2 + 0x + k

_

6×2 + 0x – 10

___________

k + 10

___________

 

Given that f(x) is completely divisible with 3×2 – 5.

Hence, the remainder = 0

k + 10 = 0

k = -10

 

Question 23 Check that the numbers given alongside the cubic polynomials below are their zeroes. Also check the relationship for the zeroes and the coefficients in each case:

(i) 2×3+x2-5x+2; -1/2, 1, -2

Answer 23 (i):

Given that,

p(x) = 2×3+x2-5x+2

zeroes for p(x) = 1/2, 1, -2

therefore,

p(1/2) = 2(1/2)3+(1/2)2-5(1/2)+2 = (1/4)+(1/4)-(5/2)+2 = 0

p(1) = 2(1)3+(1)2-5(1)+2 = 0

p(-2) = 2(-2)3+(-2)2-5(-2)+2 = 0

So, proved 1/2, 1, -2 are the zeroes for 2×3+x2-5x+2.

Then, comparing the given polynomial using general expression, we have;

∴ ax3+bx2+cx+d = 2×3+x2-5x+2

a=2, b=1, c= -5 and d = 2

As you know, if α, β, γ are the zeroes for the cubic polynomial ax3+bx2+cx+d , then;

α +β+γ = –b/a

αβ+βγ+γα = c/a

α βγ = – d/a.

As a result, putting the values of zeroes for the polynomial,

α+β+γ = ½+1+(-2) = -1/2 = –b/a

αβ+βγ+γα = (1/2×1)+(1 ×-2)+(-2×1/2) = -5/2 = c/a

α β γ = ½×1×(-2) = -2/2 = -d/a

here, the relationship between the zeroes and the coefficients are satisfied.

(ii) x3-4×2+5x-2 ;2, 1, 1

Answer (ii):

Given that,

p(x) = x3-4×2+5x-2

zeros for p(x) are 2,1,1.

Therefore, p(2)= 23-4(2)2+5(2)-2 = 0

p(1) = 13-(4×12 )+(5×1)-2 = 0

Hence proved, 2, 1, 1 are the zeroes of x3-4×2+5x-2

Then, comparing the given polynomial with general expression, we get;

Therefore, ax3+bx2+cx+d = x3-4×2+5x-2

a = 1, b = -4, c = 5 and d = -2

we observe, if α, β, γ are the zeroes of the cubic polynomial ax3+bx2+cx+d , then;

α + β + γ = –b/a

αβ + βγ + γα = c/a

α β γ = – d/a.

Hence , putting the values of zeroes for the polynomial,

α +β+γ = 2+1+1 = 4 = -(-4)/1 = –b/a

αβ+βγ+γα = 2×1+1×1+1×2 = 5 = 5/1= c/a

αβγ = 2×1×1 = 2 = -(-2)/1 = -d/a

So, the relationship for the zeroes and the coefficients are satisfied.

 

Question 24: Find out the quadratic polynomial each with the given numbers having the sum and product of its zeroes, respectively.

(i) 1/4, -1

(ii) 1, 1

(iii) 4, 1

Answer 24:

(i) From the formulas for sum and product of zeroes, we get,

Sum for zeroes = α + β

Product for zeroes = αβ

Given that,

Sum for zeroes = 1/4

Product for zeroes = -1

Hence, when α and β are zeroes of any quadratic polynomial, the polynomial can be written as:-

x2 – (α + β)x + αβ

= x2 – (1/4)x + (-1)

= 4×2 – x – 4

Therefore, 4×2 – x – 4 is the required quadratic polynomial.

(ii) Given that,

Sum for zeroes = 1 = α + β

Product for zeroes = 1 = αβ

So, if α and β are zeroes of any quadratic polynomial, then the polynomial can be written as:-

x2 – (α + β)x + αβ

= x2 – x + 1

Hence, x2 – x + 1 is the quadratic polynomial.

(iii) Given that,

Sum for zeroes, α + β = 4

Product for zeroes, αβ = 1

So, if α and β are zeroes of any quadratic polynomial, then the polynomial can be written as:-

x2 – (α + β)x + αβ

= x2 – 4x + 1

Hence, x2 – 4x +1 is the quadratic polynomial.

 

Question 25. Find out a cubic polynomial having the sum, sum of the product for its zeroes taken two at a time, and the product for its zeroes as 2, –7, –14 respectively.

Answer 25:

Consider that the cubic polynomial is ax3+bx2+cx+d and the values for the zeroes of the polynomials be α, β, γ.

According to the given question,

α+β+γ = -b/a = 2/1

αβ +βγ+γα = c/a = -7/1

α βγ = -d/a = -14/1

Therefore, from the above three expressions we get the values for the coefficient of the polynomial is.

a = 1, b = -2, c = -7, d = 14

So, the cubic polynomial is x3-2×2-7x+14

 

Question 26: Divide the polynomial f(x) = 3×2 – x3 – 3x + 5 with the polynomial g(x) = x – 1 – x2 and check the division algorithm.

Answer 26:

Given,

For f(x) = 3×2 – x3– 3x + 5

And

g(x) = x – 1 – x2

Dividing for f(x) = 3×2 – x3 – 3x + 5 with g(x) = x – 1 – x2

 

x2 – 3x + 2

________________

– x2+ x – 1 )– x3 + 3×2 – 3x + 5

_

– x3. + x2 .– x

____________

2×2 – 2x + 5

_

2×2 – 2x + 2

 

________

3

Where,

Quotient = q(x) = x – 2

Remainder = r(x) = 3

For division method of polynomials,

Dividend = (Quotient × Divisor) + Remainder

For

[q(x) × g(x)] + r(x) = (x – 2)(x – 1 – x2) + 3

= x2 – x – x3 – 2x + 2 + 2×2 + 3

= 3×2 – x3 – 3x + 5

= f(x)

Therefore, the division algorithm is verified.

 

Question 27. When the zeroes of the polynomial x3-3×2+x+1 are a – b, a, a + b, find out a and b.

Answer 27:

You are given that the polynomial,

p(x) = x3-3×2+x+1

zeroes are given, a – b, a, a + b

Then, comparing the given polynomial with general expression, we find;

∴px3+qx2+rx+s = x3-3×2+x+1

p = 1, q = -3, r = 1 and s = 1

Sum for zeroes = a – b + a + a + b

-q/p = 3a

Substituting the values q and p.

-(-3)/1 = 3a

a=1

So, the zeroes are 1-b, 1, 1+b.

Then, product of zeroes = 1(1-b)(1+b)

-s/p = 1-b2

-1/1 = 1-b2

b2 = 1+1 = 2

b = ±√2

We conclude that 1-√2, 1 ,1+√2 are the zeroes of x3-3×2+x+1.

 

Question 28: Find out a quadratic polynomial whose zeroes are reciprocals for the zeroes of the polynomial of f(x) = ax2 + bx + c, a ≠ 0, c ≠ 0.

Answer 28:

Assume α and β be the zeroes for the polynomial f(x) = ax2 + bx + c.

Thus, α + β = -b/a

αβ = c/a

As per the given, 1/α and 1/β are the zeroes of the required quadratic polynomial.

Then, the sum for zeroes = (1/α) + (1/β)

= (α + β)/αβ

= (-b/a)/ (c/a)

= -b/c

Product for two zeroes = (1/α) (1/β)

= 1/αβ

= 1/(c/a)

= a/c

The required quadratic polynomial is given by

= k[x2 + (sum of zeroes)x + (product of zeroes)]

= k[x2 – (-b/c)x + (a/c)]

= k[ x2 + (b/c)x + a/c]

= cx2 + bx + a

 

Question 29. If two zeroes of the polynomial x4-6×3-26×2+138x-35 are 2 ±√3, find out other zeroes.

Answer 29:

If this is a polynomial equation of degree 4, thus there will be a total of 4 roots.

Assume f(x) = x4-6×3-26×2+138x-35

If 2 +√3 and 2-√3 are zeroes of the given polynomial f(x).

∴ [x−(2+√3)] [x−(2-√3)] = 0

(x−2−√3)(x−2+√3) = 0

On multiplying the given equation we observe,

x2-4x+1, this is a factor of the given polynomial f(x).

Then, if we divide f(x) by g(x), the quotient will also be a factor for f(x) and the remainder will have 0.

For, x4-6×3-26×2+138x-35 = (x2-4x+1)(x2 –2x−35)

Then, on further factorizing (x2–2x−35) we get,

x2–(7−5)x −35 = x2– 7x+5x+35 = 0

x(x −7)+5(x−7) = 0

(x+5)(x−7) = 0

Hence, its zeroes are given by:

x= −5 and x = 7.

Thus, all four zeroes of the given polynomial equation are: 2+√3 , 2-√3, −5 and 7.

 

Question 30: When 4 is a zero for the cubic polynomial x3 – 3×2 – 10x + 24, find out its other two zeroes.

Answer 30:

Given that the cubic polynomial

For p(x) = x3 – 3×2 – 10x + 24

4 is a zero of p(x).

Therefore, (x – 4) is the factor of p(x).

When you divide the given polynomial by (x – 4).

Where, the quotient

= x2 + x – 6

= x2 + 3x – 2x – 6

= x(x + 3) – 2(x + 3)

= (x – 2)(x + 3)

Hence, the other two zeroes of the given cubic polynomial are 2 and -3.

 

Question 31. For each of the following, find out a quadratic polynomial whose sum and product respectively of the zeroes are as given. Also find out the zeroes of these polynomials by the factorisation.

(i) (–8/3), 4/3

(ii) 21/8, 5/16

(iii) -2√3, -9

(iv) (-3/(2√5)), -½

Answer 31:

(i) As we know,

Sum of the zeroes = – 8/3

And

Product of the zeroes = 4/3

Thus, P(x) = x2 – (sum of the zeroes) + (product of the zeroes)

Now, P(x)= x2 – (-8x)/3 + 4/3

Thus, P(x)= 3×2 + 8x + 4

Using splitting the middle term algorithm,

3×2 + 8x + 4 = 0

3×2 + (6x + 2x) + 4 = 0

3×2 + 6x + 2x + 4 = 0

3x(x + 2) + 2(x + 2) = 0

(x + 2)(3x + 2) = 0

Therefore, x = -2, -2/3

(ii) As we know,

Sum of the zeroes = 21/8

And

Product of the zeroes = 5/16

Thus, P(x) = x2 – (sum of the zeroes) + (product of the zeroes)

Now, P(x)= x2 – 21x/8 + 5/16

P(x)= 16×2 – 42x + 5

Using splitting the middle algorithm,

16×2 – 42x + 5 = 0

16×2 – (2x + 40x) + 5 = 0

16×2 – 2x – 40x + 5 = 0

2x (8x – 1) – 5(8x – 1) = 0

(8x – 1)(2x – 5) = 0

Therefore, x = 1/8, 5/2

(iii) As we know,

Sum of the zeroes = – 2√3

And

Product of the zeroes = – 9

Thus, P(x) = x2 – (sum of the zeroes) + (product of the zeroes)

Now, P(x) = x2 – (-2√3x) – 9

Using splitting the middle term algorithm

x2 + 2√3x – 9 = 0

x2 + (3√3x – √3x) – 9 = 0

x(x + 3√3) – √3(x + 3√3) = 0

(x – √3)(x + 3√3) = 0

Therefore, x = √3, -3√3

(iv) Sum of the zeroes = -3/2√5x

And

Product of the zeroes = – ½

Thus, P(x) = x2 – (sum of the zeroes) + (product of the zeroes)

Now, P(x)= x2 -(-3/2√5x) – ½

Thus, P(x)= 2√5×2 + 3x – √5

Using splitting the middle term algorithm,

2√5×2 + 3x – √5 = 0

2√5×2 + (5x – 2x) – √5 = 0

2√5×2 – 5x + 2x – √5 = 0

√5x (2x + √5) – (2x + √5) = 0

(2x + √5)(√5x – 1) = 0

Therefore, x = 1/√5, -√5/2

 

Question 32: Compute the zeroes of the polynomial 4×2 – 4x – 8. Also, establish a relationship for the zeroes and coefficients.

Answer 32:

Assume the given polynomial be p(x) = 4×2 – 4x – 8

To get the zeroes, take p(x) = 0

Then, factorise the equation 4×2 – 4x – 8 = 0

4×2 – 4x – 8 = 0

4(x2 – x – 2) = 0

x2 – x – 2 = 0

x2 – 2x + x – 2 = 0

x(x – 2) + 1(x – 2) = 0

(x – 2)(x + 1) = 0

x = 2, x = -1

Therefore, the roots of 4×2 – 4x – 8 are -1 and 2.

Relation for the sum of zeroes and coefficients:

-1 + 2 = 1 = -(-4)/4 i.e. (- coefficient of x/ coefficient of x2)

Relation for the product of zeroes and coefficients:

(-1) × 2 = -2 = -8/4 i.e (constant/coefficient of x2)

 

Question 33. Given that the zeroes of the cubic polynomial x3 – 6x2 + 3x + 10 are for the form a, a + b, a + 2b for some real numbers a and b,

find out the values of a and b and the zeroes for the given polynomial.

Answer 33:

It is given that a, a+b, a+2b are roots for the polynomial x³-6x²+3x+10

Sum for the roots ⇒ a+2b+a+a+b = -coefficient of x²/ coefficient of x³

⇒ 3a+3b = -(-6)/1 = 6

⇒ 3(a+b) = 6

⇒ a+b = 2 ——— (1) b = 2-a

Product for the roots ⇒ (a+2b)(a+b)a = -constant/coefficient of x³

⇒ (a+b+b)(a+b)a = -10/1

Substituting the value for a+b=2 in it

⇒ (2+b)(2)a = -10

⇒ (2+b)2a = -10

⇒ (2+2-a)2a = -10

⇒ (4-a)2a = -10

⇒ 4a-a² = -5

⇒ a²-4a-5 = 0

⇒ a²-5a+a-5 = 0

⇒ (a-5)(a+1) = 0

a-5 = 0 or a+1 = 0

a = 5 a = -1

a = 5, -1 in (1) a+b = 2

When a = 5, 5+b=2 ⇒ b=-3

a = -1, -1+b=2 ⇒ b= 3

∴ If a=5 then b= -3

or

If a= -1 then b=3

 

Question 34. Given that √2 is a zero for the cubic polynomial 6x3 + √2 x2 – 10x – 4√2 , find out its other two zeroes.

Answer 34:

Given, √2 is one for the zero of the cubic polynomial.

Then, (x-√2) is one of the factors of the given polynomial p(x) = 6x³+√2x²-10x- 4√2.

So, by dividing p(x) by x-√2

6x³+√2x²-10x-4√2= (x-√2) (6x² +7√2x + 4)

By splitting the middle term,

We get,

(x-√2) (6x² + 4√2x + 3√2x + 4)

= (x-√2) [ 2x(3x+2√2) + √2(3x+2√2)]

= (x-√2) (2x+√2) (3x+2√2)

To get the zeroes of p(x),

Substitute p(x)= 0

(x-√2) (2x+√2) (3x+2√2)= 0

x= √2 , x= -√2/2 ,x= -2√2/3

Hence, the other two zeroes of p(x) are -√2/2 and -2√2/3