Mathematics Part i Solutions Solutions for Class 9 Maths Chapter 3 Polynomials are provided here with simple step-by-step explanations. These solutions for Polynomials are extremely popular among class 9 students for Maths Polynomials Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part i Solutions Book of class 9 Maths Chapter 3 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Part i Solutions Solutions. All Mathematics Part i Solutions Solutions for class 9 Maths are prepared by experts and are 100% accurate.
Page No 39:
Question 1:
State whether the given algebraic expressions are polynomials? Justify.
(i) (ii) (iii) (iv) (v) 10
Answer:
In an algebraic expression, if the powers of the variables are whole numbers then the algebraic expression is a polynomial.
(i)
Here, one of the powers of y is −1, which is not a whole number. So, is not a polynomial.
(ii)
Here, the power of x is , which is not a whole number. So, is not a polynomial.
(iii)
Here, the powers of the variable x are 2, 1 and 0, which are whole numbers. So, is a polynomial.
(iv)
Here, one of the powers of m is −2, which is not a whole number. So, is not a polynomial.
(v)
10 = 10 × 1 = 10x0
Here, the power of x is 0, which is a whole numbers. So, 10 is a polynomial (or constant polynomial).
Page No 39:
Question 2:
Write the coefficient of m3 in each of the given polynomial.
(i) m3 (ii) (iii)
Answer:
(i)
Coefficient of m3 = 1
(ii)
Coefficient of m3 =
(iii)
Coefficient of m3 =
Page No 39:
Question 3:
(i) Monomial with degree 7
Answer:
(i) A polynomial having only one term is called a monomial. Also, the highest power of the variable in a polynomial is called the degree of the polynomial.
3x7 is a monomial in x with degree 7.
(ii) A polynomial having only two terms is called a binomial. Also, the highest power of the variable in a polynomial is called the degree of the polynomial.
2x35 + 1 is a binomial in x with degree 35.
(iii) A polynomial having only three terms is called a trinomial. Also, the highest power of the variable in a polynomial is called the degree of the polynomial.
5x8 + 6x4 + 7x is a trinomial in x with degree 8.
Page No 40:
Question 4:
Write the degree of the given polynomials.
(i) (ii) (iii) (iv) (v) (vi) (vii) (viii)
Answer:
The highest power of the variable in a polynomial of one variable is called the degreee of the polynomial. Also, the highest sum of the powers of the variables in each term of the polynomial in more than one variable is the degree of the polynomial.
(i)
The degree of the polynomial is 0.
(ii)
The degree of the polynomial x0 is 0.
(iii)
The degree of the polynomial x2 is 2.
(iv)
The degree of the polynomial is 10.
(v)
The degree of the polynomial is 1.
(vi)
The degree of the polynomial is 5.
(vii)
The sum of the powers of the variables in the polynomial are 1 + 1 + 1 = 3 and 1 + 1 = 2.
The degree of the polynomial is 3.
(viii)
The sum of the powers of the variables in the polynomial are 3 + 7 = 10, 5 + 1 = 6 and 1 + 1 = 2.
The degree of the polynomial is 10.
Page No 40:
Question 5:
Classify the following polynomials as linear, quadratic and cubic polynomial.
Answer:
(i)
The degree of the polynomial 2x2 + 3x + 1 is 2.
So, the polynomial 2x2 + 3x + 1 is a quadratic polynomial.
(ii)
The degree of the polynomial 5p is 1.
So, the polynomial 5p is a linear polynomial.
(iii)
The degree of the polynomial is 1.
So, the polynomial is a linear polynomial.
(iv)
The degree of the polynomial is 3.
So, the polynomial is a cubic polynomial.
(v)
The degree of the polynomial a2 is 2.
So, the polynomial a2 is a quadratic polynomial.
(vi)
The degree of the polynomial 3r3 is 3.
So, the polynomial 3r3 is a cubic polynomial.
Page No 40:
Question 6:
Write the following polynomials in standard form.
(i) (ii)
Answer:
A polynomial written in either descending or ascending powers of its variable is called the standard form of the polynomial.
(i)
The given polynomial is m3 + 3 + 5m.
The standard form of the polynomial is 3 + 5m + m3 or m3 + 5m + 3.
(ii)
The given polynomial is .
The standard form of the polynomial is or .
Page No 40:
Question 7:
Write the following polynomials in coefficient form.
(i) (ii) (iii) (iv)
Answer:
(i)
The coefficient form of the polynomial is (1, 0, 0, −2).
(ii)
5y = 5y + 0
The coefficient form of the polynomial is (5, 0).
(iii)
The coefficient form of the polynomial is (2, 0, −3, 0, 7).
(iv)
The coefficient form of the polynomial is .
Page No 40:
Question 8:
Answer:
(i)
The coefficient form of the polynomial is (1, 2, 3).
Therefore, the index form of the polynomial is x2 + 2x + 3.
(ii)
The coefficient form of the polynomial is (5, 0, 0, 0, −1).
Therefore, the index form of the polynomial is 5x4 + 0x3 + 0x2 + 0x − 1 or 5x4 − 1.
(iii)
The coefficient form of the polynomial is (−2, 2, −2, 2).
Therefore, the index form of the polynomial is −2x3 + 2x2 −2x + 2.
Page No 40:
Question 9:
Write the appropriate polynomials in the boxes.
Answer:
Page No 43:
Question 1:
Answer:
(i)
Initial number of trees in the village = a
Increase in the number of trees every year = b
∴ Number of trees in the village after x years
= Initial number of trees in the village + Increase in the number of trees every year × x
= a + bx
Thus, the number of trees after x years is a + bx.
(ii)
Number of students in each row = y
Number of rows = x
∴ Total number of students in the parade = Number of students in each row × Number of rows = y × x = yx = xy
Thus, there are in all xy students for the parade.
(iii)
Digit at the tens place = m
Digit at the units place = n
∴ Two digit number = Digit at the tens place × 10 + Digit at the units place = m × 10 + n = 10m + n
Thus, the polynomial which represents the two digit number is 10m + n.
Page No 43:
Question 2:
Add the given polynomials.
(i) (ii) (iii)
Answer:
(i)
(ii)
(iii)
Page No 43:
Question 3:
Subtract the second polynomial from the first.
(i) (ii)
Answer:
(i)
(ii)
Page No 43:
Question 4:
Multiply the given polynomials.
(i) (ii) (iii)
Answer:
(i)
(ii)
(iii)
Page No 43:
Question 5:
(i) (ii)
Answer:
(i)
Using long division method,
Dividend = Divisor ×â Quotient + Remainder
(ii)
Using long division method,
Dividend = Divisor ×â Quotient + Remainder
Page No 43:
Question 6:
Answer:
Lenght of the rectangular farm = (2a2 + 3b2) m
Breadth of the rectangular farm = (a2 + b2) m
Side of the square plot = (a2 − b2) m
∴ Area of the remaining part of the farm
= Total area of the farm − Area of the square plot
= Lenght of the rectangular farm × Breadth of the rectangular farm − (Side of the square plot)2
= (2a2 + 3b2) × (a2 + b2) − (a2 − b2)2
= 2a2(a2 + b2) + 3b2(a2 + b2) − (a4 + b4 − 2a2b2)
= 2a4 + 2a2b2 + 3a2b2 + 3b4 − a4 − b4 + 2a2b2
= 2a4 − a4 + 2a2b2 + 3a2b2 + 2a2b2 + 3b4 − b4
= (a4 + 7a2b2 + 2b4) m2
Thus, the area of the remaining part of the farm is (a4 + 7a2b2 + 2b4) m2.
Page No 46:
Question 1:
(i) (ii) (iii)
(iv) (v) (vi)
Answer:
(i)
Synthetic Division:
Dividend =
Divisor =
Opposite of −5 = 5
The coefficient form of the quotient is (2, 7).
∴ Quotient = 2m + 7 and Remainder = 45
Linear Method:
(ii)
Synthetic Division:
Dividend =
Divisor =
Opposite of 2 = −2
The coefficient form of the quotient is (1, 0, 3, −2).
∴ Quotient = x3 + 3x − 2 and Remainder = 9
Linear Method:
(iii)
Synthetic Division:
Dividend =
Divisor =
Opposite of −6 = 6
The coefficient form of the quotient is (1, 6, 36).
∴ Quotient = y2 + 6y + 36 and Remainder = 0
Linear Method:
(iv)
Synthetic Division:
Dividend =
Divisor =
Opposite of 3 = −3
The coefficient form of the quotient is (2, −3, 7, −17).
∴ Quotient = 2x3 − 3x2 + 7x − 17 and Remainder = 51
Linear Method:
(v)
Synthetic Division:
Dividend =
Divisor =
Opposite of 4 = −4
The coefficient form of the quotient is (1, −4, 13, −52).
∴ Quotient = x3 − 4x2 + 13x − 52 and Remainder = 200
Linear Method:
(vi)
Synthetic Division:
Dividend =
Divisor =
Opposite of −1 = 1
The coefficient form of the quotient is (1, −2, 3).
∴ Quotient = y2 − 2y + 3 and Remainder = 2
Linear Method:
Page No 48:
Question 1:
For = 0 find the value of the polynomial .
Answer:
Let .
Hence, for x = 0 the value of the polynomial is 5.
Page No 48:
Question 2:
If then find .
Answer:
Page No 48:
Question 3:
If then
Answer:
.....(1)
Also,
.....(2)
Adding (1) and (2), we get
Page No 48:
Question 4:
If then find
Answer:
Page No 53:
Question 1:
Find the value of the polynomial using given values for .
(i) (ii) (iii) x = 0
Answer:
Let .
(i)
Thus, the value of polynomial for x = 3 is −41.
(ii)
Thus, the value of polynomial for x = −1 is 7.
(iii)
Thus, the value of polynomial for x = 0 is 7.
Page No 53:
Question 2:
For each of the following polynomial, find and .
(i) (ii) (iii)
Answer:
(i)
(ii)
(iii)
Page No 53:
Question 3:
If the value of the polynomial is 12 for , then find the value of a.
Answer:
Let
For m = 2, p(2) = 12.
Thus, the value of a is 0.
Page No 53:
Question 4:
For the polynomial if then find m.
Answer:
Let .
Thus, the value of m is 2.
Page No 53:
Question 5:
Divide the first polynomial by the second polynomial and find the remainder using factor theorem .
(i) (ii) (iii)
Answer:
(i)
By synthetic division:
Dividend =
Divisor = x + 1
Opposite of 1 = −1
The coefficient form of the quotient is (1, −8).
∴ Quotient = x − 8
Remainder = 17
By remainder theorem:
Let .
Divisor = x + 1
By remainder theorem,
Remainder = p(−1) =
(ii)
By synthetic division:
Dividend =
Divisor = x − a
Opposite of −a = a
The coefficient form of the quotient is (2, 2a − 2, 2a2â − a).
∴ Quotient = 2x2 + (2a − 2)x + 2a2â − a
Remainder = 2a3â − a2 − a
By remainder theorem:
Let .
Divisor = x − a
By remainder theorem,
Remainder = p(a) =
(iii)
By synthetic division:
Dividend =
Divisor = m − 3
Opposite of −3 = 3
The coefficient form of the quotient is (54, 180, 513).
∴ Quotient = 54x2 + 180x + 513
Remainder = 1544
By remainder theorem:
Let .
Divisor = m − 3
By remainder theorem,
Remainder = p(3) =
Page No 53:
Question 6:
If the polynomial is divided by y + 2 and the remainder is 50 then find the value of m.
Answer:
Let .
When the polynomial is divided by (y + 2), the remainder is 50. This means that the value of the polynomial when y = −2 is 50.
By remainder theorem,
Remainder = p(−2) = 50
Thus, the value of m is 92.
Page No 53:
Question 7:
Use factor theorem to determine whether x + 3 is factor of x 2 + 2x − 3 or not.
Answer:
Let p(x) = x2 + 2x − 3.
Divisor = x + 3
So, by factor theorem, (x + 3) is a factor of x2 + 2x − 3.
Page No 53:
Question 8:
If ( x ) is a factor of then find the value of m.
Answer:
Let .
It is given that (x − 2) is a factor of .
Thus, the value of m is 2.
Page No 53:
Question 9:
By using factor theorem in the following examples, determine whether q ( x ) is a factor p ( x ) or not.
(i)
(ii)
Answer:
(i)
Divisor =
Since p(1) ≠ 0, so by factor theorem is not a factor of polynomial .
(ii)
Divisor =
So, by factor theorem is a factor of polynomial .
Page No 53:
Question 10:
If ( x31 + 31) is divided by (x + 1) then find the remainder.
Answer:
Let p(x) = x31 + 31.
Divisor = x + 1
By remainder theorem, we have
Remainder = p(−1) = (−1)31 + 31 = −1 + 31 = 30
Thus, the remainder when (x31 + 31) is divided by (x + 1) is 30.
Page No 53:
Question 11:
Show that m 1 is a factor of m21 1 and m22 1.
Answer:
Let p(m) = m21 − 1 and q(m) = m22 − 1.
Divisor = m − 1
Now,
p(1) = (1)21 − 1 = 1 − 1 = 0
Therefore, by factor theorem (m − 1) is a factor of p(m) = m21 − 1.
Also,
q(1) = (1)22 − 1 = 1 − 1 = 0
Therefore, by factor theorem (m − 1) is a factor of q(m) = m22 − 1.
Hence, (m − 1) is a factor of m21 − 1 and m22 − 1.
Page No 53:
Question 12:
If and both are the factors of the polynomial nx2 − 5x + m, then show that
Answer:
Let p(x) = nx2 − 5x + m.
It given that (x − 2) and are the factors of the polynomial p(x) = nx2 − 5x + m.
∴ By factor theorem, p(2) = 0 and .
Also,
From (1) and (2), we have
4n + m = n + 4m
⇒ 4n − n = 4m − m
⇒ 3n = 3m
⇒ n = m
Putting n = m in (1), we have
4m + m = 10
⇒ 5m = 10
⇒ m = 2
∴ n = m = 2
Page No 53:
Question 13:
(i) If then .
(ii) then .
Answer:
(i)
(ii)
Page No 54:
Question 1:
Find the factors of the polynomials given below.
(i) 2x2 + x – 1
(ii) 2m2 + 5m – 3
(iii) 12x2 + 61x + 77
(iv) 3y2 – 2y – 1
(v)
(vi)
Answer:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
Page No 55:
Question 2:
Factorize the following polynomials.
(i) (x2 – x)2 – 8 (x2 – x) + 12
(ii) (x – 5)2 – (5x – 25) – 24
(iii) (x2 – 6x)2 – 8 (x2 – 6x + 8) – 64
(iv) (x2 – 2x + 3) (x2 – 2x + 5) – 35
(v) (y + 2) (y – 3) (y + 8) (y + 3) + 56
(vi) (y2 + 5y) (y2 + 5y – 2) – 24
(vii) (x – 3) (x – 4)2 (x – 5) – 6
Answer:
(i)
(x2 – x)2 – 8(x2 – x) + 12
Let x2 – x = z.
(ii)
(x – 5)2 – (5x – 25) – 24
= (x – 5)2 – 5(x – 5) – 24
Let x – 5 = z.
(iii)
(x2 – 6x)2 – 8(x2 – 6x + 8) – 64
= (x2 – 6x)2 – 8(x2 – 6x) – 64 – 64
= (x2 – 6x)2 – 8(x2 – 6x) – 128
Let x2 – 6x = z.
(iv)
(x2 – 2x + 3)(x2 – 2x + 5) – 35
Let x2 – 2x = z.
(v)
(y + 2)(y – 3)(y + 8)(y + 3) + 56
= (y + 2)(y + 3)(y + 8)(y – 3) + 56
= (y2 + 5y + 6)(y2 + 5y – 24) + 56
Let y2 + 5y = z.
(vi)
(y2 + 5y)(y2 + 5y – 2) – 24
Let y2 + 5y = z.
(vii)
(x – 3)(x – 4)2(x – 5) – 6
= (x – 3)(x – 5)(x – 4)2 – 6
= (x2 – 8x + 15)(x2 – 8x + 16) – 6
Let x2 – 8x = z.
Page No 55:
Question 1:
(A) (B) (C) (D)
(ii) What is the degree of the polynomial ?
(A) (B) 5 (C) 2 (D) 0
(iii) What is the degree of the 0 polynomial ?
(iv) What is the degree of the polynomial 2x2 + 5x3 + 7?
(A) 3 (B) 2 (C) 5 (D) 7
(v) What is the coefficient form of ?
(A) (1, 1) (B) (3,1) (C) (1, 0, 0, 1) (D) (1, 3, 1)
(vi) then
(A) 3 (B) (C) (D)
(vii) When , what is the value of the polynomial 2x3 + 2x ?
(A) 4 (B) 2 (C) (D)
(viii) If , what is a factor of the polynomial then find the value of m.
(A) 2 (B) 2 (C) (D) 3
(ix) Multiply ( x23) (2x 7x 3 + 4) and write the degree of the product.
(A) 5 (B) 3 (C) 2 (D) 0
(x) Which of the following is a linear polynomial ?
Answer:
(i)
In an algebraic expression, if the powers of the variables are whole numbers then the algebraic expression is a polynomial.
In the expression , the power of the variable x is 2, which is a whole number. So, the expression is a polynomial.
Hence, the correct answer is option (D).
(ii)
The degree of the polynomial is 0.
Hence, the correct answer is option (D).
(iii)
The degree of 0 polynomial is not defined.
Hence, the correct answer is option (C).
(iv)
The highest power of the variable in a polynomial is called the degree of the polynomial.
The degree of the polynomial 2x2 + 5x3 + 7 is 3.
Hence, the correct answer is option (A).
(v)
The coefficient form of the polynomial is (1, 0, 0, −1).
Hence, the correct answer is option (C).
(vi)
Hence, the correct answer is option (A).
(vii)
Let p(x) = 2x3 + 2x
∴ p(−1) = 2 × (−1)3 + 2 × (−1) = 2 × (−1) − 2 = −2 − 2 = −4
Thus, the value of the polynomial when x = −1 is −4.
Hence, the correct answer is option (D).
(viii)
Let p(x) = 3x2 + mx.
(x − 1) is a factor of p(x).
∴ p(1) = 0
⇒ 3 × (1)2 + m × 1 = 0
⇒ 3 + m = 0
⇒ m = −3
Hence, the correct answer is option (C).
(ix)
Thus, the degree of the resultant polynomial is 5.
Hence, the correct answer is option (A).
(x)
A polynomial with degree one is called a linear polynomial.
Thus, the polynomial x + 5 is a linear polynomial.
Hence, the correct answer is option (A).
Page No 56:
Question 2:
Write the degree of the polynomial for each of the following.
Answer:
The highest power of the variable in a polynomial in one variable is called the degree of the polynomial.
(i)
The degree of the polynomial 5 + 3x4 is 4.
(ii)
7 = 7 × 1 = 7x0
The degree of the constant polynomial 7 is 0.
(iii)
The degree of the polynomial ax7 + bx9 is 9.
Page No 56:
Question 3:
Answer:
A polynomial written in either descending or ascending power of its variable is called the standard form of the polynomial.
(i)
The given polynomial is 4x2 + 7x4 − x3 − x + 9.
The standard form of the polynomial is 7x4− x3 + 4x2 − x + 9 or 9 − x + 4x2 − x3 + 7x4.
(ii)
The given polynomial is p + 2p3 + 10p2 + 5p4 − 8.
The standard form of the polynomial is 5p4 + 2p3 + 10p2 + p − 8 or −8 + p + 10p2 + 2p3 + 5p4.
Page No 56:
Question 4:
Answer:
(i)
Therefore, the given polynomial in coefficient form is (1, 0, 0, 0, 16).
(ii)
Therefore, the given polynomial in coefficient form is (1, 0, 0, 2, 3, 15).
Page No 56:
Question 5:
Answer:
(i)
The coefficient form of the polynomial is (3, −2, 0, 7, 18).
Therefore, the index form the polynomial is 3x4 − 2x3 + 0x2 + 7x + 18 or 3x4 − 2x3 + 7x + 18.
(ii)
The coefficient form of the polynomial is (6, 1, 0, 7).
Therefore, the index form the polynomial is 6x3 + x2 + 0x + 7 or 6x3 + x2 + 7.
(iii)
The coefficient form of the polynomial is (4, 5, −3, 0).
Therefore, the index form the polynomial is 4x3 + 5x2 − 3x + 0 or 4x3 + 5x2 − 3x.
Page No 56:
Question 6:
Answer:
(i)
(ii)
Page No 56:
Question 7:
Answer:
(i)
(ii)
Page No 56:
Question 8:
(ii) (5m32)(m2 m + 3)
Answer:
(i)
(ii)
Page No 56:
Question 9:
Answer:
Dividend =
Divisor =
Opposite of −3 = 3
The coefficient form of the quotient is (3, 1, 4).
∴ Quotient = 3x2 + x + 4 and Remainder = 19
Page No 56:
Question 10:
For which the value of m, x + 3 is the factor of the polynomial x3 2mx + 21?
Answer:
Let .
(x + 3) is the factor of the polynomial .
Thus, the value of m is 1.
Page No 56:
Question 11:
At the end of the year 2016, the population of villages Kovad, Varud, Chikhali is 5x2 3 y2 , 7 y2 + 2 xy and 9 x2 + 4 xy respectively. At the beginning of the year 2017, x2 + xy y2 , 5 xy and 3 x2 + xy persons from each of the three villages respectively went to another village for education then what is the remaining total population of these three villages ?
Answer:
Total population of the three villages
Total number of persons who went to another village for education
∴ Remaining total population of the three villages = Total population of the three villages − Total number of persons who went to another village for education
Thus, the remaining total population of these three villages is 10x2 + 5y2 − xy.
Page No 56:
Question 12:
Polynomials bx2 + x + 5 and bx32x + 5 are divided by polynomial x3 and the remainders are m and n respectively. If m n = 0 then find the value of b.
Answer:
Let and .
The remainder when is divided by (x − 3) is m.
By remainder theorem,
Remainder =
The remainder when is divided by (x − 3) is n.
By remainder theorem,
Remainder =
Now,
Thus, the value of b is .
Page No 56:
Question 13:
Simplify
(8m2 + 3m6) (9m 7) + (3m2 2m + 4)
Answer:
Page No 56:
Question 14:
Which polynomial is to be subtracted from x2 + 13x + 7 to get the polynomial 3x2 + 5x - 4?
Answer:
Let p(x) be the polynomial which is to be subtracted from x2 + 13x + 7 to get the polynomial 3x2 + 5x − 4.
∴ (x2 + 13x + 7) − p(x) = 3x2 + 5x − 4
⇒ p(x) = (x2 + 13x + 7) − (3x2 + 5x − 4)
⇒ p(x) = x2 + 13x + 7 − 3x2 − 5x + 4
⇒ p(x) = x2 − 3x2 + 13x − 5x + 7 + 4
⇒ p(x) = −2x2 + 8x + 11
Thus, the required polynomial is −2x2 + 8x + 11.
Page No 56:
Question 15:
Which polynomial is to be added to 4m + 2n + 3 to get the polynomial 6m + 3n + 10?
Answer:
The required polynomial can be obtained by subtracting the polynomial 4m + 2n + 3 from 6m + 3n + 10.
∴ Required polynomial
= (6m + 3n + 10) − (4m + 2n + 3)
= 6m + 3n + 10 − 4m − 2n − 3
= 6m − 4m + 3n − 2n + 10 − 3
= 2m + n + 7
Thus, the polynomial 2m + n + 7 is to be added to 4m + 2n + 3 to get the polynomial 6m + 3n + 10.
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