RS Aggarwal (2021 2022) Solutions for Class 9 Maths Chapter 2 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 RS Aggarwal (2021 2022) Book of class 9 Maths Chapter 2 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RS Aggarwal (2021 2022) Solutions. All RS Aggarwal (2021 2022) Solutions for class 9 Maths are prepared by experts and are 100% accurate.
Page No 74:
Question 1:
Answer:
(i) is an expression having only non-negative integral powers of x. So, it is a polynomial. Also, the highest power of x is 5, so, it is a polynomial of degree 5.
(ii) is an expression having only non-negative integral powers of y. So, it is a polynomial. Also, the highest power of y is 3, so, it is a polynomial of degree 3.
(iii) is an expression having only non-negative integral powers of t. So, it is a polynomial. Also, the highest power of t is 2, so, it is a polynomial of degree 2.
(iv) is an expression having only non-negative integral power of x. So, it is a polynomial. Also, the highest power of x is 100, so, it is a polynomial of degree 100.
(v) is an expression having only non-negative integral powers of x. So, it is a polynomial. Also, the highest power of x is 2, so, it is a polynomial of degree 2.
(vi) is an expression having negative integral powers of x. So, it is not a polynomial.
(vii) Clearly, 1 is a constant polynomial of degree 0.
(viii) Clearly, is a constant polynomial of degree 0.
(ix)
This is an expression having negative integral power of x i.e. −2. So, it is not a polynomial.
(x) is an expression having only non-negative integral power of x. So, it is a polynomial. Also, the highest power of x is 2, so, it is a polynomial of degree 2.
(xi) is an expression having negative integral power of x. So, it is not a polynomial.
(xii)
In this expression, the power of x is which is a fraction. Since it is an expression having fractional power of x, so, it is not a polynomial.
(xiii) is an expression having only non-negative integral powers of x. So, it is a polynomial. Also, the highest power of x is 2, so, it is a polynomial of degree 2.
(xiv)
In this expression, one of the powers of x is which is a fraction. Since it is an expression having fractional power of x, so, it is not a polynomial.
(xv)
In this expression, one of the powers of x is which is a fraction. Since it is an expression having fractional power of x, so, it is not a polynomial.
Page No 75:
Question 2:
(i) is an expression having only non-negative integral powers of x. So, it is a polynomial. Also, the highest power of x is 5, so, it is a polynomial of degree 5.
(ii) is an expression having only non-negative integral powers of y. So, it is a polynomial. Also, the highest power of y is 3, so, it is a polynomial of degree 3.
(iii) is an expression having only non-negative integral powers of t. So, it is a polynomial. Also, the highest power of t is 2, so, it is a polynomial of degree 2.
(iv) is an expression having only non-negative integral power of x. So, it is a polynomial. Also, the highest power of x is 100, so, it is a polynomial of degree 100.
(v) is an expression having only non-negative integral powers of x. So, it is a polynomial. Also, the highest power of x is 2, so, it is a polynomial of degree 2.
(vi) is an expression having negative integral powers of x. So, it is not a polynomial.
(vii) Clearly, 1 is a constant polynomial of degree 0.
(viii) Clearly, is a constant polynomial of degree 0.
(ix)
This is an expression having negative integral power of x i.e. −2. So, it is not a polynomial.
(x) is an expression having only non-negative integral power of x. So, it is a polynomial. Also, the highest power of x is 2, so, it is a polynomial of degree 2.
(xi) is an expression having negative integral power of x. So, it is not a polynomial.
(xii)
In this expression, the power of x is which is a fraction. Since it is an expression having fractional power of x, so, it is not a polynomial.
(xiii) is an expression having only non-negative integral powers of x. So, it is a polynomial. Also, the highest power of x is 2, so, it is a polynomial of degree 2.
(xiv)
In this expression, one of the powers of x is which is a fraction. Since it is an expression having fractional power of x, so, it is not a polynomial.
(xv)
In this expression, one of the powers of x is which is a fraction. Since it is an expression having fractional power of x, so, it is not a polynomial.
Answer:
(i) –7 + x is a polynomial with degree 1. So, it is a linear polynomial.
(ii) 6y is a polynomial with degree 1. So, it is a linear polynomial.
(iii) –z3 is a polynomial with degree 3. So, it is a cubic polynomial.
(iv) 1 – y – y3 is a polynomial with degree 3. So, it is a cubic polynomial.
(v) x – x3 + x4 is a polynomial with degree 4. So, it is a quartic polynomial.
(vi) 1 + x + x2 is a polynomial with degree 2. So, it is a quadratic polynomial.
(vii) – 6x2 is a polynomial with degree 2. So, it is a quadratic polynomial.
(viii) –13 is a polynomial with degree 0. So, it is a constant polynomial.
(ix) – p is a polynomial with degree 1. So, it is a linear polynomial.
Page No 75:
Question 3:
(i) –7 + x is a polynomial with degree 1. So, it is a linear polynomial.
(ii) 6y is a polynomial with degree 1. So, it is a linear polynomial.
(iii) –z3 is a polynomial with degree 3. So, it is a cubic polynomial.
(iv) 1 – y – y3 is a polynomial with degree 3. So, it is a cubic polynomial.
(v) x – x3 + x4 is a polynomial with degree 4. So, it is a quartic polynomial.
(vi) 1 + x + x2 is a polynomial with degree 2. So, it is a quadratic polynomial.
(vii) – 6x2 is a polynomial with degree 2. So, it is a quadratic polynomial.
(viii) –13 is a polynomial with degree 0. So, it is a constant polynomial.
(ix) – p is a polynomial with degree 1. So, it is a linear polynomial.
Answer:
(i) The coefficient of x3 in is −5.
(ii) The coefficient of x in is .
(iii) 2x – 3 + x3 = – 3 + 2x + 0x2 + x3
The coefficient of x2 in 2x – 3 + x3 is 0.
(iv) The coefficient of x in is .
(v) The constant term in is .
Page No 75:
Question 4:
(i) The coefficient of x3 in is −5.
(ii) The coefficient of x in is .
(iii) 2x – 3 + x3 = – 3 + 2x + 0x2 + x3
The coefficient of x2 in 2x – 3 + x3 is 0.
(iv) The coefficient of x in is .
(v) The constant term in is .
Answer:
(i)
Here, the highest power of x is 2. So, the degree of the polynomial is 2.
(ii) y2(y – y3) = y3 – y5
Here, the highest power of y is 5. So, the degree of the polynomial is 5.
(iii) (3x – 2)(2x3 + 3x2) = 6x4 + 9x3 – 4x3 – 6x2 = 6x4 + 5x3 – 6x2
Here, the highest power of x is 4. So, the degree of the polynomial is 4.
(iv)
Here, the highest power of x is 1. So, the degree of the polynomial is 1.
(v) – 8
–8 is a constant polynomial. So, the degree of the polynomial is 0.
(vi) x–2(x4 + x2) = x2 + x0 = x2 + 1
Here, the highest power of x is 2. So, the degree of the polynomial is 2.
Page No 75:
Question 5:
(i)
Here, the highest power of x is 2. So, the degree of the polynomial is 2.
(ii) y2(y – y3) = y3 – y5
Here, the highest power of y is 5. So, the degree of the polynomial is 5.
(iii) (3x – 2)(2x3 + 3x2) = 6x4 + 9x3 – 4x3 – 6x2 = 6x4 + 5x3 – 6x2
Here, the highest power of x is 4. So, the degree of the polynomial is 4.
(iv)
Here, the highest power of x is 1. So, the degree of the polynomial is 1.
(v) – 8
–8 is a constant polynomial. So, the degree of the polynomial is 0.
(vi) x–2(x4 + x2) = x2 + x0 = x2 + 1
Here, the highest power of x is 2. So, the degree of the polynomial is 2.
Answer:
(i) A polynomial having one term is called a monomial. Since the degree of required monomial is 5, so the highest power of x in the monomial should be 5.
An example of a monomial of degree 5 is 2x5.
(ii) A polynomial having two terms is called a binomial. Since the degree of required binomial is 8, so the highest power of x in the binomial should be 8.
An example of a binomial of degree 8 is 2x8 − 3x.
(iii) A polynomial having three terms is called a trinomial. Since the degree of required trinomial is 4, so the highest power of x in the trinomial should be 4.
An example of a trinomial of degree 4 is 2x4 − 3x + 5.
(iv) A polynomial having one term is called a monomial. Since the degree of required monomial is 0, so the highest power of x in the monomial should be 0.
An example of a monomial of degree 0 is 5.
Page No 75:
Question 6:
(i) A polynomial having one term is called a monomial. Since the degree of required monomial is 5, so the highest power of x in the monomial should be 5.
An example of a monomial of degree 5 is 2x5.
(ii) A polynomial having two terms is called a binomial. Since the degree of required binomial is 8, so the highest power of x in the binomial should be 8.
An example of a binomial of degree 8 is 2x8 − 3x.
(iii) A polynomial having three terms is called a trinomial. Since the degree of required trinomial is 4, so the highest power of x in the trinomial should be 4.
An example of a trinomial of degree 4 is 2x4 − 3x + 5.
(iv) A polynomial having one term is called a monomial. Since the degree of required monomial is 0, so the highest power of x in the monomial should be 0.
An example of a monomial of degree 0 is 5.
Answer:
A polynomial written either in ascending or descending powers of a variable is called the standard form of a polynomial.
(i) is a polynomial in standard form as the powers of x are in ascending order.
(ii) is a polynomial in standard form as the powers of y are in ascending order.
(iii) is a polynomial in standard form as the powers of x are in ascending order.
(iv) is a polynomial in standard form as the powers of t are in ascending order.
Page No 78:
Question 1:
A polynomial written either in ascending or descending powers of a variable is called the standard form of a polynomial.
(i) is a polynomial in standard form as the powers of x are in ascending order.
(ii) is a polynomial in standard form as the powers of y are in ascending order.
(iii) is a polynomial in standard form as the powers of x are in ascending order.
(iv) is a polynomial in standard form as the powers of t are in ascending order.
Answer:
Page No 78:
Question 2:
Answer:
Page No 78:
Question 3:
Answer:
Page No 78:
Question 4:
Answer:
.....(1)
Putting x = 0 in (1), we get
Thus, x = 0 is a zero of p(x).
Putting x = 1 in (1), we get
Thus, x = 1 is a zero of p(x).
Putting x = 2 in (1), we get
Thus, x = 2 is a zero of p(x).
Page No 78:
Question 5:
.....(1)
Putting x = 0 in (1), we get
Thus, x = 0 is a zero of p(x).
Putting x = 1 in (1), we get
Thus, x = 1 is a zero of p(x).
Putting x = 2 in (1), we get
Thus, x = 2 is a zero of p(x).
Answer:
p(x) = x3 + x2 – 9x – 9 .....(1)
Putting x = 0 in (1), we get
p(0) = 03 + 02 – 9 × 0 – 9 = 0 + 0 – 0 – 9 = –9 ≠ 0
Thus, x = 0 is not a zero of p(x).
Putting x = 3 in (1), we get
p(3) = 33 + 32 – 9 × 3 – 9 = 27 + 9 – 27 – 9 = 0
Thus, x = 3 is a zero of p(x).
Putting x = –3 in (1), we get
p(–3) = (–3)3 + (–3)2 – 9 × (–3) – 9 = –27 + 9 + 27 – 9 = 0
Thus, x = –3 is a zero of p(x).
Putting x = –1 in (1), we get
p(–1) = (–1)3 + (–1)2 – 9 × (–1) – 9 = –1 + 1 + 9 – 9 = 0
Thus, x = –1 is a zero of p(x).
Page No 78:
Question 6:
p(x) = x3 + x2 – 9x – 9 .....(1)
Putting x = 0 in (1), we get
p(0) = 03 + 02 – 9 × 0 – 9 = 0 + 0 – 0 – 9 = –9 ≠ 0
Thus, x = 0 is not a zero of p(x).
Putting x = 3 in (1), we get
p(3) = 33 + 32 – 9 × 3 – 9 = 27 + 9 – 27 – 9 = 0
Thus, x = 3 is a zero of p(x).
Putting x = –3 in (1), we get
p(–3) = (–3)3 + (–3)2 – 9 × (–3) – 9 = –27 + 9 + 27 – 9 = 0
Thus, x = –3 is a zero of p(x).
Putting x = –1 in (1), we get
p(–1) = (–1)3 + (–1)2 – 9 × (–1) – 9 = –1 + 1 + 9 – 9 = 0
Thus, x = –1 is a zero of p(x).
Answer:
= 0
Hence, 4 is the zero of the given polynomial.
Hence, 3 is the zero of the given polynomial.
Hence, is the zero of the given polynomial.
Hence, is the zero of the given polynomial.
Page No 79:
Question 7:
= 0
Hence, 4 is the zero of the given polynomial.
Hence, 3 is the zero of the given polynomial.
Hence, is the zero of the given polynomial.
Hence, is the zero of the given polynomial.
Answer:
Also,
Hence, 1 and 2 are the zeroes of the given polynomial.
Also,
Hence, 2 and are the zeroes of the given polynomial.
Also,
Hence, 0 and 3 are the zeroes of the given polynomial.
Page No 79:
Question 8:
Also,
Hence, 1 and 2 are the zeroes of the given polynomial.
Also,
Hence, 2 and are the zeroes of the given polynomial.
Also,
Hence, 0 and 3 are the zeroes of the given polynomial.
Answer:
Page No 79:
Question 9:
Answer:
It is given that 2 and 0 are the zeroes of the polynomial .
∴ f(2) = 0
Also,
f(0) = 0
Putting b = 0 in (1), we get
Thus, the values of a and b are 2 and 0, respectively.
Page No 84:
Question 1:
It is given that 2 and 0 are the zeroes of the polynomial .
∴ f(2) = 0
Also,
f(0) = 0
Putting b = 0 in (1), we get
Thus, the values of a and b are 2 and 0, respectively.
Answer:
Let f(x) = x4 + 1 and g(x) = x – 1.
Quotient = x3 + x2 + x + 1
Remainder = 2
Verification:
Putting x = 1 in f(x), we get
f(1) = 14 + 1 = 1 + 1 = 2 = Remainder, when f(x) = x4 + 1 is divided by g(x) = x – 1
Page No 84:
Question 2:
Let f(x) = x4 + 1 and g(x) = x – 1.
Quotient = x3 + x2 + x + 1
Remainder = 2
Verification:
Putting x = 1 in f(x), we get
f(1) = 14 + 1 = 1 + 1 = 2 = Remainder, when f(x) = x4 + 1 is divided by g(x) = x – 1
Answer:
and
![](/img/site_content/ask-answer/loader.gif)
Quotient =
Remainder = 92
Verification:
Divisor × Quotient + Remainder
= Dividend
Hence verified.
Page No 84:
Question 3:
and
![](/img/site_content/ask-answer/loader.gif)
Quotient =
Remainder = 92
Verification:
Divisor × Quotient + Remainder
= Dividend
Hence verified.
Answer:
â
By remainder theorem, when p(x) is divided by (x − 1), then the remainder = p(1).
Putting x = 1 in p(x), we get
∴ Remainder = 7
Thus, the remainder when p(x) is divided by g(x) is 7.
Page No 84:
Question 4:
â
By remainder theorem, when p(x) is divided by (x − 1), then the remainder = p(1).
Putting x = 1 in p(x), we get
∴ Remainder = 7
Thus, the remainder when p(x) is divided by g(x) is 7.
Answer:
â
By remainder theorem, when p(x) is divided by (x − 3), then the remainder = p(3).
Putting x = 3 in p(x), we get
∴ Remainder = 5
Thus, the remainder when p(x) is divided by g(x) is 5.
Page No 84:
Question 5:
â
By remainder theorem, when p(x) is divided by (x − 3), then the remainder = p(3).
Putting x = 3 in p(x), we get
∴ Remainder = 5
Thus, the remainder when p(x) is divided by g(x) is 5.
Answer:
â
By remainder theorem, when p(x) is divided by (x − 2), then the remainder = p(2).
Putting x = 2 in p(x), we get
∴ Remainder = 6
Thus, the remainder when p(x) is divided by g(x) is 6.
Page No 84:
Question 6:
â
By remainder theorem, when p(x) is divided by (x − 2), then the remainder = p(2).
Putting x = 2 in p(x), we get
∴ Remainder = 6
Thus, the remainder when p(x) is divided by g(x) is 6.
Answer:
â
By remainder theorem, when p(x) is divided by (2x − 3), then the remainder = .
Putting in p(x), we get
∴ Remainder = 3
Thus, the remainder when p(x) is divided by g(x) is 3.
Page No 84:
Question 7:
â
By remainder theorem, when p(x) is divided by (2x − 3), then the remainder = .
Putting in p(x), we get
∴ Remainder = 3
Thus, the remainder when p(x) is divided by g(x) is 3.
Answer:
â
By remainder theorem, when p(x) is divided by (x + 1), then the remainder = p(−1).
Putting x = −1 in p(x), we get
∴ Remainder = 4
Thus, the remainder when p(x) is divided by g(x) is 4.
Page No 84:
Question 8:
â
By remainder theorem, when p(x) is divided by (x + 1), then the remainder = p(−1).
Putting x = −1 in p(x), we get
∴ Remainder = 4
Thus, the remainder when p(x) is divided by g(x) is 4.
Answer:
â
By remainder theorem, when p(x) is divided by (x + 2), then the remainder = p(−2).
Putting x = −2 in p(x), we get
∴ Remainder = 6
Thus, the remainder when p(x) is divided by g(x) is 6.
Page No 84:
Question 9:
â
By remainder theorem, when p(x) is divided by (x + 2), then the remainder = p(−2).
Putting x = −2 in p(x), we get
∴ Remainder = 6
Thus, the remainder when p(x) is divided by g(x) is 6.
Answer:
â
By remainder theorem, when p(x) is divided by (3x + 2), then the remainder = .
Putting in p(x), we get
∴ Remainder = 7
Thus, the remainder when p(x) is divided by g(x) is 7.
Page No 84:
Question 10:
â
By remainder theorem, when p(x) is divided by (3x + 2), then the remainder = .
Putting in p(x), we get
∴ Remainder = 7
Thus, the remainder when p(x) is divided by g(x) is 7.
Answer:
â
By remainder theorem, when p(x) is divided by , then the remainder = .
Putting in p(x), we get
∴ Remainder =
Thus, the remainder when p(x) is divided by g(x) is .
Page No 84:
Question 11:
â
By remainder theorem, when p(x) is divided by , then the remainder = .
Putting in p(x), we get
∴ Remainder =
Thus, the remainder when p(x) is divided by g(x) is .
Answer:
â
By remainder theorem, when p(x) is divided by , then the remainder = .
Putting in p(x), we get
∴ Remainder = 3
Thus, the remainder when p(x) is divided by g(x) is 3.
Page No 84:
Question 12:
â
By remainder theorem, when p(x) is divided by , then the remainder = .
Putting in p(x), we get
∴ Remainder = 3
Thus, the remainder when p(x) is divided by g(x) is 3.
Answer:
â
By remainder theorem, when p(x) is divided by (x − a), then the remainder = p(a).
Putting x = a in p(x), we get
∴ Remainder = 5a
Thus, the remainder when p(x) is divided by g(x) is 5a.
Page No 84:
Question 13:
â
By remainder theorem, when p(x) is divided by (x − a), then the remainder = p(a).
Putting x = a in p(x), we get
∴ Remainder = 5a
Thus, the remainder when p(x) is divided by g(x) is 5a.
Answer:
Let and .
By remainder theorem, when f(x) is divided by (x – 2), then the remainder = f(2).
Putting x = 2 in f(x), we get
By remainder theorem, when g(x) is divided by (x – 2), then the remainder = g(2).
Putting x = 2 in g(x), we get
It is given that,
Thus, the value of a is 8.
Page No 84:
Question 14:
Let and .
By remainder theorem, when f(x) is divided by (x – 2), then the remainder = f(2).
Putting x = 2 in f(x), we get
By remainder theorem, when g(x) is divided by (x – 2), then the remainder = g(2).
Putting x = 2 in g(x), we get
It is given that,
Thus, the value of a is 8.
Answer:
Thus, we have:
And,
Now,
By putting the value of b, we get the value of a, i.e., 5.
∴ a = 5 and b = 8
Now,
Also,
Thus, we have:
Page No 85:
Question 15:
Thus, we have:
And,
Now,
By putting the value of b, we get the value of a, i.e., 5.
∴ a = 5 and b = 8
Now,
Also,
Thus, we have:
Answer:
Putting x = 2 in p(x), we get
Therefore, by factor theorem, (x − 2) is not a factor of p(x).
Hence, p(x) is not a multiple of g(x).
Page No 85:
Question 16:
Putting x = 2 in p(x), we get
Therefore, by factor theorem, (x − 2) is not a factor of p(x).
Hence, p(x) is not a multiple of g(x).
Answer:
Putting in p(x), we get
Therefore, by factor theorem, (2x + 1) is not a factor of p(x).
Hence, g(x) is not a factor of p(x).
Page No 90:
Question 1:
Putting in p(x), we get
Therefore, by factor theorem, (2x + 1) is not a factor of p(x).
Hence, g(x) is not a factor of p(x).
Answer:
Let:
p(x) = x3 – 8
Now,
By the factor theorem, (x – 2) is a factor of the given polynomial if p(2) = 0.
Thus, we have:
Hence, (x 2) is a factor of the given polynomial.
Page No 90:
Question 2:
Let:
p(x) = x3 – 8
Now,
By the factor theorem, (x – 2) is a factor of the given polynomial if p(2) = 0.
Thus, we have:
Hence, (x 2) is a factor of the given polynomial.
Answer:
Let:
p(x) = 2x3 + 7x2 – 24x – 45
Now,
By the factor theorem, (x 3) is a factor of the given polynomial if p(3) = 0.
Thus, we have:
Hence, (x 3) is a factor of the given polynomial.
Page No 90:
Question 3:
Let:
p(x) = 2x3 + 7x2 – 24x – 45
Now,
By the factor theorem, (x 3) is a factor of the given polynomial if p(3) = 0.
Thus, we have:
Hence, (x 3) is a factor of the given polynomial.
Answer:
Let:
p(x) = 2x4 + 9x3 + 6x2 – 11x – 6
Here,
â
By the factor theorem, (x 1) is a factor of the given polynomial if p(1) = 0.
Thus, we have:
Hence, (x 1) is a factor of the given polynomial.
Page No 91:
Question 4:
Let:
p(x) = 2x4 + 9x3 + 6x2 – 11x – 6
Here,
â
By the factor theorem, (x 1) is a factor of the given polynomial if p(1) = 0.
Thus, we have:
Hence, (x 1) is a factor of the given polynomial.
Answer:
Let:
p(x) = x4 – x2 – 12
Here,
By the factor theorem, (x + 2) is a factor of the given polynomial if p (2) = 0.
Thus, we have:
Hence, (x + 2) is a factor of the given polynomial.
Page No 91:
Question 5:
Let:
p(x) = x4 – x2 – 12
Here,
By the factor theorem, (x + 2) is a factor of the given polynomial if p (2) = 0.
Thus, we have:
Hence, (x + 2) is a factor of the given polynomial.
Answer:
Putting x = −3 in p(x), we get
Therefore, by factor theorem, (x + 3) is a factor of p(x).
Hence, g(x) is a factor of p(x).
Page No 91:
Question 6:
Putting x = −3 in p(x), we get
Therefore, by factor theorem, (x + 3) is a factor of p(x).
Hence, g(x) is a factor of p(x).
Answer:
Let:
Here,
By the factor theorem, (x + 5) is a factor of the given polynomial if p (5) = 0.
Thus, we have:
Hence, (x + 5) is a factor of the given polynomial.
Page No 91:
Question 7:
Let:
Here,
By the factor theorem, (x + 5) is a factor of the given polynomial if p (5) = 0.
Thus, we have:
Hence, (x + 5) is a factor of the given polynomial.
Answer:
Let:
Here,
By the factor theorem, (2x 3) is a factor of the given polynomial if .
Thus, we have:
Hence, (2x 3) is a factor of the given polynomial.
Page No 91:
Question 8:
Let:
Here,
By the factor theorem, (2x 3) is a factor of the given polynomial if .
Thus, we have:
Hence, (2x 3) is a factor of the given polynomial.
Answer:
Putting in p(x), we get
Therefore, by factor theorem, (3x − 2) is a factor of p(x).
Hence, g(x) is a factor of p(x).
Page No 91:
Question 9:
Putting in p(x), we get
Therefore, by factor theorem, (3x − 2) is a factor of p(x).
Hence, g(x) is a factor of p(x).
Answer:
Let:
Here,
By the factor theorem, is a factor of the given polynomial if
Thus, we have:
Hence, is a factor of the given polynomial.
Page No 91:
Question 10:
Let:
Here,
By the factor theorem, is a factor of the given polynomial if
Thus, we have:
Hence, is a factor of the given polynomial.
Answer:
Let:
Here,
By the factor theorem, will be a factor of the given polynomial if = 0.
Thus, we have:
Hence, is a factor of the given polynomial.
Page No 91:
Question 11:
Let:
Here,
By the factor theorem, will be a factor of the given polynomial if = 0.
Thus, we have:
Hence, is a factor of the given polynomial.
Answer:
Let f(p) = p10 – 1 and g(p) = p11 – 1.
Putting p = 1 in f(p), we get
f(1) = 110 − 1 = 1 − 1 = 0
Therefore, by factor theorem, (p – 1) is a factor of (p10 – 1).
Now, putting p = 1 in g(p), we get
g(1) = 111 − 1 = 1 − 1 = 0
Therefore, by factor theorem, (p – 1) is a factor of (p11 – 1).
Page No 91:
Question 12:
Let f(p) = p10 – 1 and g(p) = p11 – 1.
Putting p = 1 in f(p), we get
f(1) = 110 − 1 = 1 − 1 = 0
Therefore, by factor theorem, (p – 1) is a factor of (p10 – 1).
Now, putting p = 1 in g(p), we get
g(1) = 111 − 1 = 1 − 1 = 0
Therefore, by factor theorem, (p – 1) is a factor of (p11 – 1).
Answer:
Let:
Hence, the required value of k is 12.
Page No 91:
Question 13:
Let:
Hence, the required value of k is 12.
Answer:
Let:
Hence, the required value of a is 8.
Page No 91:
Question 14:
Let:
Hence, the required value of a is 8.
Answer:
Let f(x) = ax3 + x2 – 2x + 4a – 9
It is given that (x + 1) is a factor of f(x).
Using factor theorem, we have
f(−1) = 0
Thus, the value of a is 2.
Page No 91:
Question 15:
Let f(x) = ax3 + x2 – 2x + 4a – 9
It is given that (x + 1) is a factor of f(x).
Using factor theorem, we have
f(−1) = 0
Thus, the value of a is 2.
Answer:
Let f(x) = x5 – 4a2 x3 + 2x + 2a +3
It is given that (x + 2a) is a factor of f(x).
Using factor theorem, we have
f(−2a) = 0
Thus, the value of a is .
Page No 91:
Question 16:
Let f(x) = x5 – 4a2 x3 + 2x + 2a +3
It is given that (x + 2a) is a factor of f(x).
Using factor theorem, we have
f(−2a) = 0
Thus, the value of a is .
Answer:
Let f(x) = 8x4 + 4x3 – 16x2 + 10x + m
It is given that is a factor of f(x).
Using factor theorem, we have
Thus, the value of m is −2.
Page No 91:
Question 17:
Let f(x) = 8x4 + 4x3 – 16x2 + 10x + m
It is given that is a factor of f(x).
Using factor theorem, we have
Thus, the value of m is −2.
Answer:
Let:
Now,
By the factor theorem, .
Thus, we have:
Also,
Hence,
Page No 91:
Question 18:
Let:
Now,
By the factor theorem, .
Thus, we have:
Also,
Hence,
Answer:
Let:
And,
Now, .
For this, we must have:
Thus, we have:
And,
. So, .
Hence, .
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Question 19:
Let:
And,
Now, .
For this, we must have:
Thus, we have:
And,
. So, .
Hence, .
Answer:
Let:
Now,
By the factor theorem, we can say:
Now,
Thus, we have:
Subtracting (1) from (2), we get:
a = 3
By putting the value of a in (1), we get the value of b, i.e., 1.
∴ a = 3 and b = 1
Page No 91:
Question 20:
Let:
Now,
By the factor theorem, we can say:
Now,
Thus, we have:
Subtracting (1) from (2), we get:
a = 3
By putting the value of a in (1), we get the value of b, i.e., 1.
∴ a = 3 and b = 1
Answer:
Let:
Now,
By the factor theorem, we can say:
.
Thus, we have:
∴
Also,
By the factor theorem, we can say:
.
Thus, we have:
∴
Putting the value of a, we get the value of b, i.e., 14.
∴ a = 23 and b = 14
Page No 91:
Question 21:
Let:
Now,
By the factor theorem, we can say:
.
Thus, we have:
∴
Also,
By the factor theorem, we can say:
.
Thus, we have:
∴
Putting the value of a, we get the value of b, i.e., 14.
∴ a = 23 and b = 14
Answer:
Let:
Now,
By the factor theorem, we can say:
.
Thus, we have:
∴
Also,
By the factor theorem, we can say:
.
Thus, we have:
∴
Putting the value of a, we get the value of b, i.e., 12.
∴ a = 2 and b = 12
Page No 91:
Question 22:
Let:
Now,
By the factor theorem, we can say:
.
Thus, we have:
∴
Also,
By the factor theorem, we can say:
.
Thus, we have:
∴
Putting the value of a, we get the value of b, i.e., 12.
∴ a = 2 and b = 12
Answer:
Let f(x) = px2 + 5x + r
It is given that (x – 2) is a factor of f(x).
Using factor theorem, we have
Also, is a factor of f(x).
Using factor theorem, we have
From (1) and (2), we have
Page No 91:
Question 23:
Let f(x) = px2 + 5x + r
It is given that (x – 2) is a factor of f(x).
Using factor theorem, we have
Also, is a factor of f(x).
Using factor theorem, we have
From (1) and (2), we have
Answer:
Let f(x) = 2x4 – 5x3 + 2x2 – x + 2 and g(x) = x2 – 3x + 2.
Now, f(x) will be divisible by g(x) if f(x) is exactly divisible by both (x − 1) and (x − 2).
Putting x = 1 in f(x), we get
f(1) = 2 × 14 – 5 × 13 + 2 × 12 – 1 + 2 = 2 – 5 + 2 – 1 + 2 = 0
By factor theorem, (x − 1) is a factor of f(x). So, f(x) is exactly divisible by (x − 1).
Putting x = 2 in f(x), we get
f(2) = 2 × 24 – 5 × 23 + 2 × 22 – 2 + 2 = 32 – 40 + 8 – 2 + 2 = 0
By factor theorem, (x − 2) is a factor of f(x). So, f(x) is exactly divisible by (x − 2).
Thus, f(x) is exactly divisible by both (x − 1) and (x − 2).
Hence, f(x) = 2x4 – 5x3 + 2x2 – x + 2 is exactly divisible by (x − 1)(x − 2) = x2 – 3x + 2.
Page No 91:
Question 24:
Let f(x) = 2x4 – 5x3 + 2x2 – x + 2 and g(x) = x2 – 3x + 2.
Now, f(x) will be divisible by g(x) if f(x) is exactly divisible by both (x − 1) and (x − 2).
Putting x = 1 in f(x), we get
f(1) = 2 × 14 – 5 × 13 + 2 × 12 – 1 + 2 = 2 – 5 + 2 – 1 + 2 = 0
By factor theorem, (x − 1) is a factor of f(x). So, f(x) is exactly divisible by (x − 1).
Putting x = 2 in f(x), we get
f(2) = 2 × 24 – 5 × 23 + 2 × 22 – 2 + 2 = 32 – 40 + 8 – 2 + 2 = 0
By factor theorem, (x − 2) is a factor of f(x). So, f(x) is exactly divisible by (x − 2).
Thus, f(x) is exactly divisible by both (x − 1) and (x − 2).
Hence, f(x) = 2x4 – 5x3 + 2x2 – x + 2 is exactly divisible by (x − 1)(x − 2) = x2 – 3x + 2.
Answer:
Let k be added to 2x4 – 5x3 + 2x2 – x – 3 so that the result is exactly divisible by (x – 2). Here, k is a constant.
∴ f(x) = 2x4 – 5x3 + 2x2 – x – 3 + k is exactly divisible by (x – 2).
Using factor theorem, we have
Thus, 5 must be added to 2x4 – 5x3 + 2x2 – x – 3 so that the result is exactly divisible by (x – 2).
Page No 91:
Question 25:
Let k be added to 2x4 – 5x3 + 2x2 – x – 3 so that the result is exactly divisible by (x – 2). Here, k is a constant.
∴ f(x) = 2x4 – 5x3 + 2x2 – x – 3 + k is exactly divisible by (x – 2).
Using factor theorem, we have
Thus, 5 must be added to 2x4 – 5x3 + 2x2 – x – 3 so that the result is exactly divisible by (x – 2).
Answer:
Dividing (x4 + 2x3 – 2x2 + 4x + 6) by (x2 + 2x – 3) using long division method, we have
![](/img/site_content/ask-answer/loader.gif)
Here, the remainder obtained is (2x + 9).
Thus, the remainder (2x + 9) must be subtracted from (x4 + 2x3 – 2x2 + 4x + 6) so that the result is exactly divisible by (x2 + 2x – 3).
Page No 91:
Question 26:
Dividing (x4 + 2x3 – 2x2 + 4x + 6) by (x2 + 2x – 3) using long division method, we have
![](/img/site_content/ask-answer/loader.gif)
Here, the remainder obtained is (2x + 9).
Thus, the remainder (2x + 9) must be subtracted from (x4 + 2x3 – 2x2 + 4x + 6) so that the result is exactly divisible by (x2 + 2x – 3).
Answer:
Let f(x) = xn + an
Putting x = −a in f(x), we get
f(−a) = (−a)n + an
If n is any odd positive integer, then
f(−a) = (−a)n + an = −an + an = 0
Therefore, by factor theorem, (x + a) is a factor of (xn + an) for any odd positive integer.
Page No 92:
Question 1:
Let f(x) = xn + an
Putting x = −a in f(x), we get
f(−a) = (−a)n + an
If n is any odd positive integer, then
f(−a) = (−a)n + an = −an + an = 0
Therefore, by factor theorem, (x + a) is a factor of (xn + an) for any odd positive integer.
Answer:
(c)
Clearly, is a polynomial in one variable because it has only non-negative integral powers of x.
Page No 92:
Question 2:
(c)
Clearly, is a polynomial in one variable because it has only non-negative integral powers of x.
Answer:
(d)
We have:
It a polynomial because it has only non-negative integral powers of x.
Page No 93:
Question 3:
(d)
We have:
It a polynomial because it has only non-negative integral powers of x.
Answer:
(c) y
y is a polynomial because it has a non-negative integral power 1.
Page No 93:
Question 4:
(c) y
y is a polynomial because it has a non-negative integral power 1.
Answer:
(d) −4
4 is a constant polynomial of degree zero.
Page No 93:
Question 5:
(d) −4
4 is a constant polynomial of degree zero.
Answer:
(d) 0
0 is a polynomial whose degree is not defined.
Page No 93:
Question 6:
(d) 0
0 is a polynomial whose degree is not defined.
Answer:
(d) x2 + 5x + 4
is a polynomial of degree 2. So, it is a quadratic polynomial.
Page No 93:
Question 7:
(d) x2 + 5x + 4
is a polynomial of degree 2. So, it is a quadratic polynomial.
Answer:
(b) x + 1
Clearly, is a polynomial of degree 1. So, it is a linear polynomial.
Page No 93:
Question 8:
(b) x + 1
Clearly, is a polynomial of degree 1. So, it is a linear polynomial.
Answer:
(b) x2 + 4
Clearly, is an expression having two non-zero terms. So, it is a binomial.
Page No 93:
Question 9:
(b) x2 + 4
Clearly, is an expression having two non-zero terms. So, it is a binomial.
Answer:
(d) 0
is a constant term, so it is a polynomial of degree 0.
Page No 93:
Question 10:
(d) 0
is a constant term, so it is a polynomial of degree 0.
Answer:
(c) not defined
Degree of the zero polynomial is not defined.
Page No 93:
Question 11:
(c) not defined
Degree of the zero polynomial is not defined.
Answer:
(d) not defined
Zero of the zero polynomial is not defined.
Page No 93:
Question 12:
(d) not defined
Zero of the zero polynomial is not defined.
Answer:
(d) 8
Let:
Thus, we have:
= 4 + 4
=8
Page No 93:
Question 13:
(d) 8
Let:
Thus, we have:
= 4 + 4
=8
Answer:
(b) 1
= 8 8 + 1
= 1
Page No 93:
Question 14:
(b) 1
= 8 8 + 1
= 1
Answer:
p(x) = 5x – 4x2 + 3
Putting x = –1 in p(x), we get
p(–1) = 5 × (–1) – 4 × (–1)2 + 3 = –5 – 4 + 3 = –6
Hence, the correct answer is option (d).
Page No 93:
Question 15:
p(x) = 5x – 4x2 + 3
Putting x = –1 in p(x), we get
p(–1) = 5 × (–1) – 4 × (–1)2 + 3 = –5 – 4 + 3 = –6
Hence, the correct answer is option (d).
Answer:
Let f(x) = x51 + 51
By remainder theorem, when f(x) is divided by (x + 1), then the remainder = f(−1).
Putting x = −1 in f(x), we get
f(−1) = (−1)51 + 51 = −1 + 51 = 50
∴ Remainder = 50
Thus, the remainder when (x51 + 51) is divided by (x + 1) is 50.
Hence, the correct answer is option (d).
Page No 93:
Question 16:
Let f(x) = x51 + 51
By remainder theorem, when f(x) is divided by (x + 1), then the remainder = f(−1).
Putting x = −1 in f(x), we get
f(−1) = (−1)51 + 51 = −1 + 51 = 50
∴ Remainder = 50
Thus, the remainder when (x51 + 51) is divided by (x + 1) is 50.
Hence, the correct answer is option (d).
Answer:
(c) 2
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Question 17:
(c) 2
Answer:
(d) 21
By the remainder theorem, we know that when p(x) is divided by (x 2), the remainder is p(2).
Thus, we have:
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Question 18:
(d) 21
By the remainder theorem, we know that when p(x) is divided by (x 2), the remainder is p(2).
Thus, we have:
Answer:
(d) 4
By the remainder theorem, we know that when p(x) is divided by (x + 2), the remainder is p(2).
Now, we have:
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Question 19:
(d) 4
By the remainder theorem, we know that when p(x) is divided by (x + 2), the remainder is p(2).
Now, we have:
Answer:
(c) −2
By the remainder theorem, we know that when p(x) is divided by (2x 1), the remainder is .
Now, we have:
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Question 20:
(c) −2
By the remainder theorem, we know that when p(x) is divided by (2x 1), the remainder is .
Now, we have:
Answer:
By remainder theorem, when p(x) = x3 – ax2 + x is divided by (x – a), then the remainder = p(a).
Putting x = a in p(x), we get
p(a) = a3 – a × a2 + a = a3 – a3 + a = a
∴ Remainder = a
Hence, the correct answer is option (b).
Page No 94:
Question 21:
By remainder theorem, when p(x) = x3 – ax2 + x is divided by (x – a), then the remainder = p(a).
Putting x = a in p(x), we get
p(a) = a3 – a × a2 + a = a3 – a3 + a = a
∴ Remainder = a
Hence, the correct answer is option (b).
Answer:
(c) −a
By the remainder theorem, we know that when p (x) is divided by (x + a), the remainder is p (−a).
Thus, we have:
Page No 94:
Question 22:
(c) −a
By the remainder theorem, we know that when p (x) is divided by (x + a), the remainder is p (−a).
Thus, we have:
Answer:
(b) x3 − 2x2 − x − 2
Let:
f(x) = x3 − 2x2 − x − 2
By the factor theorem, (x + 1) will be a factor of f(x) if f (1) = 0.
We have:
Hence, (x + 1) is a factor of .
Page No 94:
Question 23:
(b) x3 − 2x2 − x − 2
Let:
f(x) = x3 − 2x2 − x − 2
By the factor theorem, (x + 1) will be a factor of f(x) if f (1) = 0.
We have:
Hence, (x + 1) is a factor of .
Answer:
The zero of the polynomial p(x) can be obtained by putting p(x) = 0.
Hence, the correct answer is option (b).
Page No 94:
Question 24:
The zero of the polynomial p(x) can be obtained by putting p(x) = 0.
Hence, the correct answer is option (b).
Answer:
The given polynomial is p(x) = x2 + x – 6.
Putting x = 2 in p(x), we get
p(2) = 22 + 2 – 6 = 4 + 2 – 6 = 0
Therefore, x = 2 is a zero of the polynomial p(x).
Putting x = –3 in p(x), we get
p(–3) = (–3)2 – 3 – 6 = 9 – 9 = 0
Therefore, x = –3 is a zero of the polynomial p(x).
Thus, 2 and –3 are the zeroes of the given polynomial p(x).
Hence, the correct answer is option (c).
Page No 94:
Question 25:
The given polynomial is p(x) = x2 + x – 6.
Putting x = 2 in p(x), we get
p(2) = 22 + 2 – 6 = 4 + 2 – 6 = 0
Therefore, x = 2 is a zero of the polynomial p(x).
Putting x = –3 in p(x), we get
p(–3) = (–3)2 – 3 – 6 = 9 – 9 = 0
Therefore, x = –3 is a zero of the polynomial p(x).
Thus, 2 and –3 are the zeroes of the given polynomial p(x).
Hence, the correct answer is option (c).
Answer:
The given polynomial is p(x) = 2x2 + 5x – 3.
Putting in p(x), we get
Therefore, is a zero of the polynomial p(x).
Putting x = –3 in p(x), we get
Therefore, x = –3 is a zero of the polynomial p(x).
Thus, and –3 are the zeroes of the given polynomial p(x).
Hence, the correct answer is option (b).
Page No 94:
Question 26:
The given polynomial is p(x) = 2x2 + 5x – 3.
Putting in p(x), we get
Therefore, is a zero of the polynomial p(x).
Putting x = –3 in p(x), we get
Therefore, x = –3 is a zero of the polynomial p(x).
Thus, and –3 are the zeroes of the given polynomial p(x).
Hence, the correct answer is option (b).
Answer:
The given polynomial is p(x) = 2x2 + 7x – 4.
Putting in p(x), we get
Therefore, is a zero of the polynomial p(x).
Putting x = –4 in p(x), we get
Therefore, x = –4 is a zero of the polynomial p(x).
Thus, and –4 are the zeroes of the given polynomial p(x).
Hence, the correct answer is option (c).
Page No 94:
Question 27:
The given polynomial is p(x) = 2x2 + 7x – 4.
Putting in p(x), we get
Therefore, is a zero of the polynomial p(x).
Putting x = –4 in p(x), we get
Therefore, x = –4 is a zero of the polynomial p(x).
Thus, and –4 are the zeroes of the given polynomial p(x).
Hence, the correct answer is option (c).
Answer:
(b) 5
Page No 94:
Question 28:
(b) 5
Answer:
(b) m = 7, n = −18
Let:
Now,
(x + 2) is a factor of p(x).
So, we have p(2)=0
Now,
Also,
(x 1) is a factor of p(x).
We have:
p(1) = 0
By substituting the value of m in (i), we get n = −18.
∴ m = 7 and n = −18
Page No 94:
Question 29:
(b) m = 7, n = −18
Let:
Now,
(x + 2) is a factor of p(x).
So, we have p(2)=0
Now,
Also,
(x 1) is a factor of p(x).
We have:
p(1) = 0
By substituting the value of m in (i), we get n = −18.
∴ m = 7 and n = −18
Answer:
(a) 1
Let:
Now,
Page No 94:
Question 30:
(a) 1
Let:
Now,
Answer:
(d) −3
Let:
Now,
Page No 94:
Question 31:
(d) −3
Let:
Now,
Answer:
(b) 0, 3
Let:
Now, we have:
Page No 95:
Question 32:
(b) 0, 3
Let:
Now, we have:
Answer:
(d)
Let:
Hence, the correct answer is option D.
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