SEBA Class 9 Mathematics Chapter 2 Polynomials Solutions, SEBA Class 9 Maths Textbook Notes in English Medium, SEBA Class 9 Mathematics Chapter 2 Polynomials Solutions in English to each chapter is provided in the list so that you can easily browse throughout different chapter Assam Board SEBA Class 9 Mathematics Chapter 2 Polynomials Notes and select needs one.
SEBA Class 9 Mathematics Chapter 2 Polynomials
Also, you can read the SCERT book online in these sections Solutions by Expert Teachers as per SCERT (CBSE) Book guidelines. SEBA Class 9 Mathematics Chapter 2 Polynomials Question Answer. These solutions are part of SCERT All Subject Solutions. Here we have given SEBA Class 9 Mathematics Chapter 2 Polynomials Solutions for All Subject, You can practice these here.
Polynomials
Chapter – 2
| Exercise 2.1 |
1. Which of the following expressions are polynomials in one vari-able and which are not? State reasons for your answer.
(i) 4x² – 3x + 7
Ans: Yes, 4 x² – 3x + 7 is a polynomial of one variable. As x has degree 2 in it and only x is one variable.
(ii) y² + √2
Ans: Yes, y is only one variable.
(iii) 3√t + t√2
Ans: No as 3√t + r√2 can be written as 3t¹/² + √2 Here, the exponent of t in 3t ¹/² which is not a whole number.
(iv) y + 2/y
Ans: No, as y + 2/y can be written as y + 2 y–¹ where the exponent of y in 2/y is -1, which is not a whole number.
(v) x¹⁰ + y³ + t⁵⁰
Ans: Yes, It is a polynomial in three variables, x, y and t.
2 Write the coefficients of x² in each of the following:
(i) 2 + x² + x
Ans: We have given that the equation 2 + x² + x
We can also write 1x² + 1x + 2
Therefore, the coefficient of x² in this equation is 1.
(ii) 2 – x² + x³
Ans: We have given that the equation: 2 – x² + x³
We can also write x³ – 1x² + 2
Therefore, the coefficient of x² in this equation is -1.
(iii) π/2 x² + x
Ans: We have given that the equation π/2x² + x
Therefore, the coefficient of x² in this equation is 11/7.
(iv) √2x- 1
Ans: We have given that the equation √2x -1
We can also write ox² + √2x – 1
Therefore, the coefficient of x² in this equation is 0.
(v) (2x – 3)(x² – 3x + 1)
Ans: (2x – 3)(x² – 3x + 1)
= 2x (x²- 3x + 1) – 3(x² – 3x + 1)
= 2x³- 6x² + 2x – 3x² + 9x – 3
= 2x³ – 9x² + 11x – 3
∴ Coefficient x2 = – 9
3. Give one example each of a binomial of degree 35, and a mono-mial of degree 100.
Ans: We know that polynomials having only two terms are called binomials.
Therefore, the example of a binomial of degree 35 is ax³⁵ + b where a and b are any real number.
Again, the example of a monomial of degree 100 is ax¹⁰⁰ where a is any real number.
4. Write the degree of each of the following polynomials:
(i) 5x³ + 4x²+ 7x
Ans: We know that the highest power of variable in a polyno-mial is called degree of the polynomial, In polynomial 5x³ + 4x² + 7x
The highest power of variable x is 3.
Therefore, the degree of the polynomial 5x³ + 4x² + 7x is 3.
(ii) 4 – y²
Ans: In polynomial 4 – y², the highest power of the variable y is 2.
Therefore, the degree of the polynomial 4 – y² is 2.
(iii) 5t -√7
Ans: In polynomial 5t – √5 the highest power of the variable t is 1.
Therefore, the degree of polynomial 5t -√5 is 1.
(iv) 3
Ans: The only term here is 3 which can be written as 3x⁰ .So the highest power of the variable x is 0.
Therefore, the degree of the polynomial 3 is 0.
5 Classify the following as linear, quadratic and cubic polynomials.
(i) x²+ x
Ans: In polynomial x² + x the highest power of the variable x is 2.
So the degree of the polynomial is 2. We know that the polynomial of degree 2 is called a quadratic polynomial.
Therefore, the polynomial x² + x is a quadratic polynomial.
(ii) x – x³
Ans: In polynomial x – x³, the highest power of the variable x is 3. So the degree of the polynomial is 3.
We know that the polynomial of degree 3 is called a cubic polynomial.
Therefore, the polynomial x – x³ is a cubic polynomial.
(iii) y + y² + 4
Ans: In polynomial y + y² + 4 the highest power of the variable y is 2.
So the degree of the polynomial is 2. We know that the polynomial of degree 2 is called quadratic polynomial.
Therefore, the polynomial y + y² + 4 is a quadratic polynomial.
(iv) 1 + x
Ans: In polynomial 1 + x, the highest power of the variable x is 1. So the degree of the polynomial is 1.
We know that the polynomial of degree 1 is called a linear polynomial.
Therefore, polynomial 1 + x is a linear polynomial.
(v) 3t
Ans: In polynomial 3t, the highest power of the variable t is 1. So, the degree of the polynomial is 1.
We know that the polynomial of degree 1 is called a linear polynomial.
Therefore, polynomial 3t is a linear polynomial.
(vi) r²
Ans: In polynomial r² the highest power of the variable r is 2. So, the degree of the polynomial is 2. We know that the polynomial of degree 2 is called quadratic polynomial.
Therefore, polynomial r2 is a quadratic polynomial.
(vii) 7x²
Ans: In polynomial 7x³ the highest power of the variable x is 3. So the degree of the polynomial is 3. We know that the polynomial of degree 3 is called a cubic polynomial.
Therefore, polynomial 7x³ is a cubic polynomial.
6. Which one is a polynomial?
(a) x-2 + 3x – 1
(b) y + 1/y
(c) 3y/2 + x
(d) 2√x + 1
Ans: (c) 3y/2 + x
| Exercise 2.2 |
Q.1. Find the value of the polynomial 5x-4x²+3 at
(i) x = 0
Ans: Let P(x) = 5x – 4x² – 3
Therefore, P(0) = 5(0) – 4 (0) + 3 = 0 – 0 + 3 = 3
So, the value of P(x) at x = 0 is 3.
(ii) x = -1
Ans: Let P(x) = 5x – 4x² + 3
Therefore, P(-1) = 5(-1) – 4(-1)² + 3 = 6
So the value of P(x) at x = -1 is -6
(iii) x = 2
Ans: Let P(x) = 5x – 4x² + 3
Therefore, P(2) = 5 × 2 – 4(2)² + 3 = 10 – 16 + 3 = -3
So, the value of P(x) at x = 2 is – 3
2 Find p(¹⁰) and p(²) for each of the following polynomials:
(i) p(y) = y² – y + 1
Ans: We have given P(y) = y² – y + 1
Therefore, the value of polynomial p(y) at y = 0 is
P(0) = 0² – 0 + 1 = 1
Again, the value of polynomial p(y) at y = 1 is P(1) = 1² – 1 + 1 = 1
p(2) = 2² – 2 + 1 = 3
(ii) p(t) = 2 + t + 2r² – t³
Ans: We have given, p(t) = 2 + t + 2r² – t³
Therefore, the value of polynomial p(t) at 1 = 0 is
P(0) = 2 + 0 + 2.(0)² – (0)³ = 2
Again, the value of polynomial p(t) at t = 1 is
p(1) = 2 + 1 + 2(1)² – (1)³ = 4
Again, the value of polynomial p(t) at t = 2 is
p(2) = 2 + 2 + 2 (2)² – (2)³ = 4
(iii) p(x) = x³
Ans: We have given, p(x) = x³
Therefore, the value of polynomial p(x) at x = 0 is
P(0) = (0)³ = 0
Again the value of polynomial p(x) at x = 1 is
p(1) = (1)³ = 1
Again, the value of polynomial p(x) at x = 2 is is
P(2) = (2)³ = 8
(iv) p(x) = (x – 1) (x + 1)
Ans: We have given p(x) = (x – 1) (x + 1)
Therefore, the value of polynomial p(x) at x = 0 is
P(0) = (0 – 1) (0 + 1) = (-1) x ( + 1) = – 1
Again, the value of polynomial p(x) at x = 1 is
p(1) = (1 – 1) x (1 + 1) = 0 x 2 = 0
Again, the value of polynomial p(x) at x = 2 is
p(2) = (2 – 1) (2 + 1) = 1 x 3 = 3
3 Verify whether the following are zeros of the polynomial indi-cated against them:
(i) P(x) = 3x + 1, x = – 1/3
Ans: We have given that, p(x) = 3x + 1
Therefore, the value of polynomial p(x) at x = – 1/3 is
Yes x = – 1/3 is the zero of polynomial p(x)
(ii) P(x) = 5x − π, x = -4/5
Ans: We have given that, p(x) = 5x – π
Therefore, the value of polynomial p(x) at x = 4/5 is
No, x = 4/5 is not the zero of p(x) = 5x – π.
(iii) p(x) = x²-1, x = 1, -1
Ans: We have given that, p(x) = x² – 1
Therefore, the value of polynomial p(x) at x = 1 is
p(1) = 1²- 1 = 0
Again, the value of polynomial p(x) at x = – 1 is
p(- 1) = (- 1)²- 1 = 0
Yes, x = 1, -1 are the zero of polynomial p(x) = x² – 1
(iv) p(x) = (x + 1 ) (x – 2), x = – 1, 2
Ans: We have given that, p(x) = (x + 1) (x – 2)
Therefore, the value of polynomial p(x) at x = – 1 is
p(- 1) = (- 1 + 1) (- 1 – 2) = 0 x (- 3) = 0
Again, the value of polynomial p(x) at x = 2 is
p(2) = (2 + 1) (2 – 2) = 3 x 0 = 0
Yes x = – 1, 2 are the zero of polynomial
p(x) = x + 1 (x – 2)
(v) p(x) = x², x = 0
Ans: We have given that, p(x) = x²
Therefore, the value of polynomial p(x) at x = 0 is
p(0) = 0² = 0
Yes x = 0 is the zero of polynomial p(x) = x²
(vi) P(x) = lx + m, x = – m/1
Ans: We have given that, p(x) = lx + m
Therefore, the value of polynomial p(x) at x = – m is
Yes, x = – m/l is the zero of polynomial p(x) = lx + m
(vii) p(x) = 3x² – 1, x = 1/√3, 2 /√3
Ans: We have given that, p(x) = 3x² – 1
Therefore, the value of polynomial p(x) at x = – 1/√3 is
Again, the value of polynomial p(x) at x = 2/√3 is
Therefore, x = – 1/√3 is the zero of polynomial and p(x) and x = 2/√3
is not the zero of polynomial p(x).
(viii) p(x) = 2x + 1, x = 1/2
Ans: We have given that, p(x) = 2x + 1
Therefore, the value of polynomial p(x) at x = 1/2 is
Therefore, x = 1/2 is not the zero of polynomial p(x).
Q.4. Find the zero of the polynomial in each of the following cases:
(i) p(x) = x + 5
Ans: We have given that, p(x) = x + 5 ..(1)
To find the zero of polynomial p(x), we can take p(x) = 0…(2)
From equations (1) and (2)
x + 5 = 0
∴ x = – 5
Therefore, x = -5 is the zero of the polynomial p(x) = x + 5.
(ii) p(x) = x – 5
Ans: We have given that, p(x) = x – 5 …(1)
To find the zero of polynomial p(x) we can take p(x) = 0 …(2)
From equations (1) and (2)
x – 5 = 0
∴ x = 5
Therefore x = 5 is the zero of polynomial p(x) = x – 5.
(iii) p(x) = 2x + 5
Ans: We have given that, p(x) = 2x + 5 …(1)
To find the zero of polynomial p(x), we can take p(x) = 0 …(2)
From equations (1) and (2)
Therefore, x = – 5/2 is the zero of polynomial p(x) = 2x + 5
(iv) p(x) = 3x – 2
Ans: We have given that, p(x) = 3x – 2 …(1)
To find the zero of polynomial p(x) we can take p(x) = 0 … (2)
Therefore, x = 2/3 is the zero of polynomial p(x) = 3x – 2
(v) p(x) = 3x
Ans: We have given that, p(x) = 3x …(1)
To find the zero of polynomial p(x) we can take p(x) = 0 …(2)
From equations (1) and (2), 3x = 0 ∴ x = 0
Therefore, x = 0 is the zero of polynomial p(x) = 3x
(vi) p(x) = ax, a ≠ 0
Ans: We have given that, p(x) = ax, a ≠ 0 …(1)
To find the zero of polynomial p(x) we can take p(x) = 0 …(2)
From equations (1) and (2)
ax = 0
∴ x = 0/a = 0
Therefore, x = 0 is the zero of polynomial p(x) = ax, a ≠ 0.
(vii) p(x) = cx + d, c ≠ 0, c; d are real numbers.
Ans: We have given that, p(x) = cx + d …(1)
c ≠ 0, c, d are real numbers.
To find the zero of polynomial p(x) we can take
p(x) = 0 …(2)
From equations (1) and (2)
cx + d = 0
x = – d/c x where c ≠ 0x and are real numbers
Therefore, x = – d/c, where c, d and c,d are real numbers is the solution of polynomial p(x) = cx + d.
| Exercise 2.3 |
1. Find the remainder when x³+ 3x²+ 3x+1 is divided by
(i) x + 1
(ii) x – 1/2
(iii) x
(iv) x + π
(v) 5 + 2x
Ans: Let p(x) = x³ + 3x² + 3x + 1
(i) x + 1 = 0 Rightarrow x = – 1
∴ Remainder = p(- 1) = (- 1)³ + 3 (- 1)² + 3(- 1) + 1 = – 1 + 3 – 3 + 1 = 0
(iii) x.
Remainder = (0)³ + 3(0)² + 3(0) + 1 = 1
(iv) x + π
x + π = 0 ⇒ x = – π
∴ Remainder = (-π)³ + 3(−π)² + 3(−π) + 1 = − π³ + 3π² − 3π + 1
(v) 5 + 2x
5 + 2x = 0 ⇒ 2x = – 5 ⇒ x = – 5/2
∴ Remainder =
2. Find the remainder when x³ – ax² + 6x – a is divided by x – a
Ans: Let p(x) = x³ – ax² + 6x – a
x – a = 0
∴ Remainder = (a)³ – a(a)² + 6(a) – a = a³ – a³ + 6a – a = 5a
3. Check whether 7 + 3x is a factor of 3x³ + 7x
Ans: 7 + 3x will be a factor of 3x³ + 7x only if 7 + 3x divides 3x³ + 7x leaving no remainder.
Let p(x) = 3x³ + 7x Now, 7 + 3x = 0 3x = -7 ⇒ x = – 7/3
∴ Remainder =
∴ 7 + 3x is not a factor of 3x³ + 7x
4. Find the quotient and reminder.
(i) x³ – 4x² + 2x + 5 is divided by x – 2
Ans:
∴ Quotient = x² – x – 2
Reminder = 1
(ii) 4x³ – 2x² – 3 is divided by 4x² – 1
Ans:
∴ Quotient = 2x – 1
Reminder = 2x – 4
(iii) 3x³ – 5x² + 10x – 3 is divided by 3x + 1
Ans:
∴ Quotient = x² – 2x + 4
Reminder = -7
(iv) x¹¹ – 5 is divided by x + 1
Ans:
∴ Quotient = x¹⁰ – x⁹ + x⁸ – x⁷ + x⁶ – x⁵ – x⁴ – x³ + x²- x + 1
Reminder = – 6
5. (i) Divide 3x²- 2x – 40 by 3x + 10
Ans:
∴ Quotient = 2x + 10
Reminder = – 2x – 140
(ii) Divide 4 + 7x + 7x² + 2x³ by 2x + 1
Ans:
∴ Quotient = x² + 3x + 2
Reminder = 2
6. (i) Is the polynomial – 14x² – 13x + 12 is completely divisible by 2x + 3
Ans: Let, P(x) = – 14x² – 13x + 12
∴ P(x) is completely divisible by 2x + 3
∴ 2x + 3 is a factor of P(x)
(ii) Examine if x – 7 is a factor of x³+ 2x² – 3x + 4
Ans: Let, P(x) = x³ + 2x² – 3x + 4
x – 7 = 0 ⇒ x = 7
∴ P(7) = 7³ + 2 (7)² – 3×7 + 4
= 343 + 98 – 21 + 4
= 445 – 21
= 424 ≠ 0
∴ x – 7 is not a factor of P(x)
7. The polynomial ax³ + 3x² + 5x and x³ – 4x – a when divided by x – 2, equal reminders are found. Find the value of a.
Ans: Let, P(x) = ax³ + 3x² + 5x – 4
F(x) = x³ – 4x – a
x – 2 = 0 ⇒ x = 2
∴ P(2) = a (2)³ + 3(2)² + 5(2) – 4
= 8a + 12 + 10 – 4 = 8a + 18
∴ F(2) = 2³ – 4(2) – a = 8 – 8 – a = – a
According to question, 8a + 18 = – a ⇒ 9a = -18 ⇒ a = – 2
| Exercise 2.4 |
1. Which of the following polynomials has (x+1) a factor:
(i) x³ + x² + x + 1
Ans: Here, p(x) = x³ + x² + x + 1 and the zero of x + 1 is – 1
So, p(-1) = (-1)³ + (-1)² + (-1) + 1 = – 1 + 1 – 1 + 1 = 0
Here, remainder is 0, therefore x + 1 is the factor of x³ + x² + x + 1
(ii) x⁴ + x³ + x² + x + 1
Ans: Here, p(x) = x⁴ + x³ + x² + x + 1 and the zero of x + 1 is x – 1
So, p(-1) = (-1)⁴ + (-1)3 + (-1)² (-1) + 1 = 1 – 1 + 1 – 1 + 1 = 1
Here, remainder is 1, therefore x + 1 is not the factor of x⁴ + x³ + x² + x + 1
(iii) x⁴ + 3x³ + 3x² + x + 1
Ans: Here, p(x) = x⁴ + 3x³ + 3x² + x + 1 and the zero of x + 1 is – 1
So,p(-1) = (-1)⁴ + 3(-1)³ + 3(-1)² + (-1) + 1 = 1 – 3 + 3 – 1 + 1 = 1
Here, remainder is 1, therefore x+1 is not the factor of x⁴ + 3x³ + 3x² + x + 1
(iv) x³ – x² – (2 + √2)x + √2
Ans: Here, p(x) = x³ – x² – (2 + √2) (x) + √2 and the zero of x + 1 is -1
So, p(-1) = (-1)³ – (-1)² – (2 + √2) (-1) + √2
= 1 – 1 + 2 + √2 + √2 = 2√2
Here, remainder is 2√2, therefore x + 1 is not the factor of x³ – x² – (2 + √2) x + √2.
2. Use the factor theorem to determine whether x³ – x² – g(x) is a factor of p(x) in each of the following cases:
(i) p(x) = 2x³+x²-2x-1, g(x) = x+1
Ans: Here, p(x) = 2x³ + x²-2x-1 and the zero of g(x) = x+1 is -1
So, p(-1) = 2(-1)³ +(-1)²-2(-1)-1 = -2+1+2+1 = 0
Here, remainer is zero, therefore g(x) = x+1 is the factor of p(x) = 2x³+x²-2x-1
(ii) p(x) = x³-3x²+3x+1 g(x) = x+2
Ans: Here, p(x) = x³+3x²+3x+1 and the zero of g(x) = x+2 is-2.
So,p(-2) = (-2)³+3(-2)²+3(-2)+1 = -8+12-6+1 = -1
Here, remainder is -1, therefore g(x)=x+2 is not the factor of p(x) = x³+3x²+3x+1
(iii) p(x)=x³-4x²+x+6 g(x)=x-3
Ans: Here, p(x) = x³-4x²+x+6 and the zero of g(x) = x-3 is 3
So, p(3) = (3)³-4(3)²+3+6 = 27-36+3+6 = 0
Here, remainder is 0, therefore g(x) = x -3 is the factor of p(x) = x³+4x²+x+6
3. Find the value of k if x – 1 is a factor of p(x) in each of the following cases:
(i) p(x) = x²+x+k
Ans: We have given that x-1 is the factor of p(x) = x²+x+k
So, (1)²+1+k = 0 or,
1+1+k = 0
2+k = 0
∴ k = -2
Therefore, the value of k in p(x) = x²+x+k is -2
(ii) p(x)=2x² + kx+√2
Ans: We have given that x-1 is the factor of p(x) = 2x²+kx+√2
∴ p(1) = 0
Now, p(1) = 2(1)²+k(1)+ √2
So, 0 = 2+k+ √2 or k = -2-√2
or, k = -(2+√2)
Therefore the value of k in p(x) = 2x² + kx+√2 is -(2+√2)
(iii) p(x)=kx²-√2x+1
Ans: We have given that x-1 is a factor of.
p(x) = kx² -√2x+1 ∴ p(1) = 0
Now, p(1) = k(1)² -√2(1)+1
So, k-√2+1 = 0 ∴ k = √2-1
Therefore, the value of k in p(x) = kx²-√2x+1 is √2-1
(iv) p(x)=kx²-3x+k
Ans: We have given that x-1 is the factor of p(x) = kx² -3x+k
So, k-3+k = 0 or, 2k-3 = 0. k =0 ∴ K = 3/2
Therefore the value of k in p(x) = kx²-3x+k -3/2
4. Factorise:
(i) 12x²7x+1
Ans: 12x² -7x+1
= 12x²-4x-3x+1
= 4x(3x-1)-1(3x-1)
= (3x-1) (4x-1)
(ii) 2x²+7x+3
Ans: 2x² +7x+3
= 2x²+6x+x+3
= 2x(x+3)+1(x+3)
= (x+3)(2x+1)
(iii) 6x²+5x-6
Ans: 6x²+5x-6
= 6x²+9x-4x-6
= 3x(2x+3)-2(2x+3)
= (2x+3)(3x-2)
(iv) 3x²-x-4
Ans: 3x²-x-4
= 3x²-4x+3x-4
= x(3x-4)+1(3x-4)
= (3x-4) (x+1)
(v) 2x² +x-45
Ans: 2x² + x-45
= 2x²+10x-9x-45
= 2x(x+5)-9(x+5)
= (x+5) (2x-9)
(vi) y²+18y+65
Ans: y²+18y-65
= y²+13y+5y+65
= y(y+13)+5(y+13)
= (y+13) (y+5)
(vii) p² + 14p + 13
Ans: p² + 14p + 13
= p²+13p+p+13
= p(p+13)+1(p+13)
= (p+13) (p+1)
(viii) -18+11x-x²
Ans: -18+11x-x²
= -18 +9x+2x-x²
= -9(2-x)+x(2-x)
= (2-x) (x-9)
(ix) 8a²-22ab+15b²
Ans: 8a²-22ab+15b²
= 8a²-12ab-10ab+15b²
= 4a(2a-3b)-5b(2a-3b)
= (2a-3b)(4a-5b)
5. Factorise:
(i) x³-2x²-x+2
Ans: Let p(x) = x³-2x²-x+2
The factors of 2 are ±1, ±2
By trial, we find that p(2)=0
So, (x-2) must be the factor of p(x)
Therefore,
x³-2x²-x+2 = (x-2) (x²-1) = (x-2) (x+1) (x-1)
[∵ a²-b² = (a+b)(a-b)]
∵ x³-2x²-x+2 = (x-2) (x-1) (x+1)
(ii) x³-3x²-9x-5
Ans: Let p(x) = x³-3x²-9x-5
The factors of -5 are ±1, ±5
By trial, we find that p(-1) = 0
So, (x+1) is the factor of p(x)
Therefore, x³-3x²-9x-5
= (x+1)(x²-4x-5) = (x+1)[x²-5x+x-5]
= (x+1)[x(x-5)+1(x-5)]
∴ x³-3x²-9x-5 = (x+1)(x+1)(x-5)
(iii) x³ +13x²+32x+20
Ans: Let p(x) = x³ +13x² +32x+20
The factor of 20 are ±1, ±2, ±4, ±5,±10,±20
By trial we find that p(-1) = 0
So, (x+1) must be the factors of P(x).
Therefore,
x³+12x²+32x+20
= (x+1)(x²+12x+20) = (x+1)[x²+10x+2x+20]
= (x+1)[x(x+10)+2(x+10)]
∴ x³ +13x²+32x+20 = (x+1)(x+10)(x+2)
(iv) 2y³ + y²-2y-1
Ans: Let p (y) = 2y³ + y²-2y-1
The factor of -1, are ±1
By trial, we find that p(1) = 0
So, (y-1) must be the factor of 2y³ +y²-2y-1
Therefore, 2y³ + y²-2y-1 = (y-1) (2y²+3y+1)
= (y-1) [2y²+2y+y+1] = (y-1) [2x(x+1)+1(y+1)]
= 2y³ + y²-2y-1 = (y-1) (y+1) (2y+1)
(v) x³+x²-x-1
Ans: Let, P(x) = x³ + x²-x-1
P(1) = 1³+ 1²-1-1-2-2 = 0
∴ x-1 is a factor of P(x)
∴ P(x) = (x-1) (x² + 2x + 1) = (x-1) {(x)² +2. x. 1+1}²
= (x-1) (x+1)²
= (x − 1) (x + 1)(x+1)
(vi) x³+x²+x+1
Ans: Let, P(x) = x³+x²+x+1
P(-1) = (-1)³+(-1)² + (-1)+1
=−1 + 1 − 1 + 1 = 0
∴ (x+1) is a factor of P(x)
∴ P(x) = (x+1) (x² + 1)
(vii) x³+2x²-x-2
Ans: Let, P(x) = x³ + 2x²-x-2
P(-2) = (-2)³+2(-2)² – (-2)-2
= -8+8+2-2 = 0
∴ x + 2 is a factor of P(x)
∴ P(x) = (x+2) (x² – 1)
= (x+2) (x-1)(x+1)
(viii) x³+3x²-7x-6
Ans: Let, P(x) = x³+ 3x²- 7x – 6
P(2) = 2³+ 3(2)²- 7(2) – 6
= – 8 + 12 – 14 – 6 = 20 – 20 = 0
∴ x – 2 is a factor of P(x)
∴ P(x) = (x – 2)(x²+ 5x + 3)
(ix) 3x³+5x²-16x-2
Ans: Let, P(x) = 3x³ + 5x² – 16x – 12
P(- 3) = 3(- 3)³ + 5(- 3)²- 16(- 3) – 12
= – 81 + 45 + 48 – 12
= 93-93-0
∴ x + 3 is a factor of P(x)
∴ P(x) = (x + 3)(3x²- 4x – 4)
= (x + 3)(3x²- 6x + 2x – 4)
= (x + 3){3x(x – 2) + 2(x – 2)}
= (x + 3)(x – 2)(3x + 2)
6. If x + a is a common factor of the polynomials x²+ px + q and
Ans: x + a is a common factor of the polynomials x²+ px + q and x² + mx + n
Let, F(x) = x²+ Px + q
G (x) = x²+ mn + n
x+a = 0 ⇒ x = -a
Now F a) = (-a)²+p(-a)+q = a²-pa+q
G(-a) =(-a)²+m(-a)+n = a²-ma+b
∴ a²- pa + q = a²- ma + n
⇒ – p +a + ma = n – q
⇒ a(m – p) = n – q a -q
| Exercise 2.5 |
1. Use suitable identities to find the following products:
(i) (x + 4)(x + 10)
Ans: (x + 4)(x + 10)
[∵(x+a)(x+b) = x²+(a+b)x+ab]
Therefore, (x+4)(x+10) = x²+(4+10)x+4×10 = x²+14x+40
(ii) (x + 8)(x – 10)
Ans: (x+8)(x-10) or (x+8)(x+(-10))
We know that, (x+a)(x+b) = x²+(a+b)x+ab
∴ (x+8)(x+(-10)) = x² +(8+(-10))x+8x(-10)
= x²-2x-80
(iii) (3x + 4)(3x – 5)
Ans: (3x+4)(3x-5) or (3x+4) (3x+(-5))
We know that, (x+a)(x+b) = x²+(a+b)x+a.b
∴ (3x+4) (3x-5) = (3x)2 + (4+ ((5))3x+4x (-5) = 9x²-3x-20
(iv)
Ans:
We know that, (x+y)(x-y) = x²-y²
(v) (3 – 2x)(3 + 2x)
Ans: (3-2x) (3+2x)
We know that, (x-y)(x+y) = x²-y²
∴ (3-2x) (3+2x) = (3)² – (3)² – (2x)² = 9-4x²
2. Evaluate the following products without multiplying directly:
(i) 103×107
Ans: 103×107
We can also write (100+3) x (100+7)
We know that, (x+1)(x+6) = x²+(a+b)x+a.b
∴ (100+3)(100+7)
= (100)²+(3+7)×100+3×7 = 10000+1000+21 = 11021
(ii) 95×96
Ans: 95×96 = (90+5) x (90+6)
We know that, (x+a)(x+b) = x²+(a+b)x+a.b
∴ (90+5) x (90+6)
= (90)² +(5+6)×90+5×6 = 8,100+990+30 = 9,120
(iii) 104×96
Ans: 104×96
We can also write, (104+4) x (100-4)
We know that, (x + y)(x − y) = x²- y²
∴ (100+4)(100-4) = (100)²-(4)² = 10000-16 = 9,984
3. Factorise the following using appropriate identities:
(i) 9x²+6xy + y²
Ans: 9x²+6xy+ y²
We can also write, (3x)² +2.3x. y+(y)²
We know that, a²+2ab+b² = (a+b)²
∴ (3x)²+2.3x.y + (y)² = (3x+y)² = (3x+y) (3x+y)
(ii) 4y²-4y+1
Ans: (4y)2-2.2y.1+ (1)²
We can also write, (2y)² – 2.2y.1+ (1)²
We know that, x²-2xy + y² = (x-y)²
∴ (2y)²-2.2y.1+(1) = (2y-1) = (2y-1) (2y-1)
Ans:
We know that, a²-b² = (a+b)(a-b)
4. Expand each of the following using suitable identities:
(i) (x+2y+4z)²
Ans: We are given that, (x+2y+4z)²
We know that, (a+b+c)² = a²+b² +c²+2ab+2bc+2ca
∴ (x+2y+4z)² = (x)² +(2y)²+(4z)²+2.x.
2y+2.2y.4z+2.4z.x
= x²+4y²+16z²+4xy+16yz+8zx
(ii) (2x – y + z)²
Ans: We are given that, (2x-y+ z)²
We can also write, (2x + (y)+ z)²
Now, we know that, (a+b+c)²
= a²+b²+c²+2ab+2bc+2ca
∴ {2x+(-y)+z}² = (2x)²+(y)²+z²+2.2x. (-y)+2. (y). z+2. z. 2x
= 4x² + y²+z²-4xy-2yz+4zx
(iii) (-2x+3y+2z)²
Ans: We are given that, (-2x+3y+2z)²
We can also write, {(-2x)+3y+2z}²
Now, we know that,
(a+b+c)² = a²+b²+c²+2ab+2bc+2ca
Therefore, (-2x+3y+2z)² = (-2x)²+(3y)²+(2z)²+2.(-2x)(3y)+2(3y)(2z)+2.2z.(-2x)
= 4x²+9y²+4z²-12xy +12yz-8zx
(iv) (3a-7b-c)²
Ans: We are given that, (-3a-7b-c)²
We can also write, (3a +(-7b)+(-c)²
Now, we know that,
(a+b+c)² = a²+b²+c²+2ab+2bc+2ca
Therefore, (3a.+(-7b)+(-c))² = (3a)² +(-76)²+(-c)² +2.3a (-7b)+2. (-7b) (-c)+2.(c).3a
3a9a²+49b²+c²-42ab+14bc-6ca
(v) (-2x+5y-3z)²
Ans: We are given that, (-2x+5y-3z)²
We can also write, [(-2x+5y-3x)]²
Now, we know that, (a+b+c)²
= a²+b²+c²+2ab+2bc+2ca
Therefore,
[(-2x)+5y+(-32)]² = (-2x)² +(5y)²+ (-3z)² +2.(-2x). 5y+2.5y(-3z) (-2x) z+2(-3z) (-2x)
= 4x²+25y²+9z2-20xy-30yz+12zx
Ans:
Now, we know that,
(a+b+c)² = a²+b²+c²+2ab+2bc+2ca
5. Factorise:
(i) 4x²+9y²+16z² +12xy-24yz-16xz
Ans: We are given,
4x²+9y²+16z² +12xy-24yz-16xz
We can also write,
(2x)²+(3y)²+(-4z)² +2.2x. 3y+2.3y.
(-4z)+2.(-4z). 2x
Now we know that,
a²+b²+c²+2ab+2bc+2ca = (a+b+c)²
Therefore,
(2x)²+(3y)²+(-4z)²+2.2x3y+23y(-4z)+2.(-4z). 2x
= (2x+3y+(-4z))² = (2x+3y-4z)² = (2x+3y-4z)(2x+3y-4z)
(ii) 2x² + y²+8z²-2√2xy +4√2xy +4√2yz-8xz
Ans: We are given that,
2x² + y²+8z²-2√2xy +4√2yz-8xz
We can also write,
(-√2x)²+(y)² +(2√2z)²+2(-√2x). y+2. y. (2√2z)+2. (2√2z). (-√2z)
Now, we know that,
a²+b²+c²+2ab+2bc+2ca = (a+b+c)²
Therefore,
(-√2x)²+(y)²+ (2√2z)²+2.(-√2x)y+2.y.(2√2 z)+2. (2√2z)(- √2x)
= ((-√2x)+ y+ 2√2z)² = (-√2x+ y+2√2z)(-√2x + y + 2√2z)
6. Write the following cubes in expanded form:
(i) (2x + 1)³
Ans: We are given that, (2x + 1)³
We know that, (a + b)³ = a³ + b³ + c³ + 3ab + (a + b)
Therefore, (2x+1)³ = (2x)³ +(1)³ +3. (2x)(1)(2x + 1)
= 8x³ + 1 + 6x(2x + 1) = 8x³ + 12x²+ 6x + 1
(ii) (2a – 3b)³
Ans: We are (2a – 3b)³
We know that, (a – b)³ = a³ – b³ – 3ab(a – b)
∴ (2a – 3b)² = 8a³ – 27b³ – 18ab(2a – 3b)
= 8a³- 27b³- 336a²* b + 54ab²
Ans:
We know that,(a+b)³ = a³+b²+3ab(a+b)
Therefore,
Ans:
We know that, (a – b)³ = a³- b³ – 3ab(a – b)
Therefore,
7. Evaluate the following using suitable identities:
(i) (99)³
Ans: We are given, (99)³
We can also write, (100-1)³
We know that, (a-b)³ = a³-b³-3ab (a-b)
∴ (100-1)³ = (100)³ – (1)³-3.100.1(100-1)
= 10,00,000-1-30000+300 = 9,70,299
(ii) (102)³
Ans: We are given that, (102)³
Now, we know that, (a+b)³ = a³+b³+3ab (a+b)
∴ (100+2)³ = (100)³+(2)³ +3.100.2 (100+2)
= 10,00,000+8+600 (100+2)
= 10,00,000+8+60000+1200 = 10,61,208
(iii) (998)³
Ans: We are given that, (998)³
We can also write, (1000-2)³
Now, we know that, (a-b)³ = a³-b³-3ab (a-b)
∴ (1000-2)³ = (1000)³-(2)³-3.1000 2(1000-2)
= 1,00,00,000-8-6000 (100-2) = 99,40,11,992
8. Factorise each of the following:
(i) 8a³ +b³ +12a²b+6ab²
Ans: We are given that, 8a³+b³+12a²b+6ab²
We can also write, (2a)³+(b)³+6ab (2a+b)
or (2a)³+b²+3.2a.b (2a+b)
We know that, a³+b³+3ab (a+b) = (a+b)³
∴ (2a)³+(b)³ +3.2a.b (2a+b) = (2a+b)³
(ii) 8a³-b³-12a²b+6ab²
Ans: We are given that, 8a³-b³-12a²b+6ab²
We can also write, (2a)³-(b)³-6ab (2a-b)
or (2a)³-b³-3.2a.b (2a-b)
We know that, a³-b³-3ab (a-b) = (a-b)³
∴ (2a)³-(b)³-3.2a.b (2a-b) = (2a-b)³
(iii) 27-125a³-135a+225a²
Ans: We are given that, 27-125a³-135a+225a²
We can also write, (3)³-(5a)³-45a (3-5a)
or, (3)³-(5a)³-3.3.5a (3-5a)
We know that, a³-b³-3ab (a-b) = (a+b)³
∴ (3)³-(5a)³ -3.3.5a (3-5a)
= (3-5a)³ = (3-5a) (3-5a) (3-5a)
(iv) 64a³-276³-144a²b+108ab²
Ans: We are given that, 64a³-27b³-144a²b+108ab²
We can also write, (4a)³-(36)³-36ab (4a-3b)
or, (4a)³-(3b)³-3.4a. 3b (4a-3b)
Now, we know that, a³-b³-3ab (a – b) = (a-b)³
∴ (4a)³-(36)³-3.4a.3b (4a-3b) = (4a-3b)³
Ans: we are given that,
we can also write,
we know that a³-b³-3ab(a-b) = (a-b)³
9. Verify:
(i) x³+ y³ = (x + y) (x²-xy + y²)
Ans: R.H.S.= (x+y)(x²-xy + y²)
By actual multiplication we have
x³-x²y+xy²-xy²+x²y+y³ = x³ + y³ = L.HS.
(ii) x³-y³ = (x-y)(x² + xy + y²)
Ans: R.H.S.= (x+y)(x²-xy + y²)
by actual multiplication, we have,
x³-x²y+xy²-xy²+x²y+y³ = x³ + y³ = L.H.S
10. Factorise each of the following:
(i) 27y³+125z³
Ans: We are given, 27 y³+125z³
We can also write, (3y)³+(5z)³
We know that, a³+b³ = (a+b) (a²-ab+b²)
Therefore,
(3y)³ + (5z)³ = (3y+5z) ((3y)² – (3y). (5z)+(5z)²)
= (3y+5z) (9y²-15yz+25z²)
(ii) 64m³-34.3n³
Ans: We are given, 64m³-343n³
We can also write, (4m)³-(7n)³
We know that, a³-b³ = (a-b)(a²+ab+b²)
Therefore,
((4m)³-(7n)³ = (4m-7n) ((4m)²+(4m) (7n)+(7n)²)
= (4m-7n) (16m²+28 mn+49n²)
11. Factorise: 27x³+ y³+z³-9xyz
Ans: We have given, 27x³+ y³ + z³-9xyz
We can also write, (3x)³+(y)³+(x)³-3. (3x). (y). (z)
We know that, a³ +b³+c³-3abc = (a+b+c)(a²+b²+c²-ab-bc-ca)
Therefore,
(3x)³+(v)³+ (z)³-3. (3x) (y)(z) = (3x+y+z) [(3x)²+(y)² +(z)²-(3x). (y)-(y)(z)-(z) (3x)]
= (3x+y+z) (9x²+y²+z²-3xy-yz-3zx)
12. Verify that:
x² + y²+z² −3xyz=1/2(x+y+z) [(x−y)² +(v−z)² +(z−x)²]
Ans: We are given L.H.S. is x³ + y³ +z³-3xyz
We know that, x³ + y³+z³-3xyz
=(x+y+z) (x² + y²+z²-xy-yz-zx)
= (x+y+z)2×1/2(x²+y²+z²-xy-yz-zx)
= 1/2 (x+y+z) (2x²+2y²+2z² -2xy-2yz-2zx)
= 1/2(x + y + z) (( x − y)² + (y + z)² + (z−x)²)
Hence verified.
13. (i) If x+y+z = 0, show that x³+ y³+z³ = 3xyz
Ans: We know that, x³ + y³ + z³-3xyz
= (x + y + z) (x² + y²+z²-xy-yz-zx)
Here, we have given x+y+z = 0 ∴ x³+y³ +z³-3xyz
= 0x (x² + y²+z²-xy-yz-zx)
or, x³ + y³ + z³-3xyz = 0
∴ x³+y³+z³ = 3xyz
(ii) If a+b+c = 0, then show that
a²(b+c)+b²(c+a)+c²(a + b) + 3abc = 0
Ans: a+b+c = 0
∴ a+b = -c
b+c = -a
c+a = -b
L.H.S. = a²(b+c)+b²(c+a)+c²(a+b)+3abc
= a²(-a)+b²(-b) + c²(-c) + 3abc
= -a³-b³-c³+ 3abc
= -(a³+b³+c³-3abc) [∵ a+b+c = 0]
= 0 = R. H.S.
(iii) If 2a-b+c = 0, then show that 4a²b² + c² + 4ac = 0
Ans: 2a-b+c=0 ⇒ 2a+c = b
⇒ (2a+c)² = (b)² [squaring both sides]
⇒ 4a²+2.2a.c+c² = b²
⇒ 4a²-b²+c²+4ac = 0 Proved.
(iv) If a+b+c = 0, then show that
Ans: a+b+c = 0
∴ a+b = -c
b+c = -a
c+a = -b
(v) If a² + b² + c² -ab-bc-ca = 0 then show that, a = b = c
Ans: a²+b²+c²-ab-bc-ca = 0
14. Without actually calculating the cubes, find the value of each of the following:
(i) (-12)³+(7)³+(5)³
Ans: We are given that, (-12)³+(7)³+(5)³
First we check the value of x+y+z we have -12+7+5 = 0
We know that if x+y+z = 0 then x³+y³+z³ = 3xyz
∴ (-12)³+(7)³ +(5)³ = 3x(-12)x7x5 = -1260
(ii) (28)³+(-15)³+(-13)³
Ans: (28)³+(-15)³+(-13)³
First we check the value of x+y+z we have,
28+(-15)+(-13) = 0
We know that if x + y + z = 0 then, x³ + y³+z³ = 3xyz,
x³+y³+z³ = 3xyz ∴ (28)³+(-15)³+(-13)³
= 3x28x(-15)x(-13) = 16,380
15. Give possible expressions of the length and breadth of each of the following rectangles in which their areas are given.
(i) Area: 25a²-35a+12
Ans: We know that, Area of rectangle = length X breadth …(1)
But, we have given that
Area of rectangle = 35a²+35a+12
From equations (1) and (2)
lxb = 25a²-35a+12 = 25a²-35a+12 …(2)
= (5a-4) (5a-3)
∴ Length (I) = (5a-3)
(ii) Area: 35y²+13y-12
Ans: We know that, Breadth (b) = (5a-4)
Area of rectangle = length x breadth …(1)
But, we have given that
Area of rectangle = 35y²+13y-12 …(2)
From equations (1) and (2)
Length x breadth = 35y²+ 13y – 12 = (5y + 4)(7y -3)
or, Length X breadth = (5y + 4) (7y -3)
Therefore, length = (5y + 4), breadth = (7y -3)
16. What are the possible expression for the dimension of the cuboids whose volumes are given below:
(i) volume = 3x²- 12x
Ans: We know that, volume of cuboid = length X breadth X height
But we have given that,
volume of cuboid = 3x²-12x = 3x² -12x
or, length X breadth X height = 3 × x × (x – 4)
∴ length of cuboid = 3
breadth of cuboid = X height of cuboid = (x-4)
(ii) volume = 12ky² + 8ky -20k?
Ans: We know that, volume of cuboid = length x breadth X height.. (1)
But, we have given that
volume of cuboid = 12ky²+8ky-20k … (2)
From equation (1) and (2)
Length x breadth x height = 12ky² + 8ky – 20k
= 4k(3y² +2y-5) = 4k [y(3y +5) – 1(3y+ 5)]
or, length X breadth X height = 4k(3y + 5) (y-1)
∴ length of cuboid = 4k
breadth of cuboid = (3y + 5)
and height of cuboid = (y-1)
Hi! my Name is Parimal Roy. I have completed my Bachelor’s degree in Philosophy (B.A.) from Silapathar General College. Currently, I am working as an HR Manager at Dev Library. It is a website that provides study materials for students from Class 3 to 12, including SCERT and NCERT notes. It also offers resources for BA, B.Com, B.Sc, and Computer Science, along with postgraduate notes. Besides study materials, the website has novels, eBooks, health and finance articles, biographies, quotes, and more.
