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SEBA Class 9 Mathematics Chapter 2 Polynomials
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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)
