NCERT solution for class 9 maths polynomials ( Chapter 2)
Solution for Exercise 2.1
State reason for your answer.
i)\(4x^2 - 3x + 7\)
ii)\(y^2 + \sqrt{2}\)
iii)\(3\sqrt{t} + t\sqrt{2}\)
iv)\(y + 2/y\)
v)\(x^{10} + y^3 + t^{50}\)
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Answer :
i) \(4x^2 - 3x + 7\) is a polynomail in one variable. As it is having only non-negative integral powers of x.
ii)\(y^2 + \sqrt{2}\) is also a polynomial in one variable. As it is having only non-negative integral power of y.
iii)\(3\sqrt{t} + t\sqrt{2}\) is not a polynomial. As the integral power of the variable is not a whole number.
iv)\(y + 2/y\) is not a polynomial. As it having a negative integral power of y.
v) \(x^{10} + y^3 + t^{50}\) is a polynomial with three variables.
i)\(2 + x^2 + x\)
ii)\(2 - x^2 + x^3\)
iii)\(\pi{x^2}/2 + x\)
iv)\(\sqrt{2}x - 1\)
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Answer :
i) The coefficient of \(x^2\) in \(2 + x^2 + x\) is \(1\).
ii) The coefficient of \(x^2\) in \(2 - x^2 + x^3\) is \(-1\).
iii) The coefficient of \(x^2\) in \(\pi{x^2}/2 + x\) is \(\pi/2\).
iv) The coefficient of \(x^2\) in \(\sqrt{2}x - 1\) is \(0\).
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Answer :
A binomial of degree 35 = \(4y^35 + 3\sqrt{2}\) and a monomial of degree 100 = \(9y^{100}\)
i)\(5x^3 + 4x^2 + 7x\)
ii)\(4 - y^2\)
iii)\(5t - \sqrt{7}\)
iv)\(3\)
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Answer :
i)\(5x^3 + 4x^2 + 7x\) is a polynomial with highest power of variable 3.
Hence, the degree of given polynomial is 3.
ii)\(4 - y^2\) is a polynomial with highest power of variable 2.
Hence, the degree of given polynomial is 2.
iii)\(5t - \sqrt{7}\) is a polynomial with highest power of variable 1.
Hence, the degree of given polynomial is 1.
iv)\(3\) is a polynomial with highest power of variable 0.
Hence, the degree of given polynomial is 0.
i)\(x^2 + x\)
ii)\(x - x^3\)
iii)\(y + y^2 + 4\)
iv)\(1 + x\)
v)\(3t\)
vi)\(r^2\)
vii)\(7x^3\)
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Answer :
i) The degree of polynomial \(x^2 + x\) is 2. Hence, it is a quadratic polynomial.
ii) The degree of polynomial \(x - x^3\) is 3. Hence, it is a cubic polynomial.
iii) The degree of polynomial \(y + y^2 + 4\) is 2. Hence, it is a quadratic polynomial.
iv) The degree of polynomial \(1 + x\) is 1. Hence, it is a linear polynomial.
v) The degree of polynomial \(3t\) is 1. Hence, it is a linear polynomial.
vi) The degree of polynomial \(r^2\) is 2. Hence, it is a quadratic polynomial.
vii) The degree of polynomial \(7x^3\) is 3. Hence, it is a cubic polynomial.
Solution for Exercise 2.2
i) x = 0
ii) x = -1
iii) x = 2
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Answer :
Let f(x) = \(5x - 4x^2 + 3\)
i)The value of f(x) at x = 0 is :
= \(5(0) - 4(0)^2 + 3\)
= 3
f(0) = 3.
ii)The value of f(x) at x = -1 is :
= \(5(-1) - 4(-1)^2 + 3\)
= -5 - 4 + 3
= -6
f(-1) = -6.
iii)The value of f(x) at x = 2 is :
= \(5(2) - 4(2)^2 + 3\)
= 10 - 16 + 3
= -3
f(2) = -3.
i)\(p(y) = y^2 -y + 1\)
ii)\(p(t) = 2 + t + 2t^2 - t^3\)
iii)\(p(x) = x^3\)
iv)\(p(x) = (x - 1)(x + 1)\)
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Answer :
i)\(p(y) = y^2 - y + 1\)
Firstly, let us put x = 0,
Therefore, \(p(0) = 0^2 - 0 +1\)
Hence, p(0) = 1
Now, let us put x = 1,
Therefore, \(p(1) = 1^2 - 1 +1\)
Hence, p(1) = 1
Now, let us put x = 2,
Therefore, \(p(2) = 2^2 - 2 +1\)
i.e. \(p(2) = 4 - 2 +1\)
Hence, p(2) = 3
ii)\(p(t) = 2 + t + 2t^2 - t^3\)
Firstly, let us put x = 0,
Therefore, \(p(0) = 2 + 0 + 2{0}^2 - {0}^3\)
Hence, p(0) = 2
Now, let us put x = 1,
Therefore, \(p(1) = 2 + 1 + 2{1}^2 - {1}^3\)
\(p(1) = 3 + 2 - 1\)
Hence, p(1) = 4
Now, let us put x = 2,
Therefore, \(p(2) = 2 + 2 + 2{2}^2 - {2}^3\)
i.e. \(p(2) = 4 + 8 - 8\)
Hence, p(2) = 4
iii)\(p(x) = x^3\)
Firstly, let us put x = 0,
Therefore, \(p(0) = {0}^3\)
Hence, p(0) = 0
Now, let us put x = 1,
Therefore, \(p(1) = {1}^3\)
Hence, p(1) = 1
Now, let us put x = 2,
Therefore, \(p(2) = {2}^3\)
Hence, p(2) = 8
iv)\(p(x) = (x - 1)(x + 1)\)
Firstly, let us put x = 0,
Therefore, \(p(0) = (0 - 1)(0 + 1)\)
Hence, p(0) = -1
Now, let us put x = 1,
Therefore, \(p(1) = (1 - 1)(1 + 1)\)
Hence, p(1) = 0
Now, let us put x = 2,
Therefore, \(p(2) = (2 - 1)(2 + 1)\)
i.e. \(p(2) = 1 × 3\)
Hence, p(2) = 3
i)\(p(x) = 3x + 1\), x = -1/3
ii)\(p(x) = 5x - \pi\), x = 4/5
iii)\(p(x) = x^2 - 1\), x = 1, -1
iv)\(p(x) = (x + 1)(x - 2)\), x = -1, 2
v)\(p(x) = x^2\), x = 0
vi)\(p(x) = lx + m\), x = -m/l
vii)\(p(x) = 3x^2 - 1\), \(x = -1/\sqrt{3}\), \(2/\sqrt{3}\)
viii)\(p(x) = 2x + 1\), x = 1/2
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Answer :
i)\(p(x) = 3x + 1\)
To check, if x = -1/3 is zero of given polynomial then p(-1/3) should be equal to 0.
So, p(-1/3) = 3(-1/3) + 1 = -1 +1 = 0
Therefore, x = -1/3 is zero of given polynomial.
ii)\(p(x) = 5x - \pi\)
To check, if 4/5 is zero of given polynomial then p(4/5) should be equal to 0.
So, \(p(4/5) = 5(4/5) - \pi = 4 - \pi\)
Therefore, x = 4/5 is not a zero of given polynomial.
iii)\(p(x) = x^2 - 1\)
To check, if x = 1 is zero of given polynomial then p(1) should be equal to 0.
So, p(1) = \(1^2 - 1\) = 1 - 1 = 0
Therefore, x = 1 is a zero of given polynomial.
To check, if x = -1 is zero of given polynomial then p(-1) should be equal to 0.
So, p(-1) = \({-1}^2 - 1\) = 1 - 1 = 0
Therefore, x = 1 is too, a zero of given polynomial.
iv)\(p(x) = (x + 1)(x - 2)\)
To check, if x = -1 is zero of given polynomial then p(-1) should be equal to 0.
So, p(-1) = (-1 + 1)(-1 - 2) = 0 × -3 = 0
Therefore, x = -1 is zero of given polynomial.
To check, if x = 2 is zero of given polynomial then p(2) should be equal to 0.
So, p(2) = (2 + 1)(2 - 2) = 3 × 0 = 0
Therefore, x = 2 is too, a zero of given polynomial.
v)\(p(x) = x^2\)
To check, if x = 0 is zero of given polynomial then p(0) should be equal to 0.
So,\( p(0) = (0^3) = 0\)
Therefore, x = 0 is zero of given polynomial.
vi)\(p(x) = lx + m\)
To check, if x = -m/l is zero of given polynomial then p(-m/l) should be equal to 0.
So, p(-m/l) = l(-m/l) + m = -m + m = 0
Therefore, x = -m/l is zero of given polynomial.
vii)\(p(x) = 3x^2 - 1\)
To check, if\( x = -1/\sqrt{3}\) is zero of given polynomial then p(\(-1/\sqrt{3}\)) should be equal to 0.
So, p(\(-1/\sqrt{3}\)) = \(3(-1/\sqrt{3})^2 - 1\) = 1 - 1 = 0
Therefore,\( x = -1/\sqrt{3}\) is zero of given polynomial.
To check, if \(x = 2/\sqrt{3}\) is zero of given polynomial then p(\(2/\sqrt{3}\)) should be equal to 0.
So, p(\(2/\sqrt{3}\)) = \(3(2/\sqrt{3})^2 - 1\) = 3
Therefore, \(x = 2/\sqrt{3}\) is not a zero of given polynomial.
viii)\(p(x) = 2x + 1\)
To check, if x = 1/2 is zero of given polynomial then p(1/2) should be equal to 0.
So, p(1/2) = 2(1/2) + 1 = 1 + 1 = 2
Therefore, x = 1/2 is not a zero of given polynomial.
i)\(p(x) = x + 5\)
ii)\(p(x) = x - 5\)
iii)\(p(x) = 2x + 5\)
iv)\(p(x) = 3x - 2\)
v)\(p(x) = 3x\)
vi)\(p(x) = ax, a \ne 0\)
vii)\(p(x) = cx + d ,c \ne 0\), c,d, are real numbers.
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Answer :
i)We have, \(p(x) = x + 5\)
We also know that, if p(x) = 0 then x is the Zero of the polynomial.
Therefore, p(x) = 0
i.e. \(x + 5 = 0\)
i.e. x = -5
Hence, -5 is the zero of given ploynomial.
ii)We have, \(p(x) = x - 5\)
We also know that, if p(x) = 0 then x is the Zero of the polynomial.
Therefore, p(x) = 0
i.e. \(x - 5 = 0\)
i.e. x = 5
Hence, 5 is the zero of given ploynomial.
iii)We have, \(p(x) = 2x + 5\)
We also know that, if p(x) = 0 then x is the Zero of the polynomial.
Therefore, p(x) = 0
i.e. \(2x + 5 = 0\)
i.e. x = -5/2
Hence, -5/2 is the zero of given ploynomial.
iv)We have, \(p(x) = 3x - 2\)
We also know that, if p(x) = 0 then x is the Zero of the polynomial.
Therefore, p(x) = 0
i.e. \(3x - 2 = 0\)
i.e. x = 2/3
Hence, 2/3 is the zero of given ploynomial.
v)We have, \(p(x) = 3x\)
We also know that, if p(x) = 0 then x is the Zero of the polynomial.
Therefore, p(x) = 0
i.e. \(3x = 0\)
i.e. x = 0
Hence, 0 is the zero of given ploynomial.
vi)We have, \(p(x) = ax\)
We also know that, if p(x) = 0 then x is the Zero of the polynomial.
Therefore, p(x) = 0
i.e. \(ax = 0\)
i.e.\( x = 0, a \ne 0\)
Hence, 0 is the zero of given ploynomial.
vii)We have, \(p(x) = cx + d\)
We also know that, if p(x) = 0 then x is the Zero of the polynomial.
Therefore, p(x) = 0
i.e. \(cx + d = 0\)
i.e.\( x = -d/c, c \ne 0\)
Hence, -d/c is the zero of given ploynomial.
Solution for Exercise 2.3
i)\(x + 1\)
ii)\(x - 1/2\)
iii)\(x\)
iv)\(x + \pi\)
v)\(5 + 2x\)
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Answer :
Let \(p(x) = x^3 + 3x^2 + 3x + 1\)
i)The zero of \(x + 1\) is x = -1 i.e. (x + 1 = 0)
So, \(p(-1) = {-1}^3 + 3{-1}^2 + 3{-1} + 1\)
i.e. \(p(-1) = -1 + 3 - 3 +1\)
i.e.\(p(-1) = 0\)
Hence, By remainder theorem, required remainder = 0.
ii)The zero of \(x - 1/2\) is x = 1/2 i.e. (x - 1/2 = 0)
So, \(p(1/2) = {1/2}^3 + 3{1/2}^2 + 3{1/2} + 1\)
i.e. \(p(1/2) = 1/8 + 3/4 + 3/2 +1\)
i.e.\(p(1/2) = 27/8\)
Hence, By remainder theorem, required remainder = 27/8.
iii)The zero of \(x\) is x = 0 i.e. (x = 0)
So, \(p(0) = {0}^3 + 3{0}^2 + 3{0} + 1\)
i.e. \(p(-1) = 1\)
Hence, By remainder theorem, required remainder = 1.
iv)The zero of \(x + \pi\) is \(x = -\pi\) i.e. \((x + \pi = 0)\)
So, \(p(-\pi) = {-\pi}^3 + 3{-\pi}^2 + 3{-\pi} + 1\)
i.e. \(p(-\pi) = -\pi^3 + 3\pi^2 - 3\pi + 1\)
Hence, By remainder theorem, required remainder = \(-\pi^3 + 3\pi^2 - 3\pi + 1\).
v)The zero of \(5 + 2x\) is x = -5/2 i.e. (5 + 2x = 0)
So, \(p(-5/2) = {-5/2}^3 + 3{-5/2}^2 + 3{-5/2} + 1\)
i.e. \(p(-5/2) = -125/8 + 75/4 - 15/2 + 1\)
i.e. \(p(-5/2) = -27/8\)
Hence, By remainder theorem, required remainder = -27/8.
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Answer :
Let \(p(x) = x^3 - ax^2 + 6x - a\)
The zero of \(x - a\) is x = a i.e. (x - a = 0)
So, \(p(a) = {a}^3 - a{a}^2 + 6{a} - a\)
i.e. \(p(a) = a^3 - a^3 + 6a - a\)
i.e.\(p(a) = 5a\)
Hence, By remainder theorem, required remainder = 5a.
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Answer :
Let \(p(x) = 3x^3 + 7x\)
To check, whether 7 + 3x is a factor of \(3x^3 + 7x\).By factor theorem, \(7 + 3x = 0\) i.e. x = -7/3
So, \(p(-7/3) = 3(-7/3)^3 + 7(-7/3)\)
i.e. \(p(-7/3) = - 3×(343/27) - 49/3\)
i.e.\(p(-7/3) = -1470/27\)
i.e.\(p(-7/3) = -490/9\)
Hence, 7 + 3x is not the factor of \(3x^3 + 7x\) since the remainder is not zero.
Solution for Exercise 2.4
i)\(x^3 + x^2 + x + 1\)
ii)\(x^4 + x^3 + x^2 + x + 1\)
iii)\(x^4 + 3x^3 + 3x^2 + x + 1\)
iv)\(x^3 - x^2 - (2 + \sqrt{2})x + \sqrt{2}\)
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Answer :
The zero of x + 1 is –1.
i)Let \(p(x) = x^3 + x^2 + x + 1\)
So, To check, whether x + 1 is a factor of \(x^3 + x^2 + x + 1\).
By factor theorem,
\(p(–1) = {–1}^3 + {–1}^2 + {-1} + 1\)
i.e. \(p(–1) = –1 + 1 -1 + 1\)
i.e.\(p(-1) = 0\)
Hence, x + 1 is the factor of \(x^3 + x^2 + x + 1\)
ii)Let \(p(x) = x^4 + x^3 + x^2 + x + 1\)
So, To check, whether x + 1 is a factor of \(x^4 + x^3 + x^2 + x + 1\).
By factor theorem,
\(p(–1) = {-1}^4 + {–1}^3 + {–1}^2 + {-1} + 1\)
i.e. \(p(–1) = 1 – 1 + 1 - 1 + 1\)
i.e.\(p(-1) = 1\)
Hence, x + 1 is not the factor of \(x^4 + x^3 + x^2 + x + 1\)
iii)Let \(p(x) = x^4 + 3x^3 + 3x^2 + x + 1\)
So, To check, whether x + 1 is a factor of \(x^4 + 3x^3 + 3x^2 + x + 1\).
By factor theorem,
\(p(–1) = {-1}^4 + 3{–1}^3 + 3{–1}^2 + {-1} + 1\)
i.e. \(p(–1) = 1 – 3 + 3 - 1 + 1\)
i.e.\(p(-1) = 1\)
Hence, x + 1 is not the factor of \(x^4 + 3x^3 + 3x^2 + x + 1\)
iv)Let \(p(x) = x^3 - x^2 - (2 + \sqrt{2})x + \sqrt{2}\)
So, To check, whether x + 1 is a factor of \(x^3 - x^2 - (2 + \sqrt{2})x + \sqrt{2}\).
By factor theorem,
\(p(–1) = {–1}^3 - {–1}^2 - (2 + \sqrt{2}){-1} + \sqrt{2}\)
i.e. \(p(–1) = – 1 - 1 + 2 + \sqrt{2} + \sqrt{2}\)
i.e.\(p(-1) = 2\sqrt{2}\)
Hence, x + 1 is not the factor of \(x^3 - x^2 - (2 + \sqrt{2})x + \sqrt{2}\)
i) \(p(x) = 2x^3 + x^2 - 2x - 1\), \(g(x) = x + 1\)
ii)\(p(x) = x^3 + 3x^2 + 3x + 1\), \(g(x) = x + 2\)
iii)\(p(x) = x^3 - 4x^2 + x + 6\), \(g(x) = x - 3\)
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Answer :
i)The zero of g(x) = x + 1 is –1.
Let \(p(x) = 2x^3 + x^2 - 2x - 1\)
So, To check, whether x + 1 is a factor of \(2x^3 + x^2 - 2x - 1\).
By factor theorem,
\(p(–1) = 2{–1}^3 + {–1}^2 - 2{-1} - 1\)
i.e. \(p(–1) = –2 + 1 + 2 - 1\)
i.e.\(p(-1) = 0\)
Hence, g(x) = x + 1 is the factor of \(2x^3 + x^2 - 2x - 1\)
ii)The zero of \(g(x) = x + 2\) is –2.
Let \(p(x) = x^3 + 3x^2 + 3x + 1\)
So, To check, whether x + 2 is a factor of \(x^3 + 3x^2 + 3x + 1\).
By factor theorem,
\(p(–2) = {–2}^3 + 3{–2}^2 + 3{–2} + 1\)
i.e. \(p(–2) = –8 + 12 - 6 + 1\)
i.e.\(p(-2) = -1\)
Hence, \(g(x) = x + 2\) is not the factor of \(x^3 + 3x^2 + 3x + 1\)
iii)The zero of \(g(x) = x - 3\) is 3.
Let \(p(x) = x^3 - 4x^2 + x + 6\)
So, To check, whether x - 3 is a factor of \(p(x) = x^3 - 4x^2 + x + 6\).
By factor theorem,
\(p(3) = {3}^3 - 4{3}^2 + {3} + 6\)
i.e. \(p(3) = 27 - 36 + 3 + 6\)
i.e.\(p(3) = 0\)
Hence, \(g(x) = x - 3\) is the factor of \(p(x) = x^3 - 4x^2 + x + 6\)
i)\(p(x) = x^2 + x + k\)
ii)\(p(x) = 2x^2 + kx + \sqrt{2}\)
iii)\(p(x) = kx^2 - \sqrt{2}x + 1\)
iv)\(p(x) = kx^2 - 3x + k\)
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Answer :
The zero of x - 1 is 1.
i)So, as x - 1 is a factor of \(p(x) = x^2 + x + k\), therefore, \(p(1) = 0\)
i.e. \(0 = (1)^2 + 1 + k\)
i.e.\(0 = 1 + 1 + k\)
i.e.\(0 = 2 + k\)
i.e.\(k = -2\)
ii)So, as x - 1 is a factor of \(p(x) = 2x^2 + kx + \sqrt{2}\), therefore, \(p(1) = 0\)
i.e. \(0 = 2(1)^2 + k(1) + \sqrt{2}\)
i.e.\(0 = 2 + k + \sqrt{2}\)
i.e.\(k = -2 - \sqrt{2}\)
iii)So, as x - 1 is a factor of \(p(x) = kx^2 - \sqrt{2}x + 1\), therefore, \(p(1) = 0\)
i.e. \(0 = k(1)^2 - ( \sqrt{2})1 + 1\)
i.e.\(0 = k - \sqrt{2} + 1\)
i.e.\(k = \sqrt{2} - 1\)
iv)So, as x - 1 is a factor of \(p(x) = kx^2 - 3x + k\), therefore, \(p(1) = 0\)
i.e. \(0 = k(1)^2 - 3(1) + k\)
i.e.\(0 = k - 3 + k\)
i.e.\(3 = 2k\)
i.e.\(k = 3/2\)
i)\(12x^2 - 7x + 1\)
ii)\(2x^2 + 7x + 3\)
iii)\(6x^2 + 5x - 6\)
iv)\(3x^2 - x - 4\)
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Answer :
i)\(12x^2 - 7x + 1 = 12x^2 - 4x -3x + 1\)...(by splitting middle term)
i.e. \(= 4x(3x - 1) - 1(3x - 1)\)
i.e.\(=(3x - 1)(4x - 1)\)
ii)\(2x^2 + 7x + 3 = 2x^2 + 6x + x + 3\)...(by splitting middle term)
i.e. \(= 2x(x + 3) + 1(x + 3)\)
i.e.\(=(x + 3)(2x + 1)\)
iii)\(6x^2 + 5x - 6 = 6x^2 + 9x - 4x - 6\)...(by splitting middle term)
i.e. \(= 3x(2x + 3) - 2(2x + 3)\)
i.e.\(=(2x + 3)(3x - 2)\)
iv)\(3x^2 - x - 4 = 3x^2 - 4x + 3x - 4\)...(by splitting middle term)
i.e. \(= x(3x - 4) + 1(3x - 4)\)
i.e.\(=(3x - 4)(x + 1)\)
i)\(x^3 - 2x^2 - x + 2\)
ii)\(x^3 - 3x^2 -9x - 5\)
iii)\(x^3 + 13x^2 + 32x + 20\)
iv)\(2y^3 + y^2 - 2y - 1\)
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Answer :
i)Let \(p(x) = x^3 - 2x^2 - x + 2\)
Here, constant term is 2. So, finding the factors of +2, we get, ±1 and ±2.
Now, \(p(1) = (1)^3 - 2(1)^2 - (1) + 2\)
\(= 1 - 2 - 1 + 2\)
= 0
Therefore, by trial method, we find that, p(1) = 0.
Hence, x - 1 is a factor of p(x).
So, \(x^3 - 2x^2 - x + 2 = x^3 - x^2 - x^2 + x - 2x + 2\)
\(= x^2(x - 1) -x(x - 1) - 2(x - 1)\)
\(= (x - 1)(x^2 - x - 2)\)
now, splitting the middle term,
\(= (x - 1)(x^2 - 2x + x - 2)\)
\(= (x - 1)[x(x - 2) + 1(x - 2)]\)
\(= (x - 1)(x - 2)(x + 1)\)
ii)Let \(p(x) = x^3 - 3x^2 -9x - 5\)
Here, constant term is -5. So, finding the factors of -5, we get, 1 and -5 or vice versa.
Now, \(p(5) = (5)^3 - 3(5)^2 - 9(5) - 5\)
\(= 125 - 75 - 45 - 5\)
= 0
Therefore, by trial method, we find that, p(5) = 0.
Hence, x - 5 is a factor of p(x).
So, \(x^3 - 3x^2 -9x - 5 = x^3 - 5x^2 + 2x^2 - 10x + x - 5\)
\(= x^2(x - 5) + 2x(x - 5) + 1(x - 5)\)
\(= (x - 5)(x^2 + 2x + 1)\)
now, splitting the middle term,
\(= (x - 5)(x^2 + x + x + 1)\)
\(= (x - 5)[x(x + 1) + 1(x + 1)]\)
\(= (x - 5)(x + 1)^2\)
iii)Let \(p(x) = x^3 + 13x^2 + 32x + 20\)
Here, constant term is +20. So, finding the factors of 20, we get, ±1, ±2, ±4 and ±5 or vice versa.
Now, \(p(-1) = (-1)^3 + 13(-1)^2 + 32(-1) + 20\)
\(= -1 + 13 - 32 + 20\)
= 0
Therefore, by trial method, we find that, p(-1) = 0.
Hence, x + 1 is a factor of p(x).
So, \(x^3 + 13x^2 + 32x + 20 = x^3 + 2x^2 + 11x^2 + 22x + 10x + 20\)
\(= x^2(x + 2) + 11x(x + 2) + 10(x + 2)\)
\(= (x + 2)(x^2 + 11x + 10)\)
now, splitting the middle term,
\(= (x + 2)(x^2 + 10x + x + 10)\)
\(= (x + 2)[x(x + 10) + 1(x + 10)]\)
\(= (x + 2)(x + 1)(x + 10)\)
iv)Let \(p(x) = 2y^3 + y^2 - 2y - 1\)
Here, constant term is -1. So, finding the factors of -1, we get, -1.
Now, \(p(-1) = 2(-1)^3 + (-1)^2 - 2(-1) - 1\)
\(= -2 + 1 + 2 - 1\)
= 0
Therefore, by trial method, we find that, p(-1) = 0.
Hence, x + 1 is a factor of p(x).
So, \(2y^3 + y^2 - 2y - 1 = 2y^3 + 2y^2 - y^2 - y - y -1\)
\(= 2y^2(y + 1) - y(x + 1) - 1(y + 1)\)
\(= (y + 1)(2y^2 - y - 1)\)
now, splitting the middle term,
\(= (y + 1)(2y^2 - 2y + y - 1)\)
\(= (y + 1)[2y(y - 1) + 1(y - 1)]\)
\(= (y - 1)(y + 1)(2y + 1)\)
Solution for Exercise 2.5
i)\((x + 4)(x + 10)\)
ii)\((x + 8)(x - 10)\)
iii)\((3x - 4)(3x - 5)\)
iv)\((y^2 + 3/2)(y^2 - 3/2)\)
v)\((3 - 2x)(3 + 2x)\)
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Answer :
i)\((x + 4)(x + 10)\)
Using identity (iv), i.e.,\((x + a)(x + b) = x^2 + (a + b)x + ab\)
We have, \((x + 4)(x + 10) = x^2 + (4 + 10)x + (4)(10)\)
\(= x^2 + 14x + 40\)
ii)\((x + 8)(x - 10)\)
Using identity (iv), i.e.,\((x + a)(x + b) = x^2 + (a + b)x + ab\)
We have, \((x + 8)(x - 10) = x^2 + (8 - 10)x + (8)(-10)\)
\(= x^2 - 2x - 80\)
iii)\((3x + 4)(3x - 5)\)
Using identity (iv), i.e.,\((x + a)(x + b) = x^2 + (a + b)x + ab\)
We have, \((3x + 4)(3x - 5) = (3x)^2 + (4 - 5)x + (4)(5)\)
\(= 9x^2 - x + 20\)
iv)\((y^2 + 3/2)(y^2 - 3/2)\)
Using identity (iii), i.e.,\((a^2 - b^2) = (a + b)(a - b)\)
We have, \((y^2 + 3/2)(y^2 - 3/2) = (y^2)^2 - (3/2)^2\)
\(= y^4 - 9/4\)
v)\((3 - 2x)(3 + 2x)\)
Using identity (iii), i.e.,\((a^2 - b^2) = (a + b)(a - b)\)
We have, \((3 - 2x)(3 + 2x) = (3)^2 - (2x)^2)\)
\(= 9 - 4x^2\)
i)103 × 107
ii)95 × 96
iii)104 × 96
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Answer :
i)103 × 107 = (100 + 3)(100 + 7)
Using identity (iv), i.e.,\((x + a)(x + b) = x^2 + (a + b)x + ab\)
We have, \((100 + 3)(100 + 7) = {100}^2 + (3 + 7)100 + (3)(7)\)
\(= 10000 + 1000 + 21\)
\(= 11021\)
ii)95 × 96 = (100 - 5)(100 - 4)
Using identity (iv), i.e.,\((x + a)(x + b) = x^2 + (a + b)x + ab\)
We have, \((100 - 5)(100 - 4) = {100}^2 - (5 + 4)100 + (4)(5)\)
\(= 10000 - 900 + 20\)
\(= 9120\)
iii)104 × 96 = (100 + 4)(100 - 4)
Using identity (iii), i.e.,\((a^2 - b^2) = (a + b)(a - b)\)
We have, \((100 + 4)(100 - 4) = {100}^2 - {4}^2\)
\(= 10000 - 16\)
\(= 9984\)
i)\(9x^2 + 6xy + y^2\)
ii)\(4y^2 - 4y + 1\)
iii)\(x^2 - y^2/100\)
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Answer :
i)\(9x^2 + 6xy + y^2 = {3x}^2 + 2(3x)(y) + {y}^2\).....by using identity, \((a + b)^2 = a^2 + 2ab + b^2\)
= (3x + y)^2
Therefore, the factors are (3x + y)(3x + y).
ii)\(4y^2 - 4y + 1 = {2y}^2 - 2(2y)(1) + {1}^2\).....by using identity, \((a + b)^2 = a^2 + 2ab + b^2\)
= (2y - 1)^2
Therefore, the factors are (2y - 1)(2y - 1).
iii)\(x^2 - y^2/100 = (x + y/10)(x - y/10)\).....by using identity, \((a^2 - b^2) = (a + b)(a - b))\)
= (x + y/10)(x - y/10)
Therefore, the factors are (x + y/10)(x - y/10).
i)\((x + 2y + 4z)^2\)
ii)\((2x - y + z)^2\)
iii)\((-2x + 3y + 2z)^2\)
iv)\((3a - 7b - c)^2\)
v)\((-2x + 5y - 3z)^2\)
vi)\([1/4a - 1/2b + 1]^2\)
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Answer :
i)\((x + 2y + 4z)^2 = (x)^2 + (2y)^2 +(4z)^2 + 2(x)(2y) + 2(2y)(4z) + 2(x)(4z)\)
by using identity (v)
i.e., \((a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac\)
\(= x^2 + 4y^2 + 16z^2 + 4xy + 16yz + 8xz\)
ii)\((2x - y + z)^2 = (2x)^2 + (-y)^2 + (z)^2 - 2(2x)(y) - 2(y)(z) + 2(2x)(z)\)
by using identity (v)
i.e., \((a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac\)
\(= 4x^2 + y^2 + z^2 - 4xy - 2yz + 4xz\)
iii)\((-2x + 3y + 2z)^2 = (-2x)^2 + (3y)^2 +(2z)^2 - 2(2x)(3y) + 2(3y)(2z) - 2(2x)(2z)\)
by using identity (v)
i.e., \((a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac\)
\(= 4x^2 + 9y^2 + 4z^2 - 12xy + 12yz - 8xz\)
iv)\((3a - 7b - c)^2 = (3a)^2 + (-7b)^2 + (-c)^2 - 2(3a)(7b) + 2(-7b)(-c) - 2(3a)(c)\)
by using identity (v)
i.e., \((a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac\)
\(= 9a^2 + 49b^2 + c^2 - 42ab + 14bc - 6ac\)
v)\((-2x + 5y - 3z)^2 = (-2x)^2 + (5y)^2 +(-3z)^2 - 2(2x)(5y) - 2(5y)(3z) + 2(2x)(3z)\)
by using identity (v)
i.e., \((a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac\)
\(= 4x^2 + 25y^2 + 9z^2 - 20xy - 30yz + 12xz\)
vi)\([1/4a - 1/2b + 1]^2 = (1/4a)^2 + (-1/2b)^2 + (1)^2 - 2(1/4a)(1/2b) - 2(-1/2b)(1) + 2(1/4a)(1)\)
by using identity (v)
i.e., \((a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac\)
\(= a^2/16 + b^2/4 + 1 - ab/4 - b + a/2\)
i)\(4x^2 + 9y^2 + 16z^2 + 12xy - 24yz - 16xz\)
ii)\(2x^2 + y^2 + 8z^2 - 2\sqrt{2}xy + 4\sqrt{2}yz - 8xz\)
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Answer :
i)\(4x^2 + 9y^2 + 16z^2 + 12xy - 24yz - 16xz \)
\(= (2x)^2 + (3y)^2 - (4z)^2 + 2(2x)(3y) - 2(3y)(4z) - 2(2x)(4z)\)
\(=(2x + 3y - 4z)^2\)
ii)\(2x^2 + y^2 + 8z^2 - 2\sqrt{2}xy + 4\sqrt{2}yz - 8xz \)
= \(-\sqrt{2}x)^2 + (y)^2 + (2\sqrt{2}z)^2 - 2(\sqrt{2}x)(y) + 2(y)(2\sqrt{2}z) - 2(\sqrt{2}x)(2\sqrt{2}z)\)
\(=(-\sqrt{2}x + y + 2\sqrt{2}z)^2\)
i)\((2x + 1)^3\)
ii)\((2a - 3b)^3\)
iii)\([3x/2 + 1]^3\)
iv)\((x - 2y/3)^3\)
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Answer :
i)\((2x + 1)^3 = (2x)^3 + (1)^3 + 3(2x)(1)(2x + 1)\)
by using identity\((a + b)^3 = a^3 + b^3 + 3ab(a + b)\)
\(= 8x^3 + 1 + 6x(2x + 1))\)
\(= 8x^3 + 1 + 12x^2 + 6x\)
\(= 8x^3 + 12x^2 + 6x + 1\)
ii)\((2a - 3b)^3 = (2a)^3 - (3b)^3 - 3(2a)(3b)(2a - 3b)\)
by using identity \((a - b)^3 = a^3 - b^3 - 3ab(a - b)\)
\(= 8a^3 - 27b^3 - 18ab(2a - 3b))\)
\(= 8a^3 - 27b^3 - 36a^2b + 54ab^2)\)
\(= 8a^3 - 36a^2b + 54ab^2 - 27b^3\)
iii)\([3x/2 + 1]^3 = (3x/2)^3 + (1)^3 + 3(3x/2)(1)(3x/2 + 1)\)
by using identity \((a + b)^3 = a^3 + b^3 + 3ab(a + b)\)
\(= 27x^3/8 + 1 + 9x/2(3x/2 + 1))\)
\(= 27x^3/8 + 1 + 27x^2/4 + 9x/2\)
\(= 27x^3/8 + 27x^2/4 + 9x/2 + 1\)
iv)\((x - 2y/3)^3 = (x)^3 - (2y/3)^3 - 3(x)(2y/3)(x - 2y/3)\)
by using identity \((a - b)^3 = a^3 - b^3 - 3ab(a - b)\)
\(= x^3 - 8y^3/27 - 2xy(x - 2y/3)\)
\(= x^3 - 8y^3/27 - 2x^2y + 4xy^2/3)\)
\(= x^3 - 2x^2y + 4xy^2/3 + - 8y^3/27\)
i)\({99}^3\)
ii)\({102}^3\)
iii)\({998}^3\)
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Answer :
i)\({99}^3 = (100 - 1)^3 = (100)^3 - (1)^3 - 3(100)(1)(100 - 1)\)
by using identity \((a - b)^3 = a^3 - b^3 - 3ab(a - b)\)
\(= 1000000 - 1 - 300(99)\)
\(= 1000000 - 1 - 2700)\)
\(= 970299\)
ii)\({102}^3 = (100 + 2)^3 = (100)^3 + (2)^3 + 3(100)(2)(100 + 2)\)
by using identity \((a + b)^3 = a^3 + b^3 + 3ab(a + b)\)
\(= 1000000 + 8 + 600(102)\)
\(= 1000000 + 8 + 61200)\)
\(= 1061208\)
iii)\({998}^3 = (1000 - 2)^3 = (1000)^3 - (2)^3 - 3(1000)(2)(1000 - 2)\)
by using identity \((a - b)^3 = a^3 - b^3 - 3ab(a - b)\)
\(= 1000000000 - 8 - 6000(998)\)
\(= 1000000000 - 1 - 5988000)\)
\(= 994011992\)
i)\(8a^3 + b^3 + 12a^2b + 6ab^2\)
ii)\(8a^3 - b^3 - 12a^2b + 6ab^2\)
iii)\(27 - 125a^3 - 135a + 225a^2\)
iv)\(64a^3 - 27b^3 - 144a^2b + 108ab^2\)
v)\(27p^3 - 1/216 - 9p^2/2 + p/4\)
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Answer :
i)\(8a^3 + b^3 + 12a^2b + 6ab^2 \)
i.e.,\(= (2a)^3 + b^3 +3(2a)(b)(2a + b))\)
by using identity \((a + b)^3 = a^3 + b^3 + 3ab(a + b)\)
\(= (2a)^3 + b^3\)
ii)\(8a^3 - b^3 - 12a^2b + 6ab^2 \)
i.e.,\(= (2a)^3 - b^3 -3(2a)(b)(2a - b)\)
by using identity \((a - b)^3 = a^3 - b^3 - 3ab(a - b)\)
\(= (2a)^3 - b^3\)
iii)\(27 - 125a^3 - 135a + 225a^2 \)
i.e.,\(= (3)^3 - (5a)^3 -3(3)(5a)(3 - 5a)\)
by using identity \((a - b)^3 = a^3 - b^3 - 3ab(a - b)\)
\(= (3)^3 - 5a^3\)
iv)\(64a^3 - 27b^3 - 144a^2b + 108ab^2\)
i.e.,\( = (4a)^3 - (3b)^3 -3(4a)(4b)(4a - 3b))\)
by using identity \((a - b)^3 = a^3 - b^3 - 3ab(a - b)\)
\(= (4a)^3 - (3b)^3\)
v)\((27p^3 - 1/216 - 9p^2/2 + p/4\)
i.e.,\( = (3p)^3 - (1/6)^3 -3(3p)(1/6)(3p - 1/6))\)
by using identity \((a - b)^3 = a^3 - b^3 - 3ab(a - b)\)
\(= (3p)^3 - {1/6}^3\)
i)\(x^3 + y^3 = (x + y)(x^2 - xy + y^2)\)
ii)\(x^3 - y^3 = (x - y)(x^2 + xy + y^2) \)
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Answer :
i)We know that,
\((x + y)^3 = x^3 + y^3 + 3xy(x + y)\)
\(x^3 + y^3 = (x + y)^3 - 3xy(x + y)\)
\(= (x + y)[(x + y)^2 - 3xy]\)
\(= (x + y)[(x^2 + y^2 +2xy - 3xy]\)
\(= (x + y)[(x^2 + y^2 - xy]\)
\(= R.H.S\)
Hence, Proved.
ii)We know that,
\((x - y)^3 = x^3 - y^3 - 3xy(x - y)\)
\(x^3 - y^3 = (x - y)^3 + 3xy(x - y)\)
\(= (x - y)[(x - y)^2 + 3xy]\)
\(= (x - y)[(x^2 + y^2 - 2xy + 3xy]\)
\(= (x - y)[(x^2 + y^2 + xy]\)
\(= R.H.S\)
Hence, Proved.
Answer :
i)\(27y^3 + 125z^3 = (3y)^3 + (5z)^3\)
\(= (3y + 5z)[((3y)^2 + (5y)^2 - (3y)(5z)]\)
\(= (3y + 5z)[(9y^2 - 15yz + 25z^2)]\)
ii)\(64m^3 + 343n^3 = (4m)^3 + (7n)^3\)
\(= (4m + 7n)[((4m)^2 + (7n)^2 - (4m)(7n)]\)
\(= (4m + 7n)[(16m^2 - 48mn + 49n^2)]\)
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Answer :
\(27x^3 + y^3 + z^3 - 9xyz = (3x)^3 + (y)^3 + (z)^3 - 3(3x)(y)(z)\)
\(= (3x + y +z)((3x)^2 + y^2 + z^2 - 3xy - yz - z(3x)\)
i.e., by using identity \(a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)\)
We get,
\(= (3x + y +z)(9x^2 + y^2 + z^2 - 3xy - yz - 3zx)\)
Therefore, \(27x^3 + y^3 + z^3 - 9xyz = (3x + y +z)(9x^2 + y^2 + z^2 - 3xy - yz - 3zx)\)
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Answer :
We know that, \(x^3 + y^3 + z^3 - 3xyz = (x + y + z)[x^2 + y^2 + z^2 - xy - yz - zx]\)
\( = 1/2(x + y + z)[2x^2 + 2y^2 + 2z^2 - 2xy - 2yz - 2zx]\)
\( = 1/2(x + y + z)[x^2 + x^2 + y^2 + y^2 + z^2 + z^2 - 2xy - 2yz - 2zx]\)
\( = 1/2(x + y + z)[x^2 + y^2 - 2xy + y^2 + z^2 - 2yz + z^2 + x^2 - 2zx]\)
\( = 1/2(x + y + z)[(x - y)^2 + (y - z)^2 + (z - x)^2]\)
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Answer :
We know that,
\(x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)\)
by using identity \(a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)\)
As it is given that (x + y + z = 0)
i.e., \(= (0)(x^2 + y^2 + z^2 - xy - yz - zx)\)
Therefore, \(x^3 + y^3 + z^3 = 3xyz\)
Hence, proved.
i)\((-12)^3 + (5)^3 + (7))^3\)
ii)\((28)^3 + (-15)^3 + (-13)^3\)
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Answer :
We know that,
\(x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)\)
Also, if x + y + z = 0 then, \(x^3 + y^3 + z^3 = 3xyz\)
i)Here, -12 + 7 + 5 = 0
So, \((-12)^3 + (5)^3 + (7))^3 = 3(-12)(7)(5)\)
\(= -1260\)
ii)Here, 28 - 15 - 13 = 0
So, \((28)^3 + (-15)^3 + (-13))^3 = 3(-13)(28)(-15)\)
\(= 16380\)
i)Area \(25a^2 - 35a + 12\)
ii)Area \(35y^2 + 13y - 12\)
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Answer :
i)We have area of rectangle \(= 25a^2 - 35a + 12\)
\(= 25a^2 - 20a - 15a + 12\)
\(= 5a(5a - 4) - 3(5a - 4)\)
\(= (5a - 4)(5a - 3)\)
One possible answers : Length = 5a - 3, Breadth = 5a - 4.
ii)We have area of rectangle \(= 35y^2 + 13y - 12\)
\(= 35y^2 - 15y + 28y - 12\)
\(= 5y(7y - 3) + 4(7y - 3)\)
\(= (7y - 3)(5y + 4)\)
One possible answers : Length = 7y - 3, Breadth = 5y + 4.
i)Volume \(3x^2 - 12x\)
ii)Volume \(12ky^2 + 8ky - 20k\)
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Answer :
i)We have Volume of cuboid \(3x^2 - 12x = 3x(x - 4)\)
One possible expressions for the dimensions of the cuboid is 3, x and x-4.
ii)We have Volume of cuboid \(12ky^2 + 8ky - 20k\)
\(= 12ky^2 + 20ky - 12ky - 20k\)
\(= 4ky(3y - 5) - 4k(3y - 5)\)
\(= (3y - 5)(4ky - 4k)\)
\(= (3y - 5)4k(y - 1)\)
One possible expressions for the dimensions of the cuboid is 4k, 3y - 5 and y - 1.