NCERT Solutions for Maths Class 9 Chapter 2
Exercise 2.1
Question 1: Which of the following expressions are polynomials in one variable and which are not? State the reason for your answer.
i. \(4x^2-3x+7\)
ii. \(y^2+\sqrt2\)
iii. \(3\sqrt{t}+t\sqrt2\)
iv. \(y+\frac{2}{y}\)
v. \(y+2y^{-1}\)
Answer: A polynomial in one variable refers to an expression where the exponent of the variable is a whole number.
i. \(4x^2-3x+7\)
In this polynomial, only one variable is involved which \(x\) and the exponents of the variable are allwhole numbers.
Therefore, the given expression is a polynomial in one variable \(x\).
ii. \(y^2+\sqrt2\)
In this polynomial, only one variable is involved which \(y\) and the expnent of the variable is a whole number.
Therefore, the given expression is a polynomial in one variabl \(y\).
iii. \(3\sqrt{t}+t\sqrt2\)
In this exprerssion, it is given that the exponent of variabl \(t\) in term \(3\sqrt{t}\) is \(\frac{1}{2}\). This exponent is not a whole number.
Therfore, this expression is not a polynomial in one variable.
iv. \(y+\frac{2}{y}\)
We can rewrite this expression as \(y+2y{-1}\)
In this expression, it is given that the exponent of variable \(y\) in term \(2y^{-1}\) is \(-1\). This exponent is not a whole number.
Therefore, this expression is not a polynomial in one varibale.
v. \(x^{10}+y^3+t^{50}\)
In this polynomial, there are \(3\) variable involved which are \(x,\space y,\space t\).
Therefore, the given expression is not polynomial in one variable.
Question 2: Write coefficients of \(x^2\) in each of the following.
i. \(2x+x^2+x\)
ii. \(2-x^2+x^3\)
iii. \(\frac{\pi}{2}x^2+x\)
iv. \(\sqrt2x-1\)
Answer: A coefficient is an integer that is multiplied with the variable of a single term or the terms of a plynomial.
i. \(2+x^2+x\)
We can rewrite theis expression as \(2+1(x^)+x^3\)
Hence, the coefficient of \(x^2\) is \(1\).
ii. \(2-x^2+x^3\)
We can rewrite this expression as \(2-1(x^2)_x^3\)
Hence, the coefficient of \(x^2\) is \(-1\).
iii. \(\frac{\pi}{2}x^2+x\)
In the given expression, the coefficient of \(x^2\) is \(\frac{\pi}{2}\).
iv. \(\sqrt2x-1=0x^2+\sqrt2x-1\)
In the given expression, the coefficient of \(x^2\) is \(0\).
Question 3: Give one example each of a binomial of degree \(34\), and of monomial of degree \(100\).
Answer: A binomial of degree \(35\) refers to a polynomial with two terms and one of the terms have a highest degree of \(35\).
Example of \(100\).
A monomial of degree \(100\) refers to a polynomial with one term and it has a highest degree of \(100\).
Exmaple \(x^{100}\)
Question 4: Write the degree of each of the following polynomials.
i. \(5x^2=4x^2+7x\)
ii. \(4-y^2\)
iii. \(5t-\sqrt7\)
iv. 3
Answer: The degree of a polynomial refers to the highest powr of a variable in the polynomial.
i. \(5x^2+4x^2+7x\)
Here, the highest power of given variable \(x\) is \(3\).
Hence, the degree of this polynomial is \(3\).
ii. \(4-y^2\)
Here, the highest power of given variable \(y\) is \(2\).
Hence, the degree of this polynomial is \(2\)
iii. \(5t-\sqrt7\)
Here, the highest power of given variable \(t\) is \(1\).
Hence, the degree of this polynomial is \(1\).
iv. \(3\)
Here, \(3\) is constant polynomial. We the degree of a constant polynomial is always \(0\).
Hence, the degree of this polynomial is \(0\).
Question 5: Classify the following as linear, quadratic and cubic polynomial.
i. \(x^2+x\)
ii. \(x-x^3\)
iii. \(y+y^2+4\)
iv. \(1+x\)
v. \(3t\)
vi. \(r^2\)
vii. \(7x^2\space 7x^3\)
Answer: The degree of a polynomial is the highest exponential power of the variable in a polynomial equation.
A linear polynomial is a polynomial whose degree is ‘1 ’.
A quadratic polynomial is a polynomial whose degree is ‘ 2 ’.
A cubic polynomial is a polynomial whose degree is ‘ 3 ’.
i. \(x^2+x\)
The given expression has a variable \(x\) and its degree is\(2\).
Hence, it is a quadratic polynomial.
ii. \(x-x^3\)
The given expression has a variable \(x\) and its degree is \(3\).
Hence, it is a cubic polynomial.
iii. \(y+y^2+4\)
The given expression has a variable \(x\) and its degree is \(2\).
Hence, it is a quadratic polynomial.
iv. \(1+x\)
The given expression has a variable \(x\) and its degree is\(1\).
Hence, it is a linear polynomial.
v. \(3t\)
The given expression has a variable \(r\) and its degree is \(1\).
Hence, it is a linear polnomial.
vi. \(r^2\)
The given expression has a variable \(r\) and its degree is \(2\).
Hence, it is a quadratic polynomial.
vii. \(7x^2\space 7x^3\)
The given expression has a variable \(x\) and its degree is \(3\).
Exercise 2.3
1. Determine which of the following polynomials has (\(x+1\)) a factor:
i. \(x^3+x^2+x+1\)
Answer: We know that Zero of \(x+1\) is \(-1\)
Given that,
\(p(x)=x^3+x^2+x+1\)
Now, for \(x-1\)
\(p(-1)=(-1)^3+(-1)^2+(-1)+1\)
\(p(-1)=-1+1-1+1\)
\(p(-1)=0\)
Therefore, by the Factor Theorem, \(x+1\) is a factor of \(x^3+x^2+x+1\)
ii. \(x^4+x^3+x^2+x+1\)
Given that, \(p(x)=x^4+x^3+x^2+x+1\)
Now, for \(x=-1\)
\(p(-1)=(-1)^4+(-1)^3+(-1)^2+(-1)+1\)
\(p(-1)=1-1+1+1\)
\(p(-1)=1\)
Therefore, by the Factor Theorem.
\(x+1\) is not a factor of \(x^4+x^3+x^2+x+1\)
iii. \(x^4+3x^3+3x^2+x+1\)
Answer:
We know that Zero \(x+1\) is \(-1\)
Given that, \(p(x)=x^4+3x^3+3x^2+x+1\)
Now, for \(x=-1\)
\(p(-1)=(-1)^4+3(-1)^3+3(-1)^2+(-1)+1\)
\(p(-1)=1-3+3-1+1\)
\(p(-1)=1\)
Therefore, by the factor Theorem, \(x+1\) is not a factor of \(x^4+3x^3+3x^2+x+1\).
iv. \(x^3+x^2_(2+\sqrt2)x+\sqrt2\)
Answer:
We know that, Zero of \(x+1\) is \(-1\)
Given that,
\(p(x)=x^3+x^2-(2+\sqrt2)x+\sqrt2\)
Now, for \(x=-1\)
\(p(-1)=(-1)^3+(-1)^2-(2+\sqrt2)(-1)+(\sqrt2\)
\(p(-1)=-1+1+2-\sqrt2+\sqrt2\)
\(p(-1)=2+2\sqrt2\)
Therefore, by the Factor Theorem. \(x+1\) is not a factor of \(x^3+x^2-(2+\sqrt2)x+\sqrt2\)
Question 2: Use the Factor Theorem to determine whether \(g(x)\) is a factor of \(p(x)\) in each of the following cases:
i. \(p(x)=2x^3+x^2-2x-1,\space g(x)=x+1\)
We know that Zero of \(g(x)\) is \(-1\)
Now, \(p(-1)=2(-1)^3+(-1)^2-2(-1)-1\)
\(p(-1)=-2+1+2-1\)
\(p(-1)=0\)
Therefore, \(g(x)=x+1\) is a factor of \(p(x)=2x^3+x^3-2x-1\).
ii. \(p(x)=x^3+3x^2+3x+1,\space g(x)=x+2\)
Answer: Given that, \(p(x)=x^3+3x^2+3x+1\) \(g(x)=x+2\)
We know that Zero of \(g(x)\) is \(-2\)
Now, \(p(-2)=(-2)^3+3(-2)^2+3(-2)+1\)
\(p(-2)=-8+12-6+1\)
\(p(-2)=-1\)
Therefore, \(g(x)=x+2\) is not a factor of \(p(x)=x^3+3x^2+3x+1\).
iii. \(p(x)=x^3-4x^2+x+6\), \(g(x)-x-3\)
Answer: Given that, \(p(x)=x^3-4x^2+x+6\) \(g(x)=x-3\)
We know that Zero of \(g(x)\) is \(3\)
Now, \(p(3)=(3)^3-4(3)^2+(3)+6\)
\(p(3)=27-36+3+6\)
\(p(3)=0\)
Thereore, \(g(x)=x-3\) is a factor of \(p(x)=x^3-4x^2+x+6\)
Question 3: Find the value of \(k\), if \(x-1\) is factor of \(p(x)\) in each of the following cases:
i. \(p(x)=x^2+x+k\)
Answer: Given that \(x-1\) is a factor of \(p(x)=x^2+x+k\)
Thus, \(1\) is the zero of the given \(p(x)\)
\(⇒p(1)=0\)
\(⇒p(1)=(1)^2+(1)+k=0\)
\(⇒1+1+k=0\)
\(⇒k=-2\)
Therefore, the value of \(k\), if \(x-1\) is a factor of \(p(x)=x^2+x+k\) is \(-2\).
ii. \(p(x)=2x^2+kx+\sqrt2\)
Answer: Given that \(x-1\) is a factor of \(p(x)=2x^2+kx+\sqrt2\)
Thus, \(1\) is the zero of the given \(p(x)\)
\(⇒p(1)=0\)
\(⇒p(1)=2(1)^2+k(1)+\sqrt2=0\)
\(⇒2+k+\sqrt2=0\)
\(⇒k=-(2+\sqrt2)\)
Therefore, the value of \(k\). If \(x-1\) is a factor of \(p(x)=2x^2+kx+\sqrt2\) is \(-(2+\sqrt2)\)
iii. \(p(x)=kx^2-\sqrt2x+1\)
Answer:
Given that \(x-1\) is a factor of \(p(x)=kx^2-\sqrt2x+1\)
Thus, \(1\) is the zero of the given \(p(x)\)
\(⇒p(1)=0\)
\(⇒p(1)=k(1)^2-\sqrt2(1)+1=0\)
\(⇒k-\sqrt2+1=0\)
\(⇒k=\sqrt2-1\)
Therefore, the value of \(k\). If (x-1\) is a factor of \(p(x)=kx^2-\sqrt2x+1\) is \((\sqrt2-1)\).
iv. \(p(x)=kx^2+3x+k\)
Answer: Given that \(x-1\) is a factor of \(p(x)=kx^2+3x+k\)
Thus, \(1\) is the zero of the given \(p(x)\)
\(⇒p(1)=0\)
\(⇒p(1)=k(1)^2+3(1)+k=0\)
\(⇒k-3+k=0\)
\(⇒k=\frac{3}{2}\)
Therefore, the value of \(k\), if \(x-1\) is a factor of \(p(x)=kx^2+3x+lk\) is \(\frac{3}{2}\)
Question 4: Factorise:
i. \(12x^2-7x+1\)
Answer: Given that, \(p(x)=12^2-7x+1\)
Splitting the middle term
\(⇒12x^2-4x+3x+1\)
\(⇒4x(3x-1)-1(3x-1)\)
Therefore, \(12x^2-4x+3x+1=(3x-1)(4x-1)\)
ii.. \(2x^2+7x+3\)
Answer: Given that, \(p(x)=2x^2+7x+3\)
\(⇒2x^2+x+6x+3\)
\(⇒x(2x+1)+3(2x-1)\)
\(⇒(2x+1)(x+3)\)
Therefore, \(2x^2+x+6x+3=(2x+1)(x+3)\)
iii. \(6x^2+5x-6\)
Answer: Given that, \(p(x)=6x^2+5x-6\)
Splitting the middle term
\(⇒6x^2+9x-4x-6\)
\(⇒3x(2x+3)-2(2x+3)\)
\(⇒(2x+3)(3x-2)\)
Therefore, \(6x^2+9x-4x-6=(2x+3)(3x-2)\)
iv. \(3x^2-x-4\)
Answer: Given that, \(p(x)=3x^2-x-4\)
Splitting the middle term
\(⇒3x^2-4x+3x-4\)
\(⇒x(3x-4)+1(3x-4)\)
\(⇒(3x-4)(x+1)\)
Therefore, \(3x^2-4x+3x-4=(3x-4)(x+1)\)
5. Factorise:
i. \(x^3-2x^2-x+2\)
\(⇒x^3-x-2x^2+2\)
\(⇒x(x^2-1)-2(x^2-1)\)
\(⇒(x^2-1)(x-2)\)
\(⇒(x+1)(x-1)(x-2)\)
Therefore, \(x^3-2x^2-x+2=(x+1)(x-1)(x-2)\)
ii. \(x^3-3x^2-9x-5\)
Answer: Given that, \(o(x)=x^3-x^2-9x-5\)
\(⇒x^3+x^2-4x^2-4x-5x-5\)
\(⇒x^2(x+1)-4x(x+1)-5(x+1)\)
\(⇒(x+1)(x^2-4x-5)\)
\(⇒9x+1)(x^2-5x+x-5)\)
\(⇒(x+1)[x(x-5)+1(x-5)]\)
\(⇒(x+1)(x+1)(x-5)\)
Therefore, \(x^3+13x^2+32x=20=(x+1)(x+10)(x+2)\)
iii. \(x^3+13x^2+32x+20\)
Answer: Givent that, \(p(x)=x^3+13x^2+32x+20\)
\(⇒x^3+x^2+12x^2+12s+20x+20\)
\(⇒x^2(x+1)+12x(x+1)+20(x+1)\)
\(⇒(x+1)(x^2+12x+20)\)
\(⇒(x+1)(x^2+2x+10x+20)\)
\(⇒(x+1)(x^2+2x+10x+20)\)
\(⇒(x-1)[x(x+2)(10(x+2)]\)
\(⇒(x+1)(x+10)(x+2)\)
Therefore, \(x^3+13x^2+32x+20=(x+1)(x+10)(x+2)\)
iv. \(2y^3+y^2-2y-1\)
Answer: Given that, \(p(y)=2y^3+y^2-2y-1\)
\(⇒2y^3-2y^2+3y^2-3y+y-1\)
\(⇒2y^2(y-1)+3y(y-1)+1(y-1)\)
\(⇒(y-1)(2y^2+3y+1)\)
\(⇒(y-1)(2y^2+2y+y+1)\)
\(⇒(y-1)[2y(y+1)+1(y+1)]\)
\(⇒(y-1)(2y+1)(y+1)\)
Therefore, \(2yy^3+y^2-2y-1=(y-1)0(2y+1)(y+1)\)
Exercise 2.3
Question 1: Determine which of the following polynomials has \((x+1)\) a factor:
i. \(x^3+x^2+x+1\)
Answer: We know that Zero of \(x+1\) is \(-1\)
Now, for \(x=-1\)
\(p(-1)=(-1)^3+(-1)^2+(-1)+1\)
\(p(-1)=-1+1-1+1\)
\(p(-1)=0\)
Therefore, by the Factor Theorem, \(x+1\) is a factor of \(x^3+x^2+x+1\).
ii. \(x^4+x^3+x^2+x+1\)
Given that, \(p(x)=x^4+x^3+x^2+x+1\)
Answer: We know that Zero of \(x+1\) is \(-1\)
Now, for \(x=-1\)
\(p(-1)=(-1)^4+(-1)^3+(-1)^2+(-1)+1\)
\(p(-1)=1-1+1-1+1\)
\(p(-1)=1\)
Therefore, by the Factor Theorem. \(x+1\) is not a factor of \(x^4+x^3+x^2+x+1\).
iii. \(x^4+3x^3+3x^2+x+1\)
Answer: We know that Zero of \(x+1\) is \(-1\)
Given that, \(p(x)=x^4+3x^3+3x^2+x+1\)
Now, for \(x=-1\)
\(p(-1)=(-1)^4+3(-1)^3+3(-1)^2+(-1)+1\)
\(p(-1)=1-3+3-1+1\)
\(p(-1)=1\)
Therfore, ny the Factor Theorem. \(x+1\) is not a factor of \(x^4+3x^3+3x^2+x+1\)
iv. \(x^3+x^2-(2+\sqrt2)x+\sqrt2\)
Answer: We know that, Zero of \(x+1\) is \(-1\)
Given that, \(p(x)=x^3+x^2-(2+\sqrt2)x+\sqrt2\)
Now, for \(x=-1\)
\(p(-1)=(-1)^3+(-1)^2+(2+\sqrt2)(-1)+\sqrt2\)
\(p(-1)=-1+1+2+-\sqrt2+\sqrt2\)
\(p(-1)=2+2\sqrt2\)
Therefore, by the factor Theorem. \(x+1\) is not a factor of \(x^3+x^2-(2+\sqrt2)x+\sqrt2\).
Question 2: Use the Factor Theorem to determine whether \(g(x)\) is a factor of \(p(x)\) in each of the following cases.
i. \(p(x)=2x^3+x^2-2x-1\), \(g(x)=x+1\)
Answer: Given that, \(p(x)=2x^3+x^2-2x-1\) \(g(x)=x+1\)
We know that Zero of \(g(x)\) is \(-1\)
Now, \(p(-1)=2(-1)^3+(-1)^2-2(-1)-1\)
\(p(-1)=-2+1+2+-1\)
\(p(-1)=0\)
Therefore, \(g(x)=x+1\) is a factor of \(p(x)=2x^3+x^2-2x-1\).
ii. \(p(x)=x^3+3x^2+3x+1\), \(g(x)=x+2\)
Answer: Given that, \(p(x)=x^3+3x^2+3x+1\), \(g(x)=x+2\)
We know that Zero of \(g(x)\) is \(-2\).
Now, \(p(-2)=(-2)^3+3(-2)^2+3(-2)+1\)
\(p(-2)=-8+12-6+1\)
\(p(-2)=-1\)
Therefore, \(g(x)=x+2\) is not a factor of \(p(x)=x^3+3x^2+3x+1\).
iii. \(p(x)=x^3-4x^2+x+6\), \(g(x)=x-3\)
Answer: Given that, \(p(x)=x^3-4x^2+x+6\) \(g(x)=x-3\)
We know that Zero of \(g(x)\) is \(3\).
Now, \(p(3)=(3)^3-4(3)^2+(3)+6\)
\(p(3)=27-36+3+6\)
\(p(3)=0\)
Therefore, \(g(x)=x-3\) is a factor of \(p(x)=x^3-4x^2+x+6\)
Question 3: Find the value of \(k\), if \(x-1\) is a factor of \(p(x)\) in each of the following cases:
i. \(p(x)=x^2+x+k\)
Answer: Given that \(x-1\) is a fctor of \(p(x)=x^2+x+k\)
Thus, \(1\) is the zero of the given \(p(x)\)
\(⇒p(1)=0\)
\(⇒p(1)=(1)^2+(1)+k=0\)
\(⇒1+1+k=0\)
\(⇒k=-2\)
Therefore, the value of \(k\), if \(x-1\) is a factor of \(p(x)=x^2+x+k\) is \(-2\)
ii. \(p(x)=2x^2+kx+\sqrt2\)
Answer: Given that \(x-1\) is a factor of \(p(x)=2x^2+kx+\sqrt2\)
Thus, \(1\) is the zero of the given \(p(x)\)
\(⇒p(1)=0\)
\(⇒p(1)=2(1)^2+k(1)+\sqrt3=0\)
\(⇒2+k+\sqrt2=0\)
\(⇒k=-(2+\sqrt2)\)
Therefore, the value of \(k\), if \(x-1\) is a factor of \(p(x)=2x^2+kx+\sqrt2\) is \(-(2+\sqrt2)\)
iii. \(p(x)=kx^2-\sqrt2x+1\)
Answer: Given that \(x-1\) is a factor of \(p(x)=kx^2-\sqrt2x+1\)
Thus, \(1\) is the zero of the given \(p(x)\)
\(⇒p(1)=0\)
\(⇒p(1)+k(1)^2-\sqrt2(1)+1=0\)
\(⇒k=\sqrt2+1=0\)
\(⇒k=\sqrt2-1\)
Therefore, the value of \(k\), if \(x-1\) is a factor of \(p(x)=kx^2-\sqrt2x+1\) is (\sqrt2-1)\).
iv. \(p(x)=kx^2+3x+k\)
Answer: Given that \(x-1\) is a factor of \(p(x)=kx^2+3x+k\)
Thus, \(1\) is the zero of the given \(p(x)\)
\(⇒p(1)=0\)
\(⇒p(1)=k(1)^2+3(1)+k=0\)
\(⇒k-3+k=0\)
\(⇒k=\frac{3}{2}\)
Therefore, the value of \(k\), if \(x-1\) is a factor of \(p(x)=kx^2+3x+k\) is \(\frac{3}{2}\)
Question 4: Factorise.
i. \(12x^2-7x+1\)
Answer: Given that, \(p(x)=12x^2-7x+1\)
Splitting the middle term
\(⇒12x^2-4x+3x+1\)
\(⇒4x(3x-1)-1(3x-1)\)
\(⇒(3x-1)(4x-1)\)
Therefore, \(12x^2-4x+3x+1=(3x-1)(4x-1)\)
ii. \(2x^2+7x+3\)
ANswer: Given that, \(p(x)=2x^2+7x+3\)
Splitting the middle term
\(⇒2x^2+x+6x+3\)
\(⇒x(2x+1)+3(2x-1)\)
\(⇒(2x+1)(x+3)\)
Therefore, \(2x^2+x+6x+3=(2x+1)(x+3)\)
iii. \(6x^2+5x-6\)
Answer: Given that, \(p(x)=6x^2+5x-6\)
Splitting the middle term
\(⇒6x^2+9x-4x-6\)
\(⇒3x(2x+3)-2(2x+3)\)
\(⇒(2x+3)(3x-2)\)
Therefore, \(6x^2+9x-4x-6=(2x+3)(3x-2)\)
iv. \(3x^2-x-4\)
Answer: Given that, \(p(x)=3x^2-x-4\)
Splitting the middle term
\(⇒3x^2-4x+3x-4\)
\(⇒x93x-4)+1(3x-4)\)
\(⇒(3x-4)(x+1)\)
Therefore, \(3x^2-4x+3x-4=(3x-4)(x+1)\).
Question 5: Factorise:
i. \(x^3-2x^2-x+2\)
Answer:
