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: