Chapter 2: Polynomials

Ans: $ {x^2} + x + 1 $  is a Polynomial in one variable. 

Ans: $ {y^3} - 5y $  is a Polynomial in one variable.

Ans: $ xy + yz + zx $  is a Polynomial in three variables.

Ans: $ {x^2} - 2xy + {y^2} + 1 $  is a Polynomial in two variables.

Ans: Since the highest power of x is  $ 1 $ , the degree of the Polynomial  $ 2x - 1 $  is  $ 1 $ .

Ans: $  - 10 $  is a non-zero constant. A constant term does not contain any variable and its degree is always 0.

Ans: Since the highest power of x is  $ 5 $ , the degree of the Polynomial  $ {x^3} - 9x + 3{x^5} $  is  $ 5 $ .

Ans: Here, we have $ {y^3}\left( {1 - {y^4}} \right) = {y^3} - {y^7} $ . Since the highest power of y is  $ 7 $ , the degree of the Polynomial is  $ 7 $ .

Ans: $ \dfrac{{{x^3} + 2x + 1}}{5} - \dfrac{7}{2}{x^2} - {x^6} $ 

We can also write it as,

 $ \dfrac{{{x^3}}}{5} + \dfrac{2}{5}x + \dfrac{1}{5} - \dfrac{7}{2}{x^2} - {x^6} $ 

As we know that highest power of variable in a Polynomial is known as  degree of a Polynomial. In given Polynomial, the term with highest of x is  $  - {x^6} $ , and the exponent of x in this term in  $ 6 $ .

Ans: The coefficient of  $ {x^3} $ is  $ \dfrac{1}{5} $ .

Ans:  The coefficient of  $ {x^6} $  is  $  - 1 $ .

Ans: The constant term is  $ \dfrac{1}{5} $ .

Ans: Coefficient of  $ {x^2} $  in the given Polynomial is  $ 1 $ .

Ans: The given Polynomial can also be written as  $ 0{x^2} + 3x - 5 $ . So, the coefficient of  $ {x^2} $  in the Polynomial is  $ 0 $ .

Ans: We can write the given polynomial as:

 $  \Rightarrow \left( {x - 1} \right)\left( {3x - 4} \right) $ 

 $  \Rightarrow 3{x^2} - 3x - 4x + 4 $ 

 $  \Rightarrow 3{x^2} - 7x + 4 $ 

Therefore, the coefficient of  $ {x^2} $  in the given Polynomial is  $ 3 $ .

Ans: We can write the given polynomial as:

 $  \Rightarrow \left( {2x - 5} \right)\left( {2{x^2} - 3x + 1} \right) $ 

 $  \Rightarrow 4{x^3} - 6{x^2} + 2x - 10{x^2} + 15x - 5 $ 

 $  \Rightarrow 4{x^3} - 16{x^2} + 17x - 5 $ 

So, the coefficient of  $ {x^2} $  in the given Polynomial is – 16.

Ans: To do this question, we have to keep some points in mind:

(A) A Polynomial in which there is no variable term and there is only constant term, is known as a constant polynomial. 

(B) A Polynomial having degree 1 is known as a linear Polynomial. 

(C) A Polynomial of having degree 2 is known as quadratic Polynomial. 

(D) A Polynomial having degree 3 is known as cubic Polynomial. 

Cubic Polynomial is the correct answer.

Ans: Cubic Polynomial is the correct answer.

Ans: Linear Polynomial is the correct answer.

Ans: Quadratic Polynomial is the correct answer.

Ans: Constant Polynomial is the correct answer.

Ans: Linear Polynomial is the correct answer.

Ans: Cubic Polynomial is the correct answer.

Ans:  Cubic Polynomial is the correct answer.

Ans: Quadratic Polynomial is the correct answer.

Ans: Linear Polynomial is the correct answer.

Ans: A Polynomial which contains  only one term is called a monomial, a Polynomial having only two terms is called binomial, a Polynomial having only three terms is called a trinomial.

$ 5x $  is monomial of degree  $ 1 $ .

Ans: $ {x^{20}} + 5 $  is a binomial of degree  $ 20 $ . 

Ans: $2{x^2} + 2x + 1 $  is a trinomial of degree  $ 2 $ .

Ans: Let p(x) be the given Polynomial.

 $ p\left( x \right) = 3{x^2} - 4{x^2} + 7x - 5 $ 

Now, put  $ x = 3 $ 

 $ p\left( 3 \right) = 3{\left( 3 \right)^3} - 4{\left( 3 \right)^2} + 7\left( 3 \right) - 5 $ 

 $ p\left( 3 \right) = 3\left( {27} \right) - 4\left( 9 \right) + 7\left( 3 \right) - 5 $ 

 $ p\left( 3 \right) = 61 $ 

Now,

Put  $ x =  - 3 $ 

 $ p\left( { - 3} \right) = 3{\left( { - 3} \right)^3} - 4{\left( { - 3} \right)^2} + 7\left( { - 3} \right) - 5 $ 

 $ p\left( { - 3} \right) = 3\left( { - 27} \right) - 4\left( 9 \right) + 7\left( { - 3} \right) - 5 $ 

 $ p\left( { - 3} \right) =  - 81 - 36 - 21 - 5 $ 

 $ p\left( { - 3} \right) =  - 143 $ 

Ans: We have  $ p\left( x \right) = {x^2} - 4x + 3 $ 

Now, put  $ x = 2 $ 

 $ p\left( 2 \right) = \left( {{2^2} - 4 \times 2 + 3} \right) = \left( {4 - 8 + 3} \right) = \left( { - 4 + 3} \right) =  - 1 $ 

Now, put  $ x =  - 1 $ 

 $ p\left( { - 1} \right) = \left( {{{\left( { - 1} \right)}^2} - 4 \times \left( { - 1} \right) + 3} \right) = \left( {1 + 4 + 3} \right) = 8 $ 

Now, put  $ x = \dfrac{1}{2} $ 

 $ p\left( {\dfrac{1}{2}} \right) = \left( {{{\left( {\dfrac{1}{2}} \right)}^2} - 4 \times \dfrac{1}{2} + 3} \right) = \left( {\dfrac{1}{4} - 2 + 3} \right) = \left( {1 + \dfrac{1}{4}} \right) = \dfrac{5}{4} $ 

Now,

 $ p\left( 2 \right) - p\left( { - 1} \right) + p\left( {\dfrac{1}{2}} \right) =  - 1 - 8 + \dfrac{5}{4} $ 

 $ p\left( 2 \right) - p\left( { - 1} \right) + p\left( {\dfrac{1}{2}} \right) =  - 9 + \dfrac{5}{4} = \dfrac{{ - 36 + 5}}{4} =  - \dfrac{{31}}{4} $ 

Ans: We have

 $ p\left( x \right) = 10x - 4{x^2} - 3 $ 

Put,  $ x = 0 $ 

 $ p\left( 0 \right) = 10\left( 0 \right) - 4{\left( 0 \right)^2} - 3 =  - 3 $ 

Put,  $ x = 1 $ 

 $ p\left( 1 \right) = 10\left( 1 \right) - 4{\left( 1 \right)^2} - 3 = 10 - 4 - 3 = 3 $ 

Put,  $ x =  - 2 $ 

 $ p\left( { - 2} \right) = 10\left( { - 2} \right) - 4{\left( { - 2} \right)^2} - 3 =  - 20 - 16 - 3 =  - 39 $ 

Ans: We have

 $ p\left( y \right) = \left( {y + 2} \right)\left( {y - 2} \right) $ 

Put,  $ y = 0 $ 

 $ p\left( 0 \right) = \left( {0 + 2} \right)\left( {0 - 2} \right) =  - 4 $ 

Put,  $ y = 1 $ 

 $ p\left( 1 \right) = \left( {1 + 2} \right)\left( {1 - 2} \right) = \left( 3 \right)\left( { - 1} \right) =  - 3 $ 

Put,  $ y =  - 2 $ 

 $ p\left( { - 2} \right) = \left( { - 2 + 2} \right)\left( { - 2 - 2} \right) = 0 $ 

Ans: A number c can be a zero of a Polynomial p(x) if a number c is such that \[p\left( c \right){\text{ }} = {\text{ }}0\].

 Let  $ p\left( x \right) = x - 3 $ 

 $ p\left( { - 3} \right) =  - 3 - 3 =  - 6 $ 

 $ p\left( { - 3} \right) \ne 0 $ 

Hence,  $  - 3 $  is not a zero of  $ x - 3 $  .

So, the result is False.

Ans: A number c can be a zero of a Polynomial p(x) if a number c is such that \[p\left( c \right){\text{ }} = {\text{ }}0\].

Let  $ p\left( x \right) = 3x + 1 $ 

Put,  $ x =  - \dfrac{1}{3} $ 

 $ p\left( { - \dfrac{1}{3}} \right) = 3\left( { - \dfrac{1}{3}} \right) + 1 $ 

 $ p\left( { - \dfrac{1}{3}} \right) =  - 1 + 1 = 0 $ 

Hence,  $  - \dfrac{1}{3} $  is zero of  $ p\left( x \right) = 3x + 1 $ .

So, the result is True.

Ans: A number c can be a zero of a Polynomial p(x) if a number c is such that \[p\left( c \right){\text{ }} = {\text{ }}0\].

Let  $ p\left( y \right) = 4 - 5(y) $ 

Put,  $ y =  - \dfrac{4}{5} $ 

 $ p\left( { - \dfrac{4}{5}} \right) = 4 - 5\left( { - \dfrac{4}{5}} \right) $ 

 $ p\left( { - \dfrac{4}{5}} \right) = 4 + 4 = 8 $ 

Hence,  $  - \dfrac{4}{5} $  is zero of  $ p\left( y \right) = 4 - 5(y) $ .

Ans: Let  $ p\left( t \right) = {t^2} - 2t $ 

Put, $ t = 0 $ 

 $ p\left( 0 \right) = {\left( 0 \right)^2} - 2\left( 0 \right) $ 

 $ p\left( 0 \right) = 0 $ 

Put, $ t = 2 $ 

 $ p\left( 2 \right) = {\left( 2 \right)^2} - 2\left( 2 \right) = 4 - 4 = 0 $ 

Hence,  $ 0 $  and  $ 2 $  are zeroes of the Polynomial  $ {t^2} - 2t $ .

So, the result is True.

Ans: Let  $ p\left( y \right) = {y^2} + y - 6 $ 

Put,  $ y =  - 3 $ 

 $ p\left( { - 3} \right) = {\left( { - 3} \right)^2} + \left( { - 3} \right) - 6 $ 

 $ p\left( { - 3} \right) = 9 - 9 = 0 $ 

Hence,  $  - 3 $   is a zero of the Polynomial  $ {y^2} + y - 6 $ .

So, the result is True.