Chapter 2: Polynomials

Ans: Option (D) is the correct option.

Here we will use the remainder theorem.

If any Polynomial  $ f(x) $  is divided by  $ x - h $ , then the remainder will be  $ f\left( h \right) $ 

Here given Polynomial is  $ {x^{51}} + 51 $  and divided by  $ x + 1 $ 

So, by using the remainder theorem we can say that

Remainder=  $ f\left( { - 1} \right) $  since the divisor is  $ x + 1 $ 

Therefore we can write it as

\[ \Rightarrow f(x) = {x^{51}} + 51\]

\[ \Rightarrow f( - 1) = {( - 1)^{51}} + 51\]

\[ \Rightarrow f( - 1) =  - 1 + 51\]

\[\therefore f( - 1) = 50\]

Ans: Option (C) is correct.

Let  $ p\left( x \right) = 2{x^2} + kx $ 

Since, \[\left( {x{\text{ }} + {\text{ }}1} \right)\] is a factor of \[p\left( x \right)\], then

Using, factor theorem

\[ \Rightarrow p\left( { - 1} \right) = 0\]

\[ \Rightarrow 2{\left( { - 1} \right)^2} + {\text{ }}k\left( { - 1} \right){\text{ }} = {\text{ }}0\]

\[ \Rightarrow \;2 - k{\text{ }} = {\text{ }}0\]

\[ \Rightarrow \;k = {\text{ }}2\]

Therefore, the value of k is  $ 2 $ .

Ans: (B) is the correct option.

 If  $ x + 1 $  is the factor of the Polynomial, then  $ x =  - 1 $  is the root of the Polynomial.

(A) Let  $ p\left( x \right) = {x^3} + {x^2} - x + 1 $ 

Now, put  $ x =  - 1 $ .

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

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

 $ p\left( { - 1} \right) = 2 $ 

Therefore,  $ p\left( { - 1} \right) \ne 0 $ .

So,  $ x + 1 $  is not a factor of the  $ p\left( x \right) $ .

(B) Let  $ p\left( x \right) = {x^3} + {x^2} + x + 1 $ 

Now, put  $ x =  - 1 $ .

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

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

 $ p\left( { - 1} \right) = 0 $ 

Therefore,  $ p\left( { - 1} \right) = 0 $ .

So,  $ x + 1 $  is a factor of the  $ p\left( x \right) $ .

(C) Let  $ p\left( x \right) = {x^4} + {x^3} + {x^2} + 1 $ 

Now, put  $ x =  - 1 $ .

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

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

 $ p\left( { - 1} \right) = 2 $ 

Therefore,  $ p\left( { - 1} \right) \ne 0 $ .

So,  $ x + 1 $  is not a factor of the  $ p\left( x \right) $ .

(D) Let  $ p\left( x \right) = {x^4} + 3{x^3} + 3{x^2} + x + 1 $ 

Now, put  $ x =  - 1 $ .

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

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

 $ p\left( { - 1} \right) = 1 $ 

Therefore,  $ p\left( { - 1} \right) \ne 0 $ .

So,  $ x + 1 $  is not the factor of the  $ p\left( x \right) $ .

Ans: (D) is the correct option.

 $  \Rightarrow \left( {25{x^2} - 1} \right) + {\left( {1 + 5x} \right)^2} $ 

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

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

Now, taking common  $ \left( {5x + 1} \right) $ 

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

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

 $  \Rightarrow \left( {5x + 1} \right)\left( {10x} \right) $ 

Therefore, one of the factor of  $ \left( {25{x^2} - 1} \right) + {\left( {1 + 5x} \right)^2} $  is  $ 10x $ .

Ans: (D) is the correct option.

Here we will use the identity  $ {a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right) $.

We have, 

  $  \Rightarrow {249^2} - {248^2} $ 

 $  \Rightarrow \left( {249 - 248} \right)\left( {249 + 248} \right) $ 

 $  \Rightarrow \left( 1 \right)\left( {497} \right) $ 

 $  \Rightarrow 497 $ 

Ans: (B) is the correct option.

In this question, we will use middle term splitting for factorization

 $  \Rightarrow 4{x^2} + 8x + 3 $ 

We can also write it as,

 $  \Rightarrow 4{x^2} + 6x + 2x + 3 $ 

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

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

Ans: (D) is the correct option.

We have,  $ {\left( {x + y} \right)^3} - \left( {{x^3} + {y^3}} \right) $ .

Now, in this question we will use the identity  $ {\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right) $ 

 $  \Rightarrow {x^3} + {y^3} + 3xy\left( {x + y} \right) - \left( {{x^3} + {y^3}} \right) $ 

 $  \Rightarrow 3xy\left( {x + y} \right) $ 

So, we can say that  $ 3xy $  is a factor of  $ {\left( {x + y} \right)^3} - \left( {{x^3} + {y^3}} \right) $ .

Ans: (D) is the correct option.

Here in this question, we will use the identity  $ {\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right) $ 

Therefore, 

 $ {\left( {x + 3} \right)^3} = {x^3} + {3^3} + 3 \times x \times 3\left( {x + 3} \right) $ 

 $ {\left( {x + 3} \right)^3} = {x^3} + 27 + 9{x^2} + 27x $ 

Therefore, the coefficient of x is  $ 27 $ .

Ans: The correct option is (C).

Here, it is given that   $ \dfrac{x}{y} + \dfrac{y}{x} =  - 1 $ 

On taking LCM, we get

 $ \dfrac{{{x^2} + {y^2}}}{{xy}} =  - 1 $ 

On cross-multiplication, we get

 $ {x^2} + {y^2} =  - xy $ 

Now, we know that

 $ {x^3} - {y^3} = \left( {x - y} \right)\left( {{x^2} + {y^2} + xy} \right) $ 

On putting  $ {x^2} + {y^2} =  - xy $  in above equation, we get

 $ {x^3} - {y^3} = \left( {x - y} \right)\left( { - xy + xy} \right) $ 

 $ {x^3} - {y^3} = \left( {x - y} \right)\left( 0 \right) $ 

 $ {x^3} - {y^3} = 0 $ 

Ans: (C) is the correct option.

 Here in this question, we will use the identity  $ {a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right) $ 

 $  \Rightarrow 49{x^2} - b = \left( {7x + \dfrac{1}{2}} \right)\left( {7x - \dfrac{1}{2}} \right) $ 

 $  \Rightarrow 49{x^2} - b = {\left( {7x} \right)^2} - {\left( {\dfrac{1}{2}} \right)^2} $ 

 $  \Rightarrow 49{x^2} - b = 49{x^2} - {\left( {\dfrac{1}{2}} \right)^2} $ 

On comparing, we get 

 $ b = {\left( {\dfrac{1}{2}} \right)^2} $ 

 $ b = \dfrac{1}{4} $