Chapter 2: Polynomials

Ans:

Given: \[{{\text{x}}^{\text{2}}}-\text{2x}-\text{8}\].

Now factorize the given polynomial to get the roots.

\[\Rightarrow \text{(x}-\text{4)(x+2)}\]

The value of \[{{\text{x}}^{\text{2}}}-\text{2x}-\text{8}\] is zero.

when \[\text{x}-\text{4=0}\] or \[\text{x+2=0}\]. i.e., \[\text{x = 4}\] or \[\text{x = }-\text{2}\]

Therefore, the zeroes of \[{{\text{x}}^{\text{2}}}-\text{2x}-\text{8}\] are \[\text{4}\] and \[-2\].

Now, Sum of zeroes\[\text{=4}-\text{2= 2 =}-\dfrac{\text{2}}{\text{1}}\text{=}-\dfrac{\text{Coefficient of x}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}\]

$ \therefore Sum\,of\,zeroes=-\dfrac{Coefficient\,of\,x}{Coefficient\,of\,{{x}^{2}}} $

Product of zeroes \[\text{=4 }\!\!\times\!\!\text{ (}-\text{2)=}-\text{8=}\dfrac{\text{(}-\text{8)}}{\text{1}}\text{=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}\]

$ \therefore \Pr oduct\,of\,zeroes=\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}} $ .

Ans:

Given: \[\text{4}{{\text{s}}^{\text{2}}}-\text{4s+1}\]

Now factorize the given polynomial to get the roots.

\[\Rightarrow {{\text{(2s}-\text{1)}}^{\text{2}}}\]

The value of \[\text{4}{{\text{s}}^{\text{2}}}-\text{4s+1}\] is zero.

when \[\text{2s}-\text{1=0}\], \[\text{2s}-\text{1=0}\]. i.e., \[\text{s =}\dfrac{\text{1}}{\text{2}}\] and \[\text{s =}\dfrac{\text{1}}{\text{2}}\]

Therefore, the zeroes of \[\text{4}{{\text{s}}^{\text{2}}}-\text{4s+1}\] are \[\dfrac{\text{1}}{\text{2}}\] and \[\dfrac{\text{1}}{\text{2}}\].

Now, Sum of zeroes\[\text{=}\dfrac{\text{1}}{\text{2}}\text{+}\dfrac{\text{1}}{\text{2}}\text{=1=}\dfrac{\text{(}-\text{4)}}{\text{4}}\text{=}\dfrac{-\text{(Coefficient of s)}}{\text{Coefficient of }{{\text{s}}^{\text{2}}}}\]

\[\therefore \text{Sum of zeroes =}\dfrac{-\text{(Coefficient of s)}}{\text{Coefficient of }{{\text{s}}^{\text{2}}}}\]

Product of zeroes\[=\dfrac{\text{1}}{\text{2}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{2}}=\dfrac{\text{1}}{\text{4}}=\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{s}}^{\text{2}}}}\]

$ \therefore Product of zeroes=\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{s}}^{\text{2}}}} $ .

Ans:

Given: \[\text{6}{{\text{x}}^{\text{2}}}-\text{3}-\text{7x }\]

\[\Rightarrow \text{6}{{\text{x}}^{\text{2}}}-\text{7x}-\text{3}\]

Now factorize the given polynomial to get the roots.

\[\Rightarrow \text{(3x+1)(2x}-\text{3)}\]

The value of  \[\text{6}{{\text{x}}^{\text{2}}}-\text{3}-\text{7x }\] is zero.

when \[\text{3x+1=0}\] or \[\text{2x}-\text{3=0}\]. i.e., \[\text{x =}\dfrac{-\text{1}}{\text{3}}\] or \[\text{x =}\dfrac{\text{3}}{\text{2}}\].

Therefore, the zeroes of  \[\text{6}{{\text{x}}^{\text{2}}}-\text{3}-\text{7x }\] are \[\dfrac{\text{-1}}{\text{3}}\] and \[\dfrac{\text{3}}{\text{2}}\].

Now, Sum of zeroes\[\text{=}\dfrac{-\text{1}}{\text{3}}\text{+}\dfrac{\text{3}}{\text{2}}\text{=}\dfrac{\text{7}}{\text{6}}\text{=}\dfrac{-\text{(}-\text{7)}}{\text{6}}\text{=}\dfrac{-\text{(Coefficient of x)}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}\]

$ \therefore Sum of zeroes\text{=}\dfrac{-\text{(Coefficient of x)}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}} $

Product of zeroes\[\text{=}\dfrac{-\text{1}}{\text{3}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{3}}{\text{2}}\text{=}\dfrac{-\text{3}}{\text{6}}\text{=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}\]

$ \therefore \text{Product of zeroes=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}} $ 

Ans:

Given: \[\text{4}{{\text{u}}^{\text{2}}}\text{+8u}\]

\[\Rightarrow \text{4}{{\text{u}}^{\text{2}}}\text{+8u+0}\]

\[\Rightarrow \text{4u(u+2)}\]

The value of \[\text{4}{{\text{u}}^{\text{2}}}\text{+8u}\] is zero.

when \[\text{4u=0}\] or \[\text{u+2=0}\]. i.e., \[\text{u = 0}\] or \[\text{u =}-\text{2}\]

Therefore, the zeroes of \[\text{4}{{\text{u}}^{\text{2}}}\text{+8u}\] are \[\text{0}\] and \[\text{-2}\].

Now, Sum of zeroes\[\text{=0+(}-\text{2)=}-\text{2=}\dfrac{-\text{8}}{\text{4}}\text{=}\dfrac{-\text{(Coefficient of u)}}{\text{Coefficient of }{{\text{u}}^{\text{2}}}}\]

$ \therefore \text{Sum of zeroes=}\dfrac{-\text{(Coefficient of u)}}{\text{Coefficient of }{{\text{u}}^{\text{2}}}} $

Product of zeroes\[\text{=0 }\!\!\times\!\!\text{ (}-\text{2)= 0 =}\dfrac{\text{0}}{\text{4}}\text{=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{u}}^{\text{2}}}}\]

$ \therefore \text{Product of zeroes=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{u}}^{\text{2}}}} $

Ans:

Given: \[{{\text{t}}^{\text{2}}}-\text{15}\]

\[\Rightarrow {{\text{t}}^{\text{2}}}-\text{0t}-\text{15}\]

Now factorize the given polynomial to get the roots.

\[\Rightarrow \text{(t}-\sqrt{\text{15}}\text{)(t +}\sqrt{\text{15}}\text{)}\]

The value of \[{{\text{t}}^{\text{2}}}-\text{15}\] is zero. 

when \[\text{t}-\sqrt{\text{15}}\text{=0}\] or \[\text{t+}\sqrt{\text{15}}\text{=0}\], i.e., \[\text{t=}\sqrt{\text{15}}\] or \[\text{t=}-\sqrt{\text{15}}\]

Therefore, the zeroes of \[{{\text{t}}^{\text{2}}}-\text{15}\] are \[\sqrt{\text{15}}\] and \[-\sqrt{\text{15}}\].

Now, Sum of zeroes\[\text{=}\sqrt{\text{15}}\text{+(}-\sqrt{\text{15}}\text{)=0=}\dfrac{-\text{0}}{\text{1}}\text{=}\dfrac{-\text{(Coefficient of t)}}{\text{Coefficient of }{{\text{t}}^{\text{2}}}}\]

$ \therefore \text{Sum of zeroes =}\dfrac{-\text{(Coefficient of t)}}{\text{Coefficient of }{{\text{t}}^{\text{2}}}} $

Product of zeroes\[\text{=}\left( \sqrt{\text{15}} \right)\times \left( -\sqrt{\text{15}} \right)\text{=}-\text{15=}\dfrac{-\text{15}}{\text{1}}\text{=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{t}}^{\text{2}}}}\]

$ \therefore \text{Product of zeroes =}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{t}}^{\text{2}}}} $ .

Ans: 

Given: \[\text{3}{{\text{x}}^{\text{2}}}-\text{x}-\text{4}\]

Now factorize the given polynomial to get the roots.

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

The value of \[\text{3}{{\text{x}}^{\text{2}}}-\text{x}-\text{4}\] is zero.

when \[\text{3x}-\text{4=0}\] or \[\text{x+1=0}\], i.e., \[\text{x=}\dfrac{\text{4}}{\text{3}}\] or \[\text{x=}-\text{1}\]

Therefore, the zeroes of \[\text{3}{{\text{x}}^{\text{2}}}-\text{x}-\text{4}\] are \[\dfrac{\text{4}}{\text{3}}\] and \[\text{-1}\].

Now, Sum of zeroes\[\text{=}\dfrac{\text{4}}{\text{3}}\text{+(}-\text{1)=}\dfrac{\text{1}}{\text{3}}\text{=}\dfrac{-\text{(}-\text{1)}}{\text{3}}\text{=}\dfrac{-\text{(Coefficient of x)}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}\]

$ \therefore \text{Sum of zeroes=}\dfrac{-\text{(Coefficient of x)}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}} $

Product of zeroes\[\text{=}\dfrac{\text{4}}{\text{3}}\times \text{(}-\text{1)=}\dfrac{-\text{4}}{\text{3}}\text{=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}\]

$ \therefore \text{Product of zeroes =}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}} $ .

Ans:

Given: \[\dfrac{\text{1}}{\text{4}}\text{,-1}\]

Let the zeroes of polynomial be \[\text{ }\!\!\alpha\!\!\text{ }\] and \[\text{ }\!\!\beta\!\!\text{ }\].

Then, 

\[\text{ }\!\!\alpha\!\!\text{ + }\!\!\beta\!\!\text{  =}\dfrac{\text{1}}{\text{4}}\]

\[\text{ }\!\!\alpha\!\!\text{  }\!\!\beta\!\!\text{ =}-\text{1}\]

Hence, the required polynomial is \[{{\text{x}}^{\text{2}}}-\left( \alpha +\beta  \right)\text{x+}\alpha \beta \].

$ \Rightarrow {{x}^{2}}-\dfrac{1}{4}x-1 $

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

Therefore, the quadratic polynomial is \[\text{4}{{\text{x}}^{\text{2}}}-\text{x}-\text{4}\].

Ans:

Given: \[\sqrt{\text{2}}\text{,}\dfrac{\text{1}}{\text{3}}\]

Let the zeroes of polynomial be \[\text{ }\!\!\alpha\!\!\text{ }\] and \[\text{ }\!\!\beta\!\!\text{ }\].

Then,

\[\text{ }\!\!\alpha\!\!\text{ + }\!\!\beta\!\!\text{  =}\sqrt{\text{2}}\]

\[\text{ }\!\!\alpha\!\!\text{  }\!\!\beta\!\!\text{ =}\dfrac{\text{1}}{\text{3}}\]

Hence, the required polynomial is \[{{\text{x}}^{\text{2}}}-\left( \alpha +\beta  \right)\text{x+}\alpha \beta \].

\[\Rightarrow {{\text{x}}^{\text{2}}}-\sqrt{2}\text{x+}\dfrac{1}{3}\]

\[\Rightarrow 3{{\text{x}}^{\text{2}}}-3\sqrt{2}\text{x+1}\]

Therefore, the quadratic polynomial is \[3{{\text{x}}^{\text{2}}}-3\sqrt{2}\text{x+1}\].

Ans:

Given: \[\text{0,}\sqrt{\text{5}}\]

Let the zeroes of polynomial be \[\text{ }\!\!\alpha\!\!\text{ }\] and \[\text{ }\!\!\beta\!\!\text{ }\].

Then,

\[\text{ }\!\!\alpha\!\!\text{ + }\!\!\beta\!\!\text{  = 0}\]

\[\text{ }\!\!\alpha\!\!\text{  }\!\!\beta\!\!\text{ =}\sqrt{\text{5}}\]

Hence, the required polynomial is \[{{\text{x}}^{\text{2}}}-\left( \alpha +\beta  \right)\text{x+}\alpha \beta \].

\[\Rightarrow {{\text{x}}^{\text{2}}}-0\text{x+}\sqrt{5}\]

\[\Rightarrow {{\text{x}}^{\text{2}}}\text{+}\sqrt{5}\]

Therefore, the quadratic polynomial is \[{{\text{x}}^{\text{2}}}\text{+}\sqrt{\text{5}}\].

Ans:

Given: \[\text{1,1}\]

Let the zeroes of polynomial be \[\text{ }\!\!\alpha\!\!\text{ }\] and \[\text{ }\!\!\beta\!\!\text{ }\].

Then,

\[\text{ }\!\!\alpha\!\!\text{ + }\!\!\beta\!\!\text{  =1}\]

\[\text{ }\!\!\alpha\!\!\text{  }\!\!\beta\!\!\text{  =1}\]

Hence, the required polynomial is \[{{\text{x}}^{\text{2}}}-\left( \alpha +\beta  \right)\text{x+}\alpha \beta \].

\[\Rightarrow {{\text{x}}^{\text{2}}}-\text{1x+1}\]

Therefore, the quadratic polynomial is \[{{\text{x}}^{\text{2}}}-\text{x+1}\].

Ans:

Given: \[-\dfrac{\text{1}}{\text{4}}\text{,}\dfrac{\text{1}}{\text{4}}\]

Let the zeroes of polynomial be \[\text{ }\!\!\alpha\!\!\text{ }\] and \[\text{ }\!\!\beta\!\!\text{ }\].

Then,

\[\text{ }\!\!\alpha\!\!\text{ + }\!\!\beta\!\!\text{ =}-\dfrac{\text{1}}{\text{4}}\]

\[\text{ }\!\!\alpha\!\!\text{  }\!\!\beta\!\!\text{  =}\dfrac{\text{1}}{\text{4}}\]

Hence, the required polynomial is \[{{\text{x}}^{\text{2}}}-\left( \alpha +\beta  \right)\text{x+}\alpha \beta \].

$ \Rightarrow {{\text{x}}^{\text{2}}}-\left( -\dfrac{1}{4} \right)\text{x+}\dfrac{1}{4} $

$ \Rightarrow \text{4}{{\text{x}}^{\text{2}}}\text{+ x +1} $

Therefore, the quadratic polynomial is  $ \text{4}{{\text{x}}^{\text{2}}}\text{+ x +1} $ .

Ans: 

Given: \[\text{4,1}\]

Let the zeroes of polynomial be \[\text{ }\!\!\alpha\!\!\text{ }\] and \[\text{ }\!\!\beta\!\!\text{ }\].

Then,

\[\text{ }\!\!\alpha\!\!\text{ + }\!\!\beta\!\!\text{  = 4}\]

\[\text{ }\!\!\alpha\!\!\text{  }\!\!\beta\!\!\text{  =1}\] 

Hence, the required polynomial is \[{{\text{x}}^{\text{2}}}-\left( \alpha +\beta  \right)\text{x+}\alpha \beta \]. 

$ \Rightarrow {{\text{x}}^{\text{2}}}-\text{4x+1} $ 

Therefore, the quadratic polynomial is  $ {{\text{x}}^{\text{2}}}-\text{4x+1} $.