Chapter 2: Polynomials

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Q
Polynomials CBSE NCERT Solutions

Question:

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

(i)  x^{2}-2x-8               (ii) 4s^{2}-4s+1             (iii) 6x^{2}-3-7x  

(iv) 4u^{2}+8u                 (v)  t^{2} -15                       (vi) 3x^{2} -x -4

Answer:

(i)  x^{2}-2x-8                

Let P(x) = x^{2}-2x-8

Zeroes of the polynomial are the value of  where P(x) = 0.

Putting P(x) = 0 then we get

x^{2}-2x-8

We can find zeroes of the polynomial using splitting the middle term method.

In splitting the middle term method we need to find two numbers where the sum is -2 and product is -8 x 1 = -8.

Therefore, we can write,

x^{2}-4x+2x-8=0

\Rightarrow x(x-4)+2(x-4)=0

\Rightarrow (x-4)(x+2)=0

Either (x-4)=0 \quad or  (x+2)=0 

So,  x = 4 and x = -2.

Let \alpha = 4 and \beta  = -2 are the zeroes of the polynomial.

P(x) = x^{2} - 2x - 8 = 1 x^{2} + (-2)x + (-8)

If we compare P(x)  with  ax^{2} + bx +c then we get 

a = 1, b = -2 and c = -8

Now we verify the relationship between the zeroes and the coefficients in two different cases.

Case I: 

Sum \quad of \quad zeroes = - \frac{coefficient of x}{coefficient of x^{2}}

   \Rightarrow \alpha + \beta = - \frac{b}{a}

LHS = \alpha + \beta = 4 - 2 = 2

RHS =  -\frac{b}{a} = -\frac{(-2)}{1}= 2

\therefore LHS = RHS

 

Case II: 

Product \quad of \quad zeroes = \frac{constant \quad term}{coefficient of x^{2}}

\Rightarrow \alpha \times \beta = \frac{c}{a}

LHS =  \alpha \times \beta  = 4 x (-2) = - 8

RHS = \frac{c}{a} =  \frac{-8}{1} = -8  

\therefore LHS = RHS

In both cases, LHS is equal to RHS.

Therefore, the relationship between the zeroes and the coefficients are verified.

 

(ii) 4s^{2}-4s+1   

Let P(s) = 4s^{2}-4s+1 

Zeroes of the polynomial are the value of  where P(s) = 0.

Putting P(s) = 0 then we get

4s^{2}-4s+1  

We can find zeroes of the polynomial using splitting the middle term method.

In splitting the middle term method we need to find two numbers where the sum is -4 and product is 4  1 = 4.

Therefore, we can write,

4s^{2}-2s-2s+1=0

\Rightarrow 2s(2s-1)-1(2s-1))=0

\Rightarrow (2s-1)(2s-1))=0

(2s-1))=0 \Rightarrow s = \frac{1}{2}

Let \alpha = \frac{1}{2}  and  \beta = \frac{1}{2}  are the zeroes of the polynomial.

P(s ) =  4s^{2} - 4s + 1 = 4s^{2} + (-4)s + 1

If we compare P(s) with  as^{2} + bs + c then we get 

a = 4, b = -4 and c = 1

Now we verify the relationship between the zeroes and the coefficients in two different cases.

Case I: 

Sum \quad of \quad zeroes = - \frac{coefficient of x}{coefficient of x^{2}}

   \Rightarrow \alpha + \beta = - \frac{b}{a}

LHS = \alpha + \beta = \frac{1}{2} + \frac{1}{2} = 1

RHS = -\frac{b}{a} = - \frac{(-4)}{4} = 1

\therefore LHS = RHS

 

Case II: 

Product \quad of \quad zeroes = \frac{constant \quad term}{coefficient of x^{2}}

\Rightarrow \alpha \times \beta = \frac{c}{a}

LHS = \alpha \times \beta = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}

RHS = \frac{c}{a} = \frac{1}{4}

\therefore LHS = RHS

In both cases, LHS is equal to RHS.

Therefore, the relationship between the zeroes and the coefficients are verified.

 

 (iii) 6x^{2}-3-7x

 Let P(x ) = 6x^{2}-3-7x = 6x^{2}-7x-3 

Zeroes of the polynomial are the value of x where P(x ) = 0.

Putting P(x) = 0 then we get

6x^{2}-7x-3 = 0

We can find zeroes of the polynomial using splitting the middle term method.

In splitting the middle term method we need to find two numbers where the sum is -7 and product is 6  x (-3) = -18.

Therefore, we can write,

6x^{2}-9x+2x-3 = 0

\Rightarrow 3x(2x-3)+1(2x-3 )= 0

\Rightarrow (2x-3)(3x+1 )= 0

Either (2x-3)= 0 \quad or \quad (3x+1 )= 0

So, x = 3/2 and x = -1/3

Let \alpha = \frac{3}{2}     and     \beta = -\frac{1}{3}   are the zeroes of the polynomial.

P(x) = 6x^{2}-7x-3  = 6x^{2} +(-7)x+(-3)

If we compare p(x) with  ax^{2} +bx+c then we get 

a = 6, b = -7 and c = -3

Now we verify the relationship between the zeroes and the coefficients in two different cases.

Case I: 

Sum \quad of \quad zeroes = - \frac{coefficient of x}{coefficient of x^{2}}

   \Rightarrow \alpha + \beta = - \frac{b}{a}

LHS = \alpha + \beta = \frac{3}{2} - \frac{1}{3} = \frac{9-2}{6} = \frac{7}{6}

RHS = -\frac{b}{a} = - \frac{(-7)}{6} = \frac{7}{6}

\therefore LHS = RHS

Case II: 

  Product \quad of \quad zeroes = \frac{constant \quad term}{coefficient of x^{2}}

\Rightarrow \alpha \times \beta = \frac{c}{a}

LHS =  \alpha \times \beta = \frac{3}{2} \times \frac{-1}{3} = \frac{-1}{2}

RHS =  \frac{c}{a} = \frac{-3}{6} = \frac{-1}{2}

 

\therefore LHS = RHS

In both cases, LHS is equal to RHS.

Therefore, the relationship between the zeroes and the coefficients are verified.

 

(iv) 4u^{2}+8u    

Let P(u) = 4u^{2}+8u  

Zeroes of the polynomial are the value of  u where P(u) = 0.

Putting P(u) = 0 then we get

4u^{2}+8u=0

We can find zeroes of the polynomial using common factor method.

4u is the common factor in the P(u) = 4u^{2}+8u    

Therefore, we can write,

\Rightarrow 4u(u+2)=0

Either 4u  = 0 or  (u+2) = 0

So, u = 0 and u = -2.

Let \alpha  = 0 and \beta  = -2 are the zeroes of the polynomial.

P(u) = 4u^{2}+8u = 4u^{2}+8u+0

If we compare p(u) with au^{2}+bu+c then we get 

a = 4, b = 8 and c = 0

Now we verify the relationship between the zeroes and the coefficients in two different cases.

Case I: 

Sum \quad of \quad zeroes = - \frac{coefficient of u}{coefficient of u^{2}}

   \Rightarrow \alpha + \beta = - \frac{b}{a}

LHS = \alpha + \beta = 0 - 2 = -2

RHS = -\frac{b}{a} = - \frac{8}{4} = -2

\therefore LHS = RHS

Case II: 

  Product \quad of \quad zeroes = \frac{constant \quad term}{coefficient of u^{2}}

\Rightarrow \alpha \times \beta = \frac{c}{a}

LHS =  \alpha \times \beta = 0 \times (-2) = 0

RHS =  \frac{c}{a} = \frac{0}{4} = 0

 

\therefore LHS = RHS

In both cases, LHS is equal to RHS.

Therefore, the relationship between the zeroes and the coefficients are verified.

 

 

(v) t^{2}-15    

Let P(t) = t^{2}-15      

Zeroes of the polynomial are the value of  t where P(t) = 0.

Putting P(t) = 0 then we get

t^{2}-15=0

\Rightarrow t^{2} = 15

\Rightarrow t = \pm \sqrt{15}

So, t = + \sqrt{15}  and  t = - \sqrt{15}

 

Let \alpha  \sqrt{15} and \beta  = - \sqrt{15} are the zeroes of the polynomial.

P(t) = t^{2}-15 = 1t^{2}+0\times t + (-15)

If we compare p(t) with at^{2}+bt+c  then we get 

a = 1, b = 0 and c = -15

Now we verify the relationship between the zeroes and the coefficients in two different cases.

Case I: 

Sum \quad of \quad zeroes = - \frac{coefficient \quad of \quad t}{coefficient \quad of \quad t^{2}}

   \Rightarrow \alpha + \beta = - \frac{b}{a}

LHS = \alpha + \beta = \sqrt{15} - \sqrt{15} = 0

RHS = -\frac{b}{a} = - \frac{0}{1} = 0

\therefore LHS = RHS

Case II: 

  Product \quad of \quad zeroes = \frac{constant \quad term}{coefficient of t^{2}}

\Rightarrow \alpha \times \beta = \frac{c}{a}

LHS =  \alpha \times \beta = \sqrt{15} \times (-\sqrt{15}) = -15

RHS =  \frac{c}{a} = \frac{-15}{1} = -15

\therefore LHS = RHS

In both cases, LHS is equal to RHS.

Therefore, the relationship between the zeroes and the coefficients are verified.

 

(vi) 3x^{2} -x -4

Let P(x) = 3x^{2} -x -4

Zeroes of the polynomial are the value of  where P(x) = 0.

Putting P(x) = 0 then we get

3x^{2} -x -4

We can find zeroes of the polynomial using splitting the middle term method.

In splitting the middle term method we need to find two numbers where the sum is -1 and product is 3 x (-4) = -12.

Therefore, we can write,

3x^{2}-4x+3x-4=0

\Rightarrow x(3x-4)+1(3x-4)=0

\Rightarrow (3x-4)(x+1)=0

Either (3x-4)=0 \quad or  (x+1)=0 

So,  x = 4/3 and x = -1.

Let \alpha = 4/3 and \beta  = -1 are the zeroes of the polynomial.

P(x) = 3x^{2} - x - 4 = 3x^{2} + (-1)x + (-4)

If we compare P(x)  with  ax^{2} + bx +c then we get 

a = 3, b = -1 and c = -4

Now we verify the relationship between the zeroes and the coefficients in two different cases.

Case I: 

Sum \quad of \quad zeroes = - \frac{coefficient of x}{coefficient of x^{2}}

   \Rightarrow \alpha + \beta = - \frac{b}{a}

LHS = \alpha + \beta = \frac{4}{3} - 1 = \frac{4-3}{3} = \frac{1}{3}

RHS =  -\frac{b}{a} = -\frac{(-1)}{3} = \frac{1}{3}

\therefore LHS = RHS

 

Case II: 

Product \quad of \quad zeroes = \frac{constant \quad term}{coefficient of x^{2}}

\Rightarrow \alpha \times \beta = \frac{c}{a}

LHS =  \alpha \times \beta  = \frac{4}{3} \times (-1) = -\frac{4}{3}

RHS = \frac{c}{a} =  \frac{-4}{3}  

\therefore LHS = RHS

In both cases, LHS is equal to RHS.

Therefore, the relationship between the zeroes and the coefficients are verified.

 

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