Chapter 2: Polynomials

Polynomials Exercise 2.1
Polynomials Exercise 2.2

Find the zeros of the following quadratic polynomials and verify the relationship between the zeros and the coefficients:

Question 1: x2 + 3x – 10

Solution:

We have

f(x) = (x2 + 3x – 10)

= x2 + 5x – 2x – 10

= x(x + 5) – 2(x + 5)

= (x + 5)(x – 2)

Therefore, f(x) = 0 \(\Rightarrow\) (x + 5)(x – 2) = 0

\(\Rightarrow\) x + 5 = 0 or x – 2 = 0

\(\Rightarrow\) x = -5 or x = 2

Sum of zeros = (-5) + 2 = -3 = \(\frac{-3}{1}=\frac{(coefficient\, of\, x)}{(coefficient\, of\, x^{2})}\)

Product of zeros = (-5) x (2) = -10 = \(\frac{-10}{1}=\frac{Constant\, term}{coefficient\, of\, x^{2}}\)

 

Question 2: 6x2 – 7x – 3

Solution:

We have

f(x) = (6x2 – 7x – 3)

= 6x2 – 9x + 2x – 3

= 3x(2x – 3) + 1(2x – 3)

= (2x – 3)(3x + 1)

Therefore, f(x) = 0 \(\Rightarrow\) (2x – 3)(3x + 1) = 0

\(\Rightarrow\) 2x – 3 = 0 or 3x + 1 = 0

\(x=\frac{3}{2}\) or \(x=\frac{-1}{3}\)

Sum of zeros = \(\frac{3}{2}+\left ( \frac{-1}{3} \right )=\frac{7}{6}=\frac{-(coefficient\, of\, x)}{(coefficient\, of\, x^{2})}\)

Product of zeros = \(\left ( \frac{3}{2} \right )\times \frac{-1}{3}=\frac{-3}{6}=\frac{Constant\, term}{(coefficient\, of\, x^{2})}\)

 

Question 3: 4x2 – 4x – 3

Solution:

f(x) = 4x2 – 4x – 3 = 4x2 – 6x + 2x – 3

=2x(2x – 3) + (2x – 3) = (2x – 3)(2x + 1) = 0

Therefore, 2x – 3 = 0 or 2x + 1 = 0

or \(x=\frac{3}{2}\), \(x=-\frac{1}{2}\)

Sum of zeros = \(\frac{3}{2}+\left ( -\frac{1}{2} \right )=\frac{3-1}{2}=\frac{2}{2}=1=\frac{-(-4)}{4}=-\frac{(coefficient\, of\, x)}{(coefficient\, of\, x^{2})}\)

Product of zeros = \(\frac{3}{2}\times \left ( -\frac{1}{2} \right )=-\frac{3}{4}=\frac{Constant\, term}{(coefficient\, of\, x^{2})}\)

 

Question 4: 5x2 – 4 – 8x

Solution:

We have

f(x) = 5x2 – 4 – 8x = 5x2 – 8x – 4

= 5x2 – 10x + 2x – 4

= 5x(x – 2) + 2(x – 2)

f(x) = 0 \(\Rightarrow\) (x – 2)(5x + 2) = 0

\(\Rightarrow\) x – 2 = 0 or 5x + 2 = 0

Therefore, x = 2, \(\frac{-2}{5}\)

So, the zeros of f(x) are 2 and \(\frac{-2}{5}\)

Sum of zeros = \(2+\left ( \frac{-2}{5} \right )=\frac{10-2}{5}=\frac{8}{5}=-\frac{Coeff.\, of\, x}{Coeff.\, of\, x^{2}}\)

Product of zeros = \(2\times \left ( \frac{-2}{5} \right )=\frac{-4}{5}=-\frac{Constant\, term}{Coeff.\, of\, x^{2}}\)

 

Question 5: 6x2 – 3 – 7x

Solution:

We have

f(x) = 6x2 – 3 – 7x

= 6x2 – 9x + 2x – 3

= 3x(2x – 3) + (2x – 3)

f(x) = 0 \(\Rightarrow\) (2x – 3)(3x + 1) = 0

\(\Rightarrow\) 2x – 3 = 0 or 3x + 1 = 0

Therefore, x = \(\frac{3}{2},-\frac{1}{3}\)

Sum of zeros = \(\frac{3}{2}+\left ( \frac{-1}{3} \right )=\frac{9-2}{6}=\frac{7}{6}=-\frac{Coeff.\, of\, x}{Coeff.\, of\, x^{2}}\)

Product of zeros = \(\frac{3}{2}\times \left ( \frac{-1}{3} \right )=-\frac{1}{2}=\frac{-3}{6}=-\frac{Constant\, term}{Coeff.\, of\, x^{2}}\)

 

Question 6: 2x2 – 11x + 15

Solution:

We have

f(x) = 2x2 – 11x + 15

= 2x2 – 6x – 5x + 15

= 2x(x – 3) – 5(x – 3) = (x – 3)(2x – 5)

Now, f(x) = (x – 3)(2x – 5) = 0

Therefore, x – 3 = 0 or 2x – 5 = 0

\(\Rightarrow\) x = 3 or x = \(\frac{5}{2}\)

So zeros of f(x) are 3 and \(\frac{5}{2}\)

Sum of zeros = \(3+\frac{5}{2}=\frac{11}{2}=\frac{-(-11)}{2}=-\frac{Coeff.\, of\, x}{Coeff.\, of\, x^{2}}\)

Product of zeros = \(3\times \frac{5}{2}=\frac{15}{2}=\frac{Constant\, term}{Coeff.\, of\, x^{2}}\)

 

Question 7: x2 – 5

Solution:

We have

f(x) = x2 – 5 = (x)2 – (\(\sqrt{5}\))2

= \((x-\sqrt{5})(x+\sqrt{5})\)               [a2– b2 = (a – b)(a + b)]

f(x) = 0 \(\Rightarrow\) \((x-\sqrt{5})(x+\sqrt{5})\) = 0

Therefore, \(x-\sqrt{5}=0\) or \(x+\sqrt{5}=0\)

\(\Rightarrow\) \(x=\sqrt{5}\) or \(x=-\sqrt{5}\)

So the zeros of f(x) are \(\sqrt{5}\) and \(-\sqrt{5}\)

Sum of zeros = \(\sqrt{5}+(-\sqrt{5})=0=-\frac{Coeff.\, of\, x}{Coeff.\, of\, x^{2}}\)

Product of zeros = \((\sqrt{5})(-\sqrt{5})=-5=\frac{-5}{1}=\frac{Constant\, term}{Coeff.\, of\, x^{2}}\)

 

Question 8: 8x2 – 4

Solution:

Let

f(x) = 8x2 – 4 = 4(2x2 – 1) = \(4 [ (\sqrt{2x})^{2} -1^{2} ]\)

= \(4(\sqrt{2x}-1)(\sqrt{2x}+1)\)                  [a2– b2 = (a – b)(a + b)]

f(x) = 0 \(\Rightarrow\) \((\sqrt{2x}-1)(\sqrt{2x}+1)=0\)

Therefore, \(\sqrt{2x}-1=0\) or \(\sqrt{2x}+1=0\)

Therefore, \(x=\frac{1}{\sqrt{2}}\) or \(x=-\frac{1}{\sqrt{2}}\)

So, the zeros of f(x) are \(\frac{1}{\sqrt{2}}\) and \(-\frac{1}{\sqrt{2}}\)

Sum of zeros = \(\left ( \frac{1}{\sqrt{2}} \right )+\left (-\frac{1}{\sqrt{2}} \right )=0=\frac{0}{8}=\frac{Coeff.\, of\, x}{Coeff.\, of\, x^{2}}\)

Product of zeros = \(\left ( \frac{1}{\sqrt{2}} \right )\times \left ( -\frac{1}{\sqrt{2}} \right )=-\frac{1}{2}=-\frac{4}{8}=\frac{Constant\, term}{Coeff.\, of\, x^{2}}\)

 

Question 9: 5u2 + 10u

Solution:

Let, f(x) = 5u2 + 10u = 5u(u + 2)

f(x) = 0 \(\Rightarrow\) 5u(u + 2) = 0

Therefore, u = 0 or u + 2 = 0

\(\Rightarrow\) u = 0 or u = -2

Sum of zeros = 0 + (-2) = -2 = \(\frac{-10}{5}\) = \(-\frac{Coeff.\, of\, x}{Coeff.\, of\, x^{2}}\)

Product of zeros = 0 x (-2) = 0 = \(\frac{0}{5}=\frac{Constant\, term}{Coeff.\, of\, x^{2}}\)

 

Question 10: Find the quadratic polynomial whose zeros are 2 and -6. Verify the relation between the coefficients and the zeros of the polynomials.

Solution:

If zeros are denoted by \(\alpha\) and \(\beta\) then

\(\alpha\) + \(\beta\) = 2 + (-6) = -4 or \(\alpha\)\(\beta\) = 2 x (-6) = -12

Therefore, Quadratic polynomial is

x2 – (\(\alpha +\beta\))x + \(\alpha \beta\) = x2 – (-4x) + (-12) = x2 + 4x – 12

Sum of zeros = \(-\frac{Coeff.\, of\, x}{Coeff.\, of\, x^{2}}=\frac{-4}{1}=-4\)

Also, \(\alpha +\beta\) = 2 + (-6) = -4

Product of zeros = \(\frac{Constant\, term}{Coeff.\, of\, x^{2}}=\frac{-12}{1}=-12\)

Also, \(\alpha \beta\) = 2 x (-6) = -12

 

Question 11: Find the quadratic polynomial whose zeros are \(\frac{2}{3}\) and \(\frac{-1}{4}\). Verify the relation between the coefficients and the zeros of the polynomial.

Solution:

Let \(\alpha\) + \(\beta\) are the zeros then

\(\alpha\) + \(\beta\) = \(\frac{2}{3}+\left ( -\frac{1}{4} \right )=\frac{8-3}{12}=\frac{5}{12}\)

\(\alpha \beta =\frac{2}{3}\times \left ( -\frac{1}{4} \right )=-\frac{2}{12}=-\frac{1}{6}\)

Therefore, Quadratic polynomial whose zeros are \(\alpha\), \(\beta\) is

x2 – (\(\alpha +\beta\))x + \(\alpha \beta\) = x2 – \(\left ( \frac{5}{12} \right )\)x + \(\left ( -\frac{1}{6} \right )\)

= \(\frac{1}{12}\left ( 12x^{^{2}}-5x-2 \right )\)

Sum of zeros = \(-\frac{Coeff.\, of\, x}{Coeff.\, of\, x^{2}}=-\frac{-5}{12}=\frac{5}{12}\)

Also sum of zeros = \(\frac{2}{3}+\left ( -\frac{1}{4} \right )=\frac{5}{12}\)

Product of zeros = \(\frac{Constant\, term}{Coeff.\, of\, x^{2}}=\frac{-2}{12}=\frac{-1}{6}\)

Also Product of zeros = \(\frac{2}{3}\times \left ( -\frac{1}{4} \right )=\frac{-2}{12}=\frac{-1}{6}\)

 

Question 12: Find the quadratic polynomial, sum of whose zeros is 8 and their product is 12. Hence, find the zeros of the polynomial.

Solution:

Now, \(\alpha +\beta\) = 8 and  \(\alpha \beta\) = 12

f(x) = x2 – (\(\alpha +\beta\))x + \(\alpha \beta\)

= x2 – 8x + 12

Therefore, Required polynomial is x2 – 8x + 12

Also f(x) = x2 – 8x + 12 = x2 – 6x – 2x = 12

= x(x – 6) – 2(x – 6)

= (x – 6)(x – 2)

f(x) = 0 \(\Rightarrow\) (x – 6)(x – 2) = 0

Therefore, x – 6 = 0 or x – 2 = 0

i.e. , x = 6 or x = 2

Therefore, Zeros of polynomial are 6 and 2.

 

Question 13: Find the quadratic polynomial, sum of whose zeros is -5 and their product is 6. Hence, find the zeros of the polynomial.

Solution:

Let \(\alpha\), \(\beta\) be the zeros of required quadratic polynomial f(x)

We have,

\(\alpha\) + \(\beta\) = -5 and \(\alpha \beta\) = 6

Therefore, f(x) = x2 – (\(\alpha\) + \(\beta\))x + \(\alpha \beta\)

= x2 – (-5)x + 6 = x2 + 5x + 6

So, the required polynomial is x2 + 5x + 6

Now f(x) = x2 + 5x + 6 = x2 + 3x + 2x + 6

= x(x + 3) + 2(x + 3)

= (x + 3)(x + 2)

f(x) = 0 \(\Rightarrow\) either x + 3 = 0 or x + 2 = 0

\(\Rightarrow\) either x = -3 or x = -2

Therefore, Zeros of the polynomials are -3 and -2

 

Question 14: Find the quadratic polynomial, the sum of whose zeros is \(\left ( \frac{5}{2} \right )\) and their product is 1. Hence, find the zeros of the polynomial.

Solution:

Let \(\alpha\), \(\beta\) be the zeros of required quadratic polynomial f(x)

We have,

\(\alpha\) + \(\beta\) = \(\frac{5}{2}\), \(\alpha \beta\) = 1

Now, f(x) = x2 – (\(\alpha\) + \(\beta\))x + \(\alpha \beta\)

= x2 –  \(\frac{5}{2}\)x + 1 = \(\frac{1}{2}(2x^{2}-5x+2)\)

The polynomial whose zeros are \(\alpha\), \(\beta\) is 2x2 – 5x + 2

Further, f(x) = \(\frac{1}{2}(2x^{2}-5x+2)\) = \(\frac{1}{2}(2x^{2}-4x-x+2)\)

= \(\frac{1}{2}\left [ 2x(x-2)-(x-2) \right ]\)

= \(\frac{1}{2}(x-2)(2x-1)\)

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

Therefore, for that x – 2 = 0 or 2x – 1 = 0

i.e., Either x = 2 or x = ½

Therefore, Zeros of polynomial are 2 and ½

 

Question 15:

Find the quadratic polynomial, sum of whose zeros is 0 and their product is -1. Hence, find the zeros of the polynomial.

Solution:

Let \(\alpha\), \(\beta\) be the zeros of required quadratic polynomial f(x)

We have,

\(\alpha\) + \(\beta\) = 0, \(\alpha \beta\) = -1

Therefore, Polynomial whose zeros are \(\alpha\), \(\beta\) is

f(x) = x2 – (\(\alpha\) + \(\beta\))x + \(\alpha \beta\)

= x2 – 0.x + (-1) = x2 – 1

Therefore, Required polynomial is x2 – 1

Now f(x) = x2 – 1 = (x – 1)(x + 1)

f(x) = 0 \(\Rightarrow\) (x – 1)(x + 1) = 0

Therefore, Either x – 1 = 0 or x + 1 = 0

i.e., Either x = 1 or x = -1

Therefore, Zeros of the polynomial are 1 and -1

 

Question 16: Find the quadratic polynomial, sum of whose zeros is \(\sqrt{2}\) and their product is -12. Hence, find the zeros of the polynomial.

Solution:

Let \(\alpha\), \(\beta\) be the zeros of required quadratic polynomial f(x)

We have,

\(\alpha\) + \(\beta\) = \(\sqrt{2}\), \(\alpha \beta\) = -12

Therefore, Polynomial whose zeros are \(\alpha\), \(\beta\) is

f(x) = x2 – (\(\alpha\) + \(\beta\))x + \(\alpha \beta\)

= x2 – \(\sqrt{2}\)x – 12

Therefore, Polynomial required is x2 – \(\sqrt{2}\)x – 12

For that, f(x) = x2 – \(\sqrt{2}\)x – 12

= x2 – 3\(\sqrt{2}\)x + 2\(\sqrt{2}\)x – 12

= x(x – 3\(\sqrt{2}\))x + 2\(\sqrt{2}\)(x – 3\(\sqrt{2}\))

= (x – 3\(\sqrt{2}\))(x + 2\(\sqrt{2}\))

f(x) = 0 \(\Rightarrow\) (x – 3\(\sqrt{2}\))(x + 2\(\sqrt{2}\))

\(\Rightarrow\) Either x – 3\(\sqrt{2}\) = 0 or x + 2\(\sqrt{2}\) = 0

Therefore, Either x = 3\(\sqrt{2}\) or x = -2\(\sqrt{2}\)

 

Question 17: Find the quadratic polynomial, the sum of whose zeros is 6 and their product is 4.

Solution:

\(\alpha\) and \(\beta\) are the zeros of polynomial f(x) such that

\(\alpha\) + \(\beta\) = 6, \(\alpha \beta\) = 4

The polynomial f(x) whose zeros are \(\alpha\), \(\beta\) is

x2 – (\(\alpha\) + \(\beta\))x + \(\alpha \beta\)

= x2 – 6x + 4

 

EXERCISE 2B

 

Question 1: Verify that 3, -2, 1 are the zeros of the cubic polynomial p(x) = x3 – 2x2 – 5x + 6 and verify the relation between its zeros and coefficients.

Solution:

p(x) = x3 – 2x2 – 5x + 6

Therefore, p(3) = (3)3 – 2(3)2 – 5(3) + 6

= 27 – 18 – 15 + 6 = 0

p(-2) = (-2)3 – 2(-2)2 – 5(-2) + 6

= -8 – 8 + 10 + 6 = 0

p(1) = (1)3 – 2(1)2 – 5(1) + 6

= 1 – 2 – 5 + 6 = 0

Thus, 3, -2, 1 are the zeros of p(x) = x3 – 2x2 – 5x + 6

Therefore, \(\alpha\) = 3, \(\beta\) = -2 and \(\gamma\) = 1

Comparing the given polynomial with

p(x) = ax3 + bx2 + cx + d,

We get, a = 1, b = -2, c = -5 and d = 6

Now, (\(\alpha\) + \(\beta\) + \(\gamma\)) = (3 – 2 + 1) = 2 = \(-\frac{b}{a}\)

(\(\alpha \beta +\beta \gamma +\gamma \alpha\)) = [3 x (-2) + (-2) x 1 + 1 x 3]

= (-6 – 2 + 3) = -5 = \(\frac{c}{a}\)

and \(\alpha \beta \gamma\) = [3 x (-2) x 1] = -6 = \(\frac{-d}{a}\)

 

Question 2: Verify that 5, -2, and \(\frac{1}{3}\) are the zeros of the cubic polynomial p(x) = 3x3 – 10x2 – 27x + 10 and verify the relation between its zeros and coefficients.

Solution:

p(x) = 3x3 – 10x2 – 27x + 10

Therefore, p(5) = 3(5)3 – 10(5)2 – 27(5) + 10

= 375 – 250 – 135 + 10 = 0

p(-2) = 3(-2)3 – 10(-2)2 – 27(-2) + 10

= -24 – 40 + 54 + 10 = 0

p(\(\frac{1}{3}\)) = 3(\(\frac{1}{3}\))3 – 10(\(\frac{1}{3}\))2 – 27(\(\frac{1}{3}\)) + 10

= \(\frac{1}{9}-\frac{10}{9}-9+10=0\)

Therefore, Then, 5, -2, \(\frac{1}{3}\) zeros of

p(x) = 3x3 – 10x2 – 27x + 10

Therefore, \(\alpha\) = 5, \(\beta\) = -2, \(\gamma\) = \(\frac{1}{3}\)

Comparing the given polynomial with

p(x) = ax3 + bx2 + cx + d

We get a = 3, b = -10, c = -27 and d = 10

Now, (\(\alpha\) + \(\beta\) + \(\gamma\)) = \(\left ( 5-2+\frac{1}{3} \right )=\frac{10}{3}=\frac{-b}{a}\)

(\(\alpha \beta +\beta \gamma +\gamma \alpha\)) = \(\left [ 5\times (-2)+(-1)\times \frac{1}{3}+\frac{1}{3\times 5} \right ]=\left ( -10-\frac{2}{3}+\frac{5}{3} \right )=\frac{-27}{3}=\frac{c}{a}\)

and \(\alpha \beta \gamma\) = \(\left [ 5\times (-2)\times \frac{1}{3} \right ]=\frac{-10}{3}=\frac{-d}{a}\)

 

Question 3: Find the cubic polynomial whose zeros are -2, -3 and -1.

Solution:

Required polynomial = [x – (-2)][x – (-3)][x – (-1)]

= (x + 2)(x + 3)(x + 1)

= (x2 + 5x + 6)(x + 1)

= x3 + x2 + 5x2 + 5x + 6x + 6

= x3 + 6x + 11x + 6

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