An expression with plenty of variables and coefficients which also contains indeterminates and involves various operations such as multiplication, addition, and subtraction. It comes from a Greek word, where â€˜Polyâ€™ means many and â€˜Nomialâ€™ means many. Some of the different polynomial functions are used in different fields of subjects such as Physics, Social Science and Economics. The different types of Polynomials are Trigonometric Polynomials, Matrix Polynomials, and Laurent Polynomials.

The different polynomials with unique terms are:

- Monomial: One term only
- Binomial: Only two terms
- Trinomial: Only three terms

Learn more about RS Aggarwal Class 10 Solutions Chapter 2 Polynomials is available below:

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

Question 1: x^{2} + 3x â€“ 10

Solution:

We have

f(x) = (x^{2} + 3x â€“ 10)

= x^{2} + 5x â€“ 2x â€“ 10

= x(x + 5) â€“ 2(x + 5)

= (x + 5)(x â€“ 2)

Therefore, f(x) = 0 \(\Rightarrow\) (x + 5)(x â€“ 2) = 0

\(\Rightarrow\) x + 5 = 0 or x â€“ 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: 6x^{2} â€“ 7x â€“ 3

Solution:

We have

f(x) = (6x^{2} â€“ 7x â€“ 3)

= 6x^{2} â€“ 9x + 2x â€“ 3

= 3x(2x â€“ 3) + 1(2x â€“ 3)

= (2x â€“ 3)(3x + 1)

Therefore, f(x) = 0 \(\Rightarrow\) (2x â€“ 3)(3x + 1) = 0

\(\Rightarrow\) 2x â€“ 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: 4x^{2} â€“ 4x â€“ 3

Solution:

f(x) = 4x^{2} â€“ 4x â€“ 3 = 4x^{2} â€“ 6x + 2x â€“ 3

=2x(2x â€“ 3) + (2x â€“ 3) = (2x â€“ 3)(2x + 1) = 0

Therefore, 2x â€“ 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: 5x^{2} â€“ 4 â€“ 8x

Solution:

We have

f(x) = 5x^{2} â€“ 4 â€“ 8x = 5x^{2} â€“ 8x â€“ 4

= 5x^{2} â€“ 10x + 2x â€“ 4

= 5x(x â€“ 2) + 2(x â€“ 2)

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

\(\Rightarrow\) x â€“ 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: 6x^{2} â€“ 3 â€“ 7x

Solution:

We have

f(x) = 6x^{2} â€“ 3 â€“ 7x

= 6x^{2} â€“ 9x + 2x â€“ 3

= 3x(2x â€“ 3) + (2x â€“ 3)

f(x) = 0 \(\Rightarrow\) (2x â€“ 3)(3x + 1) = 0

\(\Rightarrow\) 2x â€“ 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: 2x^{2} â€“ 11x + 15

Solution:

We have

f(x) = 2x^{2} â€“ 11x + 15

= 2x^{2} â€“ 6x â€“ 5x + 15

= 2x(x â€“ 3) â€“ 5(x â€“ 3) = (x â€“ 3)(2x â€“ 5)

Now, f(x) = (x â€“ 3)(2x â€“ 5) = 0

Therefore, x â€“ 3 = 0 or 2x â€“ 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: x^{2} – 5

Solution:

We have

f(x) = x^{2} â€“ 5 = (x)^{2} â€“ (\(\sqrt{5}\))^{2}

= \((x-\sqrt{5})(x+\sqrt{5})\)Â Â Â Â Â Â Â Â Â Â Â Â Â Â [a^{2}– b^{2} = (a â€“ 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: 8x^{2} – 4

Solution:

Let

f(x) = 8x^{2} â€“ 4 = 4(2x^{2} â€“ 1) = \(4 [ (\sqrt{2x})^{2} -1^{2} ]\)

= \(4(\sqrt{2x}-1)(\sqrt{2x}+1)\)Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [a^{2}– b^{2} = (a â€“ 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: 5u^{2} + 10u

Solution:

Let, f(x) = 5u^{2} + 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

x^{2} â€“ (\(\alpha +\beta\))x + \(\alpha \beta\) = x^{2} â€“ (-4x) + (-12) = x^{2} + 4x â€“ 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

x^{2} â€“ (\(\alpha +\beta\))x + \(\alpha \beta\) = x^{2} â€“ \(\left ( \frac{5}{12} \right )\)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 Â \(\alpha \beta\) = 12

f(x) = x^{2} â€“ (\(\alpha +\beta\))x + \(\alpha \beta\)

= x^{2} â€“ 8x + 12

Therefore, Required polynomial is x^{2} â€“ 8x + 12

Also f(x) = x^{2} â€“ 8x + 12 = x^{2} â€“ 6x â€“ 2x = 12

= x(x â€“ 6) â€“ 2(x â€“ 6)

= (x â€“ 6)(x â€“ 2)

f(x) = 0 \(\Rightarrow\) (x â€“ 6)(x â€“ 2) = 0

Therefore, x â€“ 6 = 0 or x â€“ 2 = 0

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

Therefore, Zeros of polynomial are 6 and 2.

Question 13: Find the quadratic polynomial, the 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) = x^{2} â€“ (\(\alpha\) + \(\beta\))x + \(\alpha \beta\)

= x^{2} â€“ (-5)x + 6 = x^{2} + 5x + 6

So, the required polynomial is x^{2} + 5x + 6

Now f(x) = x^{2} + 5x + 6 = x^{2} + 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) = x^{2} â€“ (\(\alpha\) + \(\beta\))x + \(\alpha \beta\)

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

The polynomial whose zeros are \(\alpha\), \(\beta\) is 2x^{2} â€“ 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 â€“ 2 = 0 or 2x â€“ 1 = 0

i.e., Either x = 2 or x = Â½

Therefore, Zeros of polynomial are 2 and Â½

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) = x^{2} â€“ (\(\alpha\) + \(\beta\))x + \(\alpha \beta\)

= x^{2} â€“ 0.x + (-1) = x^{2} â€“ 1

Therefore, Required polynomial is x^{2} â€“ 1

Now f(x) = x^{2} â€“ 1 = (x â€“ 1)(x + 1)

f(x) = 0 \(\Rightarrow\) (x â€“ 1)(x + 1) = 0

Therefore, Either x â€“ 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) = x^{2} â€“ (\(\alpha\) + \(\beta\))x + \(\alpha \beta\)

= x^{2} – \(\sqrt{2}\)x â€“ 12

Therefore, Polynomial required is x^{2} – \(\sqrt{2}\)x â€“ 12

For that, f(x) = x^{2} – \(\sqrt{2}\)x â€“ 12

= x^{2} – 3\(\sqrt{2}\)x + 2\(\sqrt{2}\)x â€“ 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

x^{2} â€“ (\(\alpha\) + \(\beta\))x + \(\alpha \beta\)

= x^{2} â€“ 6x + 4

## EXERCISE 2B

Question 1: Verify that 3, -2, 1 are the zeros of the cubic polynomial p(x) = x^{3} â€“ 2x^{2} â€“ 5x + 6 and verify the relation between its zeros and coefficients.

Solution:

p(x) = x^{3} â€“ 2x^{2} â€“ 5x + 6

Therefore, p(3) = (3)^{3} â€“ 2(3)^{2} â€“ 5(3) + 6

= 27 â€“ 18 â€“ 15 + 6 = 0

p(-2) = (-2)^{3} â€“ 2(-2)^{2} â€“ 5(-2) + 6

= -8 â€“ 8 + 10 + 6 = 0

p(1) = (1)^{3} â€“ 2(1)^{2} â€“ 5(1) + 6

= 1 â€“ 2 â€“ 5 + 6 = 0

Thus, 3, -2, 1 are the zeros of p(x) = x^{3} â€“ 2x^{2} â€“ 5x + 6

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

Comparing the given polynomial with

p(x) = ax^{3} + bx^{2} + cx + d,

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

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

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

= (-6 â€“ 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) = 3x^{3} â€“ 10x^{2} â€“ 27x + 10 and verify the relation between its zeros and coefficients.

Solution:

p(x) = 3x^{3} â€“ 10x^{2} â€“ 27x + 10

Therefore, p(5) = 3(5)^{3} â€“ 10(5)^{2} â€“ 27(5) + 10

= 375 â€“ 250 â€“ 135 + 10 = 0

p(-2) = 3(-2)^{3} â€“ 10(-2)^{2} â€“ 27(-2) + 10

= -24 â€“ 40 + 54 + 10 = 0

p(\(\frac{1}{3}\)) = 3(\(\frac{1}{3}\))^{3} â€“ 10(\(\frac{1}{3}\))^{2} â€“ 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) = 3x^{3} â€“ 10x^{2} â€“ 27x + 10

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

Comparing the given polynomial with

p(x) = ax^{3} + bx^{2} + 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 â€“ (-2)][x â€“ (-3)][x â€“ (-1)]

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

= (x^{2} + 5x + 6)(x + 1)

= x^{3} + x^{2} + 5x^{2} + 5x + 6x + 6

= x^{3} + 6x + 11x + 6