# RS Aggarwal Solutions Class 10 Polynomials

## Polynomials Solution RS Aggarwal Chapter 2

All these RS Aggarwal class 10 solutions Chapter 2 Polynomials are solved by Byju's top ranked professors as per CBSE guidelines.

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

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 = ??

Therefore, Zeros of polynomial are 2 and ??

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

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## EXERCISE 2B

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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