CBSE Class 10 Mathematics Polynomials Worksheet Set 05

Question. Find the zeroes of the quadratic polynomial \( 6x^2 - 13x + 6 \) and verify the relation between the zeroes and its coefficients.
Answer: We have, \( 6x^2 - 13x + 6 = 6x^2 - 4x - 9x + 6 \)
\( = 2x (3x - 2) - 3 (3x - 2) = (3x - 2) (2x - 3) \)
So, to find zeroes of polynomial: \( 6x^2 - 13x + 6 \) will be 0, hence \( (3x - 2) = 0 \) and \( (2x - 3) = 0 \)
So \( x = \frac{2}{3} \) and \( x = \frac{3}{2} \)
Therefore, the zeroes of: \( 6x^2 - 13x + 6 \) are \( \frac{2}{3} \) and \( \frac{3}{2} \).
Sum of the zeroes \( = \frac{2}{3} + \frac{3}{2} = \frac{13}{6} = \frac{-(-13)}{6} = \frac{-\text{coefficient of } x}{\text{coefficient of } x^2} \)
Product of the zeroes \( = \frac{2}{3} \times \frac{3}{2} = \frac{6}{6} = \frac{\text{constant term}}{\text{coefficient of } x^2} \).
Verified.

 

Question. Find the zeroes of the quadratic polynomial \( 4x^2 - 9 \) and verify the relation between the zeroes and its coefficients.
Answer: We have, \( 4x^2 - 9 = (2x)^2 - 3^2 = (2x - 3) (2x + 3) \)
So, the value of \( 4x^2 - 9 \) is 0, when \( 2x - 3 = 0 \) or \( 2x + 3 = 0 \)
i.e., when \( x = \frac{3}{2} \) or \( x = -\frac{3}{2} \)
Therefore, the zeroes of \( 4x^2 - 9 \) are \( \frac{3}{2} \) and \( -\frac{3}{2} \).
Sum of the zeroes \( = \frac{3}{2} - \frac{3}{2} = 0 = \frac{-(0)}{4} = \frac{-\text{coefficient of } x}{\text{coefficient of } x^2} \)
Product of the zeroes \( = \left( \frac{3}{2} \right) \left( -\frac{3}{2} \right) = \frac{-9}{4} = \frac{\text{constant term}}{\text{coefficient of } x^2} \).
Verified.

 

Question. Find the zeroes of the quadratic polynomials and verify a relationship between zeroes and its 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) \( x^2 - 2x - 8 = x^2 - 4x + 2x - 8 = x(x - 4) + 2(x - 4) = (x - 4)(x + 2) \)
So, the value of \( x^2 - 2x - 8 \) is zero when \( x - 4 = 0 \) or \( x + 2 = 0 \), i.e., when \( x = 4 \) or \( x = -2 \).
So, the zeroes of \( x^2 - 2x - 8 \) are \( 4, -2 \).
Sum of the zeroes \( = 4 - 2 = 2 = \frac{-(-2)}{1} = \frac{-\text{coefficient of } x}{\text{coefficient of } x^2} = 2 \)
Product of the zeroes \( = 4(-2) = -8 = \frac{-8}{1} = \frac{\text{constant term}}{\text{coefficient of } x^2} = -8 \). Verified.
(ii) \( 4s^2 - 4s + 1 = 4s^2 - 2s - 2s + 1 \)
\( = 2s (2s - 1) - 1 (2s - 1) \)
\( = (2s - 1) (2s - 1) = (2s - 1)^2 \)
So, the value of \( 4s^2 - 4s + 1 \) is zero when \( 2s - 1 = 0 \), or \( s = \frac{1}{2} \)
Zeroes of the polynomial are \( \frac{1}{2}, \frac{1}{2} \)
Sum of the zeroes \( = \frac{1}{2} + \frac{1}{2} = 1 = \frac{-(-4)}{4} = \frac{-\text{coefficient of } s}{\text{coefficient of } s^2} = 1 \)
Product of the zeroes \( = \left( \frac{1}{2} \right) \left( \frac{1}{2} \right) = \frac{1}{4} = \frac{\text{constant term}}{\text{coefficient of } s^2} = \frac{1}{4} \). Verified.
(iii) We have : \( 6x^2 - 3 - 7x = 6x^2 - 7x - 3 = 6x^2 - 9x + 2x - 3 \)
\( = 3x (2x - 3) + 1 (2x - 3) = (3x + 1) (2x - 3) \)
The value of \( 6x^2 - 3 - 7x \) is 0, when the value of \( (3x + 1)(2x - 3) \) is 0, i.e., when \( 3x + 1 = 0 \) or \( 2x - 3 = 0 \), i.e., when \( x = -\frac{1}{3} \) and \( x = \frac{3}{2} \).
\( \therefore \) The zeroes of \( 6x^2 - 3 - 7x \) are \( -\frac{1}{3} \) and \( \frac{3}{2} \).
Therefore, sum of the zeroes \( = -\frac{1}{3} + \frac{3}{2} = \frac{7}{6} = \frac{-(-7)}{6} = \frac{-\text{Coefficient of } x}{\text{Coefficient of } x^2} = \frac{7}{6} \)
and product of zeroes \( = \left( -\frac{1}{3} \right) \left( \frac{3}{2} \right) = \frac{-3}{6} = \frac{\text{Constant term}}{\text{Coefficient of } x^2} = \frac{-3}{6} \). Verified.
(iv) We have : \( 4u^2 + 8u = 4u(u + 2) \)
The value of \( 4u^2 + 8u \) is 0, when the value of \( 4u(u + 2) = 0 \), i.e., when \( u = 0 \) or \( u + 2 = 0 \), i.e., when \( u = 0 \) or \( u = -2 \).
\( \therefore \) The zeroes of \( 4u^2 + 8u \) are 0 and -2.
Therefore, sum of the zeroes \( = 0 + (-2) = -2 = \frac{-8}{4} = \frac{-\text{Coefficient of } u}{\text{Coefficient of } u^2} = -2 \).
and product of zeroes \( = (0)(-2) = 0 = \frac{0}{4} = \frac{\text{Constant term}}{\text{Coefficient of } u^2} = 0 \). Verified.
(v) We have: \( t^2 - 15 = (t - \sqrt{15})(t + \sqrt{15}) \)
The value of \( t^2 - 15 \) is 0, when the value of \( (t - \sqrt{15})(t + \sqrt{15}) \) is 0, i.e., when \( t - \sqrt{15} = 0 \) or \( t + \sqrt{15} = 0 \), i.e., when \( t = \sqrt{15} \) or \( t = -\sqrt{15} \).
\( \therefore \) The zeroes of \( t^2 - 15 \) are \( \sqrt{15} \) and \( -\sqrt{15} \).
Therefore, sum of the zeroes \( = \sqrt{15} + (-\sqrt{15}) = 0 = \frac{-0}{1} = \frac{-\text{Coefficient of } t}{\text{Coefficient of } t^2} = 0 \)
and product of the zeroes \( = (\sqrt{15})(-\sqrt{15}) = -15 = \frac{-15}{1} = \frac{\text{Constant term}}{\text{Coefficient of } t^2} = -15 \). Verified.
(vi) We have : \( 3x^2 - x - 4 = 3x^2 + 3x - 4x - 4 = 3x(x + 1) - 4(x + 1) = (x + 1) (3x - 4) \)
The value of \( 3x^2 - x - 4 \) is 0, when the value of \( (x + 1)(3x - 4) \) is 0, i.e., when \( x + 1 = 0 \) or \( 3x - 4 = 0 \), i.e., when \( x = -1 \) or \( x = \frac{4}{3} \).
\( \therefore \) The zeroes of \( 3x^2 - x - 4 \) are -1 and \( \frac{4}{3} \).
Therefore, sum of the zeroes \( = -1 + \frac{4}{3} = \frac{-3 + 4}{3} = \frac{1}{3} = \frac{-(-1)}{3} = \frac{-\text{Coefficient of } x}{\text{Coefficient of } x^2} = \frac{1}{3} \)
and product of the zeroes \( = (-1) \left( \frac{4}{3} \right) = \frac{-4}{3} = \frac{\text{Constant term}}{\text{Coefficient of } x^2} = \frac{-4}{3} \). Verified.

 

Question. Find a quadratic polynomial whose sum of zeroes and product of zeroes are respectively.
(i) \( \frac{1}{4}, -1 \)
(ii) \( \sqrt{2}, \frac{1}{3} \)
(iii) \( 0, \sqrt{5} \)
(iv) \( 1, 1 \)
(v) \( -\frac{1}{4}, \frac{1}{4} \)
(vi) \( 4, 1 \). 
Answer: Let the polynomial be \( ax^2 + bx + c \) and its zeroes be \( \alpha \) and \( \beta \).
(i) Here, \( \alpha + \beta = \frac{1}{4} \) and \( \alpha \cdot \beta = -1 \)
Thus the polynomial formed \( = x^2 - (\text{Sum of zeroes}) x + \text{Product of zeroes} \)
\( = x^2 - \left( \frac{1}{4} \right)x - 1 = x^2 - \frac{x}{4} - 1 \)
The other polynomials are \( k \left( x^2 - \frac{x}{4} - 1 \right) \)
If \( k = 4 \), then the polynomial is \( 4x^2 - x - 4 \).
(ii) Here, \( \alpha + \beta = \sqrt{2}, \alpha \cdot \beta = \frac{1}{3} \)
Thus the polynomial formed \( = x^2 - (\text{Sum of zeroes}) x + \text{Product of zeroes} \)
\( = x^2 - (\sqrt{2})x + \frac{1}{3} \) or \( x^2 - \sqrt{2}x + \frac{1}{3} \)
Other polynomials are \( k \left( x^2 - \sqrt{2}x + \frac{1}{3} \right) \)
If \( k = 3 \), then the polynomial is \( 3x^2 - 3\sqrt{2}x + 1 \).
(iii) Here, \( \alpha + \beta = 0 \) and \( \alpha \cdot \beta = \sqrt{5} \)
Thus the polynomial formed \( = x^2 - (\text{Sum of zeroes})x + \text{Product of zeroes} \)
\( = x^2 - (0)x + \sqrt{5} = x^2 + \sqrt{5} \).
(iv) Let the polynomial be \( ax^2 + bx + c \) and its zeroes be \( \alpha \) and \( \beta \). Then,
\( \alpha + \beta = 1 = \frac{-(-1)}{1} = \frac{-b}{a} \)
\( \alpha\beta = 1 = \frac{c}{a} \)
If \( a = 1 \), then \( b = -1 \) and \( c = 1 \).
\( \therefore \) One quadratic polynomial which satisfy the given conditions is \( x^2 - x + 1 \).
(v) Let the polynomial be \( ax^2 + bx + c \) and its zeroes be \( \alpha \) and \( \beta \). Then,
\( \alpha + \beta = -\frac{1}{4} = \frac{-1}{4} = \frac{-b}{a} \)
and \( \alpha\beta = \frac{1}{4} = \frac{c}{a} \)
If \( a = 4 \), then \( b = 1 \) and \( c = 1 \).
\( \therefore \) One quadratic polynomial which satisfy the given conditions is \( 4x^2 + x + 1 \).
(vi) Let the polynomial be \( ax^2 + bx + c \) and its zeroes be \( \alpha \) and \( \beta \). Then,
\( \alpha + \beta = 4 = \frac{-(-4)}{1} = \frac{-b}{a} \)
and \( \alpha\beta = 1 = \frac{1}{1} = \frac{c}{a} \)
If \( a = 1 \), then \( b = -4 \) and \( c = 1 \).
\( \therefore \) One quadratic polynomial which satisfy the given conditions is \( x^2 - 4x + 1 \).

 

Question. Find a quadratic polynomial, the sum and product of whose zeroes are \( \sqrt{2} \) and \( -\frac{3}{2} \), respectively. Also find its zeroes.
Answer: A quadratic polynomial, the sum and product of zeroes are \( \sqrt{2} \) and \( -\frac{3}{2} \) is \( x^2 - \sqrt{2}x - \frac{3}{2} \)
\( x^2 - \sqrt{2}x - \frac{3}{2} = \frac{1}{2} [2x^2 - 2\sqrt{2}x - 3] \)
\( = \frac{1}{2} [2x^2 + \sqrt{2}x - 3\sqrt{2}x - 3] \)
\( = \frac{1}{2} [\sqrt{2}x(\sqrt{2}x + 1) - 3(\sqrt{2}x + 1)] \)
\( = \frac{1}{2} [\sqrt{2}x + 1][\sqrt{2}x - 3] \)
Hence, the zeroes are \( -\frac{1}{\sqrt{2}} \) and \( \frac{3}{\sqrt{2}} \).

 

Question. Verify that the numbers given along sides of the cubic polynomials are their zeroes. Also verify the relationship between the zeroes and the coefficients, \( x^3 + 2x^2 - x - 2; 1, -1, -2 \).
Answer: Here the polynomial \( p(x) \) is \( x^3 + 2x^2 - x - 2 \)
Value of the polynomial \( x^3 + 2x^2 - x - 2 \) when \( x = 1 \)
\( p(1) = (1)^3 + 2(1)^2 - 1 - 2 = 1 + 2 - 1 - 2 = 0 \)
So, 1 is a zero of \( p(x) \).
On putting \( x = -1 \) in the cubic polynomial \( x^3 + 2x^2 - x - 2 \)
\( p(-1) = (-1)^3 + 2(-1)^2 - (-1) - 2 = -1 + 2 + 1 - 2 = 0 \)
So, -1 is a zero of \( p(x) \).
On putting \( x = -2 \) in the cubic polynomial \( x^3 + 2x^2 - x - 2 \)
\( p(-2) = (-2)^3 + 2(-2)^2 - (-2) - 2 = -8 + 8 + 2 - 2 = 0 \)
So, -2 is a zero of \( p(x) \).
Hence, 1, -1 and -2 are the zeroes of the given polynomial.
Sum of the zeroes of \( p(x) = 1 - 1 - 2 = -2 = \frac{-2}{1} = \frac{-\text{coefficient of } x^2}{\text{coefficient of } x^3} \)
Sum of the products of zeros when two zeroes taken at a time
\( = (1)(-1) + (-1)(-2) + (1)(-2) = -1 + 2 - 2 = -1 = \frac{-1}{1} = \frac{\text{coefficient of } x}{\text{coefficient of } x^3} \)
Product of all three zeroes \( = (1)(-1)(-2) = 2 = \frac{-(-2)}{1} = -\frac{\text{constant term}}{\text{coefficient of } x^3} \). Verified.

 

Question. Verify that the numbers given along sides of the cubic polynomial are their zeroes. Also verify the relationship between the zeroes and the coefficients; \( x^3 - 27x + 54; -6, 3, 3 \).
Answer: Here the polynomial \( p(x) \) is \( x^3 - 27x + 54 \)
Value of the polynomial \( x^3 - 27x + 54 \) when \( x = -6 \)
\( p(-6) = (-6)^3 - 27 (-6) + 54 = -216 + 162 + 54 = 0 \)
So, -6 is a zero of \( p(x) \).
On putting \( x = 3 \) in the cubic polynomial \( x^3 - 27x + 54 \)
\( p(3) = (3)^3 - 27(3) + 54 = 27 - 81 + 54 = 0 \)
So, 3 is a zero of \( p(x) \).
Hence, -6, 3, 3 are the zeroes of the given cubic polynomial.
Sum of the zeroes of \( p(x) = -6 + 3 + 3 = 0 = \frac{-(0)}{1} = \frac{-\text{coefficient of } x^2}{\text{coefficient of } x^3} \)
Sum of the products of two zeroes taken at a time
\( = (-6)(3) + (3)(3) + (-6)(3) = -18 + 9 - 18 = -27 = \frac{-27}{1} = \frac{\text{coefficient of } x}{\text{coefficient of } x^3} \)
Product of all the three zeroes \( = (-6)(3)(3) = -54 = \frac{-54}{1} = -\frac{\text{constant term}}{\text{coefficient of } x^3} \). Verified.

 

Question. Verify that the numbers given along side of the polynomial are their zeroes. Also verify the relationship between the zeroes and the coefficients, \( x^4 + 2x^3 - 7x^2 - 8x + 12; -3, -2, 1, 2 \).
Answer: Here the polynomial \( p(x) \) is \( x^4 + 2x^3 - 7x^2 - 8x + 12 \)
Value of the polynomial \( x^4 + 2x^3 - 7x^2 - 8x + 12 \) when \( x = -3 \)
\( p(-3) = (-3)^4 + 2(-3)^3 - 7(-3)^2 - 8(-3) + 12 = 81 - 54 - 63 + 24 + 12 = 0 \)
So, -3 is a zero of \( p(x) \). On putting \( x = -2 \) in the given polynomial, we have
\( p(-2) = (-2)^4 + 2 (-2)^3 - 7(-2)^2 - 8(-2) + 12 = 16 - 16 - 28 + 16 + 12 = 0 \)
So, -2 is a zero of \( p(x) \). On putting \( x = 1 \) in the given polynomial, we have
\( p(1) = (1)^4 + 2 (1)^3 - 7(1)^2 - 8(1) + 12 = 1 + 2 - 7 - 8 + 12 = 0 \)
So, 1 is a zero of \( p(x) \).
On putting \( x = 2 \) in the given polynomial, we have
\( p(2) = (2)^4 + 2(2)^3 - 7(2)^2 - 8(2) + 12 = 16 + 16 - 28 - 16 + 12 = 0 \)
So, 2 is a zero of \( p(x) \).
Hence, -3, -2, 1, 2 are the zeroes of given polynomial.
Sum of the zeroes of \( p(x) = -3 - 2 + 1 + 2 = -2 = \frac{-2}{1} = \frac{-\text{coefficient of } x^3}{\text{coefficient of } x^4} \)
Sum of the products of two zeroes taken at a time
\( = (-3)(-2) + (-2)(1) + (1)(2) + (-3)(1) + (-3)(2) + (-2)(2) \)
\( = 6 - 2 + 2 - 3 - 6 - 4 = -7 = \frac{-7}{1} = \frac{\text{coefficient of } x^2}{\text{coefficient of } x^4} \)
Sum of the products of three zeroes taken at a time
\( = (-3)(-2)(1) + (-2)(1)(2) + (-3)(-2)(2) + (-3)(1)(2) \)
\( = 6 - 4 + 12 - 6 = 8 = \frac{-(-8)}{1} = -\frac{\text{coefficient of } x}{\text{coefficient of } x^4} \)
Product of all the four zeroes \( = (-3)(-2)(1)(2) = 12 = \frac{12}{1} = \frac{\text{constant term}}{\text{coefficient of } x^4} \). Verified.

 

Questions based upon finding the zeroes of a given polynomial and verifying the relations between their zeroes and coefficient of polynomial

 

Question. \( x^2 - 7x - 8 \).
Answer: \( \alpha = -1, \beta = 8, a = 1, b = -7, c = -8 \)

 

Question. \( 9x^2 - 5 \)
Answer: \( \alpha = \frac{-\sqrt{5}}{3}, \beta = \frac{\sqrt{5}}{3}, a = 9, b = 0, c = -5 \)

 

Question. \( 8y^2 - 3y \)
Answer: \( \alpha = 0, \beta = \frac{3}{8}, a = 8, b = -3, c = 0 \)

 

Question. \( 3z^2 - 2z - 1 \)
Answer: \( \alpha = -\frac{1}{3}, \beta = 1, a = 3, b = -2, c = -1 \)

 

Question. \( 5x^2 + 10x + 5 \)
Answer: \( \alpha = -1, \beta = -1, (a = 5, b = 10, c = 5) \) or \( a = 1, b = 2, c = 1 \)

 

Question. \( x^3 - x \)
Answer: \( \alpha = -1, \beta = 0, \gamma = 1, a = 1, b = 0, c = -1, d = 0 \)

 

Question. \( x^3 - 6x^2 + 11x - 6 \)
Answer: \( \alpha = 1, \beta = 2, \gamma = 3, a = 1, b = -6, c = 11, d = -6 \)

 

Question. \( y^3 - 6y^2 + 9y \)
Answer: \( \alpha = 0, \beta = 3, \gamma = 3, a = 1, b = -6, c = 9, d = 0 \)

 

Questions on formation of polynomials

Find the polynomials whose zeroes are given as under:

 

Question. \( \alpha = 2, \beta = 3 \)
Answer: \( x^2 - 5x + 6 \)

 

Question. \( \alpha = -4, \beta = 5 \)
Answer: \( x^2 - 1x - 20 \)

 

Question. \( \alpha = -1, \beta = -2 \)
Answer: \( x^2 + 3x + 2 \)

 

Question. \( \alpha = 0, \beta = 5 \)
Answer: \( x^2 - 5x \)

 

Question. \( \alpha = 4, \beta = -4 \)
Answer: \( x^2 - 16 \)

 

Question. \( \alpha = 1/2, \beta = 1/2 \)
Answer: \( x^2 - x + \frac{1}{4} \)

 

Question. \( \alpha = \frac{-5}{3}, \beta = \frac{-5}{3} \)
Answer: \( x^2 + \frac{10}{3}x + \frac{25}{9} \)

 

Questions based upon verification of given numbers are zeroes or not for given polynomial

 

Question. \( p(x) = 7x - 12, (12 / 7) \)
Answer: Yes verified: \( \alpha = \frac{12}{7}, a = 7, b = -12 \) and use \( \alpha = -\frac{b}{a} \) yes verified.

 

Question. \( p(x) = x^2 - 4x + 3 (1, 3) \)
Answer: \( \alpha = 1, \beta = 3, a = 1, b = -4, c = 3 \) and use \( \alpha + \beta = -\frac{b}{a} \) & \( \alpha\beta = \frac{c}{a} \) yes verified.

 

Question. \( p(x) = x^3 - 6x^2 + 11x - 6 (1, 2, 3) \)
Answer: \( \alpha = 1, \beta = 2, \gamma = 3, a = 1, b = -6, c = 11, d = -6 \) and use \( \alpha + \beta + \gamma = -\frac{b}{a}, \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} \) and \( \alpha\beta\gamma = -\frac{d}{a} \)

 

Question. \( 2x^3 - 7x^2 + 7x - 2, \left(\frac{1}{2}, 1, 2\right) \)
Answer: Yes verified, \( \alpha = \frac{1}{2}, \beta = 1, \gamma = 2, a = 2, b = -7, c = 7, d = -2 \) and use \( \alpha + \beta + \gamma = -\frac{b}{a}, \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} \) and \( \alpha\beta\gamma = -\frac{d}{a} \)

 

Question. If the zeroes of polynomial \( x^3 - 3x^2 + x + 1 \) are \( \alpha - \beta, \alpha \) and \( \alpha + \beta \), find \( \alpha \) & \( \beta \).
Answer: \( \alpha = 1, \beta = \pm\sqrt{2} \)

 

Question. Divide \( 3x^4 - 4x^2 + 8x - 1 \) by \( x - 2 \).
Answer:
First term of \( q(x) = \frac{3x^4}{x} = 3x^3 \)
Second term of \( q(x) = \frac{6x^3}{x} = 6x^2 \)
Third term of \( q(x) = \frac{8x^2}{x} = 8x \)
Fourth term of \( q(x) = \frac{24x}{x} = 24 \)
By long division process, we get:
Quotient = \( 3x^3 + 6x^2 + 8x + 24 \)
Remainder = 47 Ans.

 

Question. Divide the polynomial \( p(x) = x^3 - 3x^2 + 5x - 3 \) by the polynomial \( g(x) = x^2 - 2 \) and find the quotient and remainder.
Answer:
Here, dividend and divisor are both in standard forms. So, we have :
By dividing \( x^3 - 3x^2 + 5x - 3 \) by \( x^2 - 2 \), we get quotient as \( x - 3 \) and remainder as \( 7x - 9 \).
\( \therefore \) The quotient is \( x - 3 \) and the remainder is \( 7x - 9 \). Ans.

 

Question. Divide the polynomial \( p(x) = x^4 - 3x^2 + 4x + 5 \) by the polynomial \( g(x) = x^2 + 1 - x \) and find the quotient and remainder.
Answer:
Here, the dividend is already in the standard form and the divisor is not in the standard form. It can be written as \( x^2 - x + 1 \).
On dividing \( x^4 - 3x^2 + 4x + 5 \) by \( x^2 - x + 1 \), we get:
Quotient = \( x^2 + x - 3 \)
Remainder = 8
\( \therefore \) The quotient is \( x^2 + x - 3 \) and the remainder is 8. Ans.

 

Question. Divide the polynomial \( p(x) = x^4 - 5x + 6 \) by the polynomial \( g(x) = 2 - x^2 \) and find the quotient and remainder.
Answer:
We have divisor \( [-x^2 + 2] \) and dividend \( : x^4 - 5x + 6 \).
On dividing \( x^4 - 5x + 6 \) by \( -x^2 + 2 \), we get:
Quotient = \( -x^2 - 2 \)
Remainder = \( -5x + 10 \)
\( \therefore \) The quotient is \( -x^2 - 2 \) and the remainder is \( -5x + 10 \). Ans.

 

Question. Check whether the first polynomial \( t^2 - 3 \) is a factor of the second polynomial \( 2t^4 + 3t^3 - 2t^2 - 9t - 12 \) by dividing.
Answer:
Let us divide \( 2t^4 + 3t^3 - 2t^2 - 9t - 12 \) by \( t^2 - 3 \).
By long division, the remainder is 0.
Since the remainder is 0, therefore, \( t^2 - 3 \) is a factor of \( 2t^4 + 3t^3 - 2t^2 - 9t - 12 \). Ans.

 

Question. Check whether the first polynomial \( x^2 + 3x + 1 \) is a factor of the second polynomial \( 3x^4 + 5x^3 - 7x^2 + 2x + 2 \) by dividing.
Answer:
Let us divide \( 3x^4 + 5x^3 - 7x^2 + 2x + 2 \) by \( x^2 + 3x + 1 \).
By long division, the remainder is 0.
Since the remainder is 0, therefore, \( x^2 + 3x + 1 \) is a factor of \( 3x^4 + 5x^3 - 7x^2 + 2x + 2 \). Ans.

 

Question. Check whether the first polynomial \( x^3 - 3x + 1 \) is a factor of the second polynomial \( x^5 - 4x^3 + x^2 + 3x + 1 \) by dividing.
Answer:
Let us divide \( x^5 - 4x^3 + x^2 + 3x + 1 \) by \( x^3 - 3x + 1 \).
By long division, we get a remainder of 2.
Here, remainder is \( 2 (\neq 0) \). Therefore, \( x^3 - 3x + 1 \) is not a factor of \( x^5 - 4x^3 + x^2 + 3x + 1 \). Ans.

 

Question. Divide \( 5x^2 - x^3 - 3x + 5 \) by \( x - 1 - x^2 \). Also verify the division algorithm.
Answer:
Note that the given polynomials are not in standard form. To carry out division, we first write both the dividend and divisor in decreasing orders of their powers.
So, dividend = \( -x^3 + 3x^2 - 3x + 5 \) and divisor = \( -x^2 + x - 1 \).
Division process shows:
Quotient = \( x - 2 \), remainder = 3.
We stop here since degree \( (3) = 0 < 1 = \) degree \( (-x^2 + x - 1) \).
Now, Divisor \( \times \) Quotient + Remainder
\( = (-x^2 + x - 1)(x - 2) + 3 \)
\( = -x^3 + x^2 - x + 2x^2 - 2x + 2 + 3 \)
\( = -x^3 + 3x^2 - 3x + 5 \)
\( = \) Dividend
In this way, the division algorithm is verified.

 

Question. On dividing \( x^3 - 3x^2 + x + 2 \) by a polynomial \( g(x) \), the quotient and remainder were \( x - 2 \) and \( -2x + 4 \), respectively. Find \( g(x) \).
Answer:
By Division Algorithm, we know that
\( p(x) = q(x) \times g(x) + r(x) \)
Therefore, \( x^3 - 3x^2 + x + 2 = (x - 2) \times g(x) + (-2x + 4) \)

\( \implies \) \( x^3 - 3x^2 + x + 2 + 2x - 4 = (x - 2) \times g(x) \)

\( \implies \) \( g(x) = \frac{x^3 - 3x^2 + 3x - 2}{x - 2} \)
On dividing \( x^3 - 3x^2 + 3x - 2 \) by \( x - 2 \), we get \( g(x) = x^2 - x + 1 \).
Hence, \( g(x) = x^2 - x + 1 \). Ans.

 

Question. If the remainder on division of \( (x^3 + 2x^2 + kx + 3) \) by \( (x - 3) \) is 21, find the quotient and the value of 'k'. Hence, find the zeroes of the cubic polynomial \( x^3 + 2x^2 + kx - 18 \).
Answer:
Let \( p(x) = x^3 + 2x^2 + kx + 3 \)
Then, \( p(3) = 3^3 + 2 \times 3^2 + 3k + 3 = 21 \).
\( 3k = -27 \)

\( \implies \) \( k = -9 \)
Hence, the given polynomial will become \( x^3 + 2x^2 - 9x + 3 \).
Now, dividing \( x^3 + 2x^2 - 9x + 3 \) by \( x - 3 \), we get quotient \( (x^2 + 5x + 6) \) and remainder 21.
So, \( x^3 + 2x^2 - 9x + 3 = (x^2 + 5x + 6)(x - 3) + 21 \)
Now, for the polynomial \( x^3 + 2x^2 - 9x - 18 \):
\( x^3 + 2x^2 - 9x - 18 = (x - 3)(x^2 + 5x + 6) \)
\( = (x - 3)(x + 3)(x + 2) \)
So, the zeroes of \( x^3 + 2x^2 + kx - 18 \) are 3, -2, -3.

 

Question. Give examples of polynomials \( p(x), g(x), q(x) \) and \( r(x) \), which satisfy the division algorithm and:
(i) deg \( p(x) = \) deg \( q(x) \)
(ii) deg \( q(x) = \) deg \( r(x) \)
(iii) deg \( q(x) = 0 \)
Answer:
(i) Let \( q(x) = 3x^2 + 2x + 6 \), degree of \( q(x) = 2 \). Let \( p(x) = 12x^2 + 8x + 24 \), deg \( p(x) = \) deg \( q(x) = 2 \).
(ii) Let \( p(x) = x^5 + 2x^4 + 3x^3 + 5x^2 + 2 \), \( q(x) = x^2 + x + 1 \), \( r(x) = 2x^2 - 2x + 1 \). Here deg \( q(x) = \) deg \( r(x) = 2 \).
(iii) Let \( p(x) = 2x^4 + 8x^3 + 6x^2 + 4x + 12 \), \( q(x) = 2 \), deg \( q(x) = 0 \). Ans.

 

Question. Find the values of \( a \) and \( b \) so that \( x^4 + x^3 + 8x^2 + ax + b \) is divisible by \( x^2 + 1 \).
Answer:
If \( x^4 + x^3 + 8x^2 + ax + b \) is exactly divisible by \( x^2 + 1 \), then the remainder should be zero. On dividing, we get:
Remainder = \( x(a - 1) + (b - 7) \)
Now, Remainder = 0

\( \implies \) \( a - 1 = 0 \) and \( b - 7 = 0 \)

\( \implies \) \( a = 1 \) and \( b = 7 \). Ans.

 

 

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