CBSE Class 10 Maths HOTs Polynomials Set 05

Very Short Answer Type Questions

Question. Form a quadratic polynomial, the sum and product of whose zeroes are \( (-3) \) and \( 2 \) respectively. 
Answer: A general form of a quadratic polynomial is \( ax^2 + bx + c \)
Here, \( \alpha + \beta = -\frac{b}{a} = -3 \) and \( \alpha\beta = \frac{c}{a} = 2 \)
where, \( \alpha \) and \( \beta \) are the roots of given polynomial.
So, the required polynomial is \( x^2 + 3x + 2 \).

Question. Find the value of \( k \) for which the roots of the equation \( 3x^2 - 10x + k = 0 \) are reciprocal of each other. 
Answer: Given, equation is \( 3x^2 - 10x + k = 0 \), where roots are reciprocals of each other.
Let the roots be \( \alpha \) and \( \frac{1}{\alpha} \)
\( \therefore \) Product of roots = \( \frac{c}{a} \)

\( \implies \alpha \cdot \frac{1}{\alpha} = \frac{k}{3} \) [\( \because a = 3, b = -10, c = k \)]

\( \implies 1 = \frac{k}{3} \)

\( \implies k = 3 \)

Question. Determine the degree of the polynomial \( (x + 1) (x^2 - x - x^4 + 1) \).
Answer: Given polynomial in standard form is :
\( -x^5 - x^4 + x^3 + 1 \)
So, its degree is 5

Question. If the product of two zeros of the polynomial \( p(x) = 2x^3 + 6x^2 - 4x + 9 \) is 3, find the third zero of the polynomial.
Answer: If \( \alpha, \beta \), and \( \gamma \) be the three zeros of \( p(x) \). Then,
\( \alpha\beta\gamma = -\frac{9}{2} \)
Since, \( \alpha\beta = 3 \),
we get \( \gamma = -\frac{9}{2} \times \frac{1}{3} = -\frac{3}{2} \)
Thus, the third zero of \( p(x) \) is \( -\frac{3}{2} \).

Question. If \( \alpha \) and \( \beta \) are the zeros of the polynomial \( p(x) = 4x^2 - 2x - 3 \), find the value of \( \frac{1}{\alpha} + \frac{1}{\beta} \).
Answer: Here, \( \alpha + \beta = \frac{2}{4} \) or \( \frac{1}{2} \) and \( \alpha\beta = \frac{-3}{4} \).
So, \( \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{1/2}{-3/4} = -\frac{2}{3} \).

Question. If one of the zeros of polynomial \( p(x) = (k - 1)x^2 - kx + 1 \) is \( -3 \), find the value of \( k \).
Answer: Since, \( (-3) \) is a zero of \( p(x) \), we have,
\( (k - 1)(-3)^2 - k(-3) + 1 = 0 \)

\( \implies 9k - 9 + 3k + 1 = 0 \),

\( \implies 12k = 8 \)

\( \implies k = \frac{2}{3} \)

Question. If \( \alpha \) and \( \beta \) be the roots of the equation \( x^2 - 1 = 0 \), then show that \( \alpha + \beta = \frac{1}{\alpha} + \frac{1}{\beta} \).
Answer: Here, \( \alpha + \beta = \frac{0}{1} = 0 \)
\( \because \) Sum of roots = \( \left[ \frac{\text{Coefficient of } x}{\text{Coefficient of } x^2} \right] \)
Also, \( \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{0}{-1} = 0 \)
Thus, \( \alpha + \beta = \frac{1}{\alpha} + \frac{1}{\beta} \)

Short Answer(SA-I) Type Questions

Question. A teacher asked 10 of his students to write a polynomial in one variable on a paper and then to handover the paper. The following were the answer given by the students: \( 2x + 3, 3x^2 + 7x + 2, 4x^3 + 3x^2 + 2, x^3 + \sqrt{3x} + 7, 7x + \sqrt{7}, 5x^3 - 7x + 2, 2x^3 + 3 - \frac{5}{x}, 5x - \frac{1}{2}, ax^3 + bx^2 + cx + d, x + \frac{1}{x} \).
Answer the following questions:
(A) How many of the above ten are not polynomials?
(B) How many of the above ten are quadratic polynomials? 
Answer: (A) Three, namely: \( x^3 + \sqrt{3x} + 7, 2x^3 + 3 - \frac{5}{x}, x + \frac{1}{x} \)
(As they contain square roots of the variable and negative power of \( x \))
(B) One, namely \( 3x^2 + 7x + 2 \)

Question. If one of the zeroes of the quadratic polynomial \( f(x) = 4x^2 - 8kx - 9 \) is equal in magnitude but opposite in sign of the other, then find the value of \( k \).
Answer: \( f(x) = 4x^2 - 8kx - 9 \)
Let one of the zeroes of the polynomial be \( \alpha \) and the other zeroes be \( -\alpha \)
Sum of zeroes = \( -\left( \frac{b}{a} \right) = \frac{8k}{4} \)
\( \alpha + (-\alpha) = 0 \)
So, \( \frac{8k}{4} = 0 \)

\( \implies k = 0 \)

Question. Can \( (x - 5) \) be the remainder on division of a polynomial \( p(x) \) by \( (x + 8) \)? 
Answer: No. We know that we cannot divide the polynomials which have same degree. As we can see that degree of \( (x - 5) \) = degree of \( (x + 8) \). So, they are not divisible.

Question. If the zeros of the polynomial \( x^3 - 3x^2 + x + 1 \) are \( a - b, a \) and \( a + b \), then find the values of \( a \) and \( b \).
Answer: As \( (a - b), a \) and \( (a + b) \) are zeros of \( x^3 - 3x^2 + x + 1 \), we have:
\( a - b + a + a + b = 3 \)

\( \implies 3a = 3 \), or \( a = 1 \) ...(i)
\( a (a - b) + a (a + b) + (a - b) (a + b) = 1 \)

\( \implies 3a^2 - b^2 = 1 \) ...(ii)
and \( (a - b) a (a + b) = -1 \)

\( \implies a(a^2 - b^2) = -1 \) ...(iii)
From (i) and (ii), we have \( b = \pm \sqrt{2} \)
Thus, \( a = 1, b = \pm \sqrt{2} \)

Question. What number should be added to the polynomial \( x^2 - 5x + 4 \) so that \( 3 \) is the zero of the polynomial?
Answer: Let \( k \) be the number to be added to the given polynomial. Then the polynomial becomes \( x^2 - 5x + (4 + k) \).
As 3 is the zero of the polynomial, we get:
\( (3)^2 - 5(3) + (4 + k) = 0 \)

\( \implies (4 + k) = 15 - 9 \)

\( \implies 4 + k = 6 \)

\( \implies k = 2 \)
Thus, 2 is to be added to the polynomial.

Question. If the zeroes of a polynomial \( x^2 - 8x + k = 0 \), is the HCF of \( (6, 12) \), then find the value of \( k \). 
Answer: HCF of \( (6, 12) = 6 \)
So, 6 is one of the roots of the polynomial.
\( f(x) = x^2 - 8x + k = 0 \)
\( f(6) = (6)^2 - 8(6) + k = 0 \)
\( 36 - 48 + k = 0 \)
\( -12 + k = 0 \)

\( \implies k = 12 \).

Short Answer (SA-II) Type Questions

Question. Find the quadratic polynomial sum and product of whose zeroes are \( -1 \) and \( -20 \) respectively. Also, find the zeroes of the polynomial so obtained. 
Answer: Let \( \alpha \) and \( \beta \) the zeroes of the polynomial.
Given : sum of zeroes, \( \alpha + \beta = -1 \)
product of zeroes, \( \alpha\beta = -20 \)
Equation of polynomial :
\( x^2 - (\text{sum of zeroes})x + \text{product of zeroes} = 0 \)
\( \therefore x^2 - (-1)x + (-20) = 0 \)

\( \implies x^2 + x - 20 = 0 \)
On splitting the middle term,
\( x^2 + 5x - 4x - 20 = 0 \)

\( \implies x(x + 5) - 4(x + 5) = 0 \)

\( \implies (x - 4) (x + 5) = 0 \)

\( \implies x = 4, -5 \)
Hence, the zeroes of the polynomial are 4 and \( -5 \).

Question. Find a quadratic polynomial whose zeroes are reciprocals of the zeroes of the polynomial \( f(x) = ax^2 + bx + c, a \neq 0, c \neq 0 \). 
Answer: Let \( \alpha, \beta \) be the zeroes of \( f(x) = ax^2 + bx + c \). Thus
\( \alpha + \beta = -\frac{b}{a} \) and \( \alpha\beta = \frac{c}{a} \)
Now, \( \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{-b/a}{c/a} = -\frac{b}{c} \)
\( \frac{1}{\alpha} \cdot \frac{1}{\beta} = \frac{1}{\alpha\beta} = \frac{1}{c/a} = \frac{a}{c} \)
\( \therefore \) Polynomial is: \( x^2 - (\text{sum of roots})x + \text{product roots} \)
\( x^2 - \left( -\frac{b}{c} \right)x + \frac{a}{c} = 0 \)

\( \implies cx^2 + bx + a = 0 \)
So, the required polynomial is \( cx^2 + bx + a \).

Question. If the zeroes of the polynomial \( x^2 + px + q \) are double the value to the zeroes of \( 2x^2 - 5x - 3 \), find the value of \( p \) and \( q \). 
Answer: Let \( \alpha \) and \( \beta \) are zeroes of the \( 2x^2 - 5x - 3 \)
\( \alpha + \beta = -\frac{b}{a} = \frac{5}{2} \) ...(i)
\( \alpha\beta = \frac{c}{a} = -\frac{3}{2} \) ...(ii)
According to the question, \( 2\alpha \) and \( 2\beta \) are zeroes of \( x^2 + px + q \)
\( 2\alpha + 2\beta = -p \)

\( \implies 2(\alpha + \beta) = -p \)
\( 2 \left( \frac{5}{2} \right) = -p \) [from eqn. (i)]
\( p = -5 \)
\( 2\alpha \times 2\beta = q \)

\( \implies 4\alpha\beta = q \)
\( 4 \left( -\frac{3}{2} \right) = q \) [from eqn. (ii)]
\( q = -6 \)
Hence, \( p = -5 \) and \( q = -6 \).

Question. Find the value of \( k \) such that the polynomial \( x^2 - (k + 6)x + 2(2k - 1) \) has the sum of its zeros equal to half of their product. 
Answer: Given polynomial is : \( p(x) = x^2 - (k + 6) x + 2(2k - 1) \)
In the given quadratic equation:
\( a = 1 \)
\( b = -(k + 6) \)
\( c = 2(2k - 1) \)
Sum of zeroes = \( -\frac{b}{a} = k + 6 \) ...(i)
Product of zeroes = \( \frac{c}{a} = 2(2k - 1) \) ...(ii)
According to the given condition:
Sum of the zeroes = \( \frac{1}{2} \times \text{Product of zeroes} \)

\( \implies k + 6 = \frac{1}{2} \times 2 (2k - 1) \)

\( \implies k + 6 = 2k - 1 \)

\( \implies 2k - k = 6 + 1 \)

\( \implies k = 7 \)
Hence, the value of \( k \) is 7.

Question. Find the zeroes of following polynomials by factorisation method and verify relation between the zeroes and coefficients of polynomials.
(A) \( 2x^2 + \frac{7}{2}x + \frac{3}{4} \) 
(B) \( 2s^2 - (1 + 2\sqrt{2})s + \sqrt{2} \) 
(C) \( 7y^2 - \frac{11}{3}y - \frac{2}{3} \) 
Answer: (A) Let \( f(x) = 2x^2 + \frac{7}{2}x + \frac{3}{4} = 8x^2 + 14x + 3 \) (Multiplying the given equation by 4)
\( = 8x^2 + (12x + 2x) + 3 \)
\( = 8x^2 + 12x + 2x + 3 \)
\( = 4x(2x + 3) + 1(2x + 3) \)
\( = (2x + 3)(4x + 1) \)
The zeroes of \( f(x) \) are given by \( f(x) = 0 \).
So, the value of \( 2x^2 + \frac{7}{2}x + \frac{3}{4} \) is zero when \( x = -\frac{3}{2} \) or \( x = -\frac{1}{4} \)

\( \implies x = -\frac{3}{2}, -\frac{1}{4} \)
Verification:
Sum of the zeroes = \( -(\text{coefficient of } x) \div \text{coefficient of } x^2 \)
\( \alpha + \beta = -\frac{b}{a} \)
\( \left( -\frac{3}{2} \right) + \left( -\frac{1}{4} \right) = -\frac{7}{4} \)
\( -\frac{7}{4} = -\frac{7}{4} \)
Product of the zeroes = \( \frac{\text{constant term}}{\text{coefficient of } x^2} \)
\( \alpha\beta = \frac{c}{a} \)
\( \left( -\frac{3}{2} \right) \left( -\frac{1}{4} \right) = \frac{3/4}{2} = \frac{3}{8} \)
\( \frac{3}{8} = \frac{3}{8} \)
Hence, verified.
(B) Let \( f(s) = 2s^2 - (1 + 2\sqrt{2})s + \sqrt{2} \)
\( = 2s^2 - s - 2\sqrt{2}s + \sqrt{2} \)
\( = s(2s - 1) - \sqrt{2}(2s - 1) \)
\( = (2s - 1)(s - \sqrt{2}) \)
The zeroes of \( f(s) \) are given by \( f(s) = 0 \)
So, the value is zero when \( 2s^2 - (1 + 2\sqrt{2})s + \sqrt{2} = 0 \)
i.e., when \( s = \frac{1}{2} \) or \( \sqrt{2} \)

\( \implies s = \frac{1}{2}, \sqrt{2} \)
Verification:
Sum of the zeroes = \( -(\text{coefficient of } s) \div \text{coefficient of } s^2 \)
\( \alpha + \beta = -\frac{b}{a} \)
\( \frac{1}{2} + \sqrt{2} = -\frac{-(1 + 2\sqrt{2})}{2} \)
\( \frac{1 + 2\sqrt{2}}{2} = \frac{1 + 2\sqrt{2}}{2} \)
Product of the zeroes = \( \frac{\text{constant term}}{\text{coefficient of } s^2} \)
\( \alpha\beta = \frac{c}{a} \)
\( \frac{1}{2} \times \sqrt{2} = \frac{\sqrt{2}}{2} \)
\( \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \)
Hence, verified.
(C) Let \( f(y) = 7y^2 - \frac{11}{3}y - \frac{2}{3} \)
\( = 21y^2 - 11y - 2 \)
\( = 21y^2 + (3y - 14y) - 2 \)
\( = 21y^2 + 3y - 14y - 2 \)
\( = 3y(7y + 1) - 2(7y + 1) \)
\( = (7y + 1)(3y - 2) \)
The zeroes of \( f(y) \) are given by \( f(y) = 0 \)
So, the value of \( 7y^2 - \frac{11}{3}y - \frac{2}{3} \) is zero when \( y = -\frac{1}{7} \) or \( y = \frac{2}{3} \)

\( \implies y = -\frac{1}{7}, \frac{2}{3} \)
Verification:
Sum of the zeroes = \( -(\text{coefficient of } y) \div \text{coefficient of } y^2 \)
\( \alpha + \beta = -\frac{b}{a} \)
\( -\frac{1}{7} + \frac{2}{3} = -\frac{-11}{21} \)
\( \frac{11}{21} = \frac{11}{21} \)
Product of the zeroes = \( \frac{\text{constant term}}{\text{coefficient of } y^2} \)
\( \alpha\beta = \frac{c}{a} \)
\( \left( -\frac{1}{7} \right) \left( \frac{2}{3} \right) = \frac{-2/3}{7} \)
\( -\frac{2}{21} = -\frac{2}{21} \)
Hence, verified.

Question. Find a quadratic polynomial whose zeroes are 1 and -3. Verify the relation between the coefficients and zeroes of polynomial. 
Answer: Sum of zeroes, \( S = 1 + (-3) = -2 \) ...(i)
Product of zeroes, \( P = 1 \times (-3) = -3 \) ...(ii)
Quadratic polynomial \( p(x) = x^2 - Sx + P \)
\( = x^2 - (-2)x - 3 = x^2 + 2x - 3 \)
Here, \( a = 1, b = 2, c = -3 \)
\( -\frac{b}{a} = -\frac{2}{1} = -2 \)
Sum of zeroes = \( -\frac{b}{a} = -2 \) [using eqn. (i)]
Also, \( \frac{c}{a} = \frac{-3}{1} = -3 \)
Product of zeroes = \( \frac{c}{a} = -3 \) [using eqn. (ii)]
Hence, verified.

Question. If one root of the equation \( 3x^2 - 8x + 2k + 1 = 0 \) is seven times the other, find the two roots and the value of \( k \).
Answer: Let \( \alpha \) and \( 7\alpha \) be the two roots of the equation: \( 3x^2 - 8x + (2k + 1) = 0 \)
Then, \( \alpha + 7\alpha = 8\alpha = \frac{8}{3} \) .....(i)
and \( \alpha \cdot (7\alpha) = 7\alpha^2 = \frac{2k + 1}{3} \) .....(ii)
From (i) \( \alpha = \frac{1}{3} \). So, the two roots are \( \frac{1}{3} \) and \( \frac{7}{3} \)
Using \( \alpha = \frac{1}{3} \) in (ii), we have:
\( 7 \left( \frac{1}{3} \right)^2 = \frac{2k + 1}{3} \)

\( \implies 2k + 1 = \frac{7}{3} \)

\( \implies 2k = \frac{4}{3} \)

\( \implies k = \frac{2}{3} \)

Question. Without actually calculating the zeroes, form a quadratic polynomial whose zeroes are reciprocals of the zeroes of the polynomial \( 5x^2 + 2x - 3 \). 
Answer: Let \( \alpha \) and \( \beta \) be the zeroes of \( 5x^2 + 2x - 3 \)
Then, \( \alpha + \beta = -\left( \frac{b}{a} \right) = -\frac{2}{5} \) and \( \alpha\beta = \frac{c}{a} = -\frac{3}{5} \)
Now \( \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{-2/5}{-3/5} = \frac{2}{3} \)
and \( \frac{1}{\alpha} \cdot \frac{1}{\beta} = \frac{1}{\alpha\beta} = -\frac{5}{3} \)
Thus, a quadratic polynomial where zeroes are \( \frac{1}{\alpha} \) and \( \frac{1}{\beta} \) is \( x^2 - (\text{sum of roots})x + \text{product of roots} = 0 \)

\( \implies x^2 - \frac{2}{3}x - \frac{5}{3} = 0 \)
i.e., \( 3x^2 - 2x - 5 \)

Long Answer Type Questions

Question. Obtain other zeroes of the polynomial \( f(x) = 2x^4 + 3x^3 - 5x^2 - 9x - 3 \) if two of its zeroes are \( -\sqrt{3} \) and \( \sqrt{3} \). 
Answer: Since \( \sqrt{3} \) and \( -\sqrt{3} \) are zeroes of \( f(x) \), \( (x - \sqrt{3})(x + \sqrt{3}) \) i.e., \( (x^2 - 3) \) is a factor of \( f(x) \). To obtain other two zeroes, we shall determine the quotient, by dividing \( f(x) \) with \( (x^2 - 3) \)
\[ \begin{array}{r|l} x^2 - 3 & 2x^4 + 3x^3 - 5x^2 - 9x - 3 \\ & 2x^4 \quad - 6x^2 \\ \hline & 3x^3 + x^2 - 9x - 3 \\ & 3x^3 \quad - 9x \\ \hline & x^2 - 3 \\ & x^2 - 3 \\ \hline & 0 \end{array} \]
Here, quotient = \( 2x^2 + 3x + 1 \)
= \( (2x + 1) (x + 1) \)
So, the two zeroes are \( -1 \) and \( -1/2 \).

Question. Given that the zeroes of the cubic polynomial \( x^3 - 6x^2 + 3x + 10 \) are of the form \( a, a + b, a + 2b \) for some real numbers \( a \) and \( b \), find the values of \( a \) and \( b \) as well as the zeroes of the given polynomial. 
Answer: Let \( p(x) = x^3 - 6x^2 + 3x + 10 \) and \( (a), (a + b) \) and \( (a + 2b) \) are the zeroes of \( p(x) \).
We know: Sum of the zeroes = \( -(\text{coefficient of } x^2) \div \text{coefficient of } x^3 \)

\( \implies a + (a + b) + (a + 2b) = -(-6) \)

\( \implies 3a + 3b = 6 \)

\( \implies a + b = 2 \)

\( \implies a = 2 - b \) ...(i)
Product of all the zeroes = \( -(\text{constant term}) \div \text{coefficient of } x^3 \)
\( a(a + b)(a + 2b) = -10 \)
\( (2 - b) (2) (2 + b) = -10 \) [Using eqn. (i)]
\( (2 - b) (2 + b) = -5 \)
\( 4 - b^2 = -5 \)

\( \implies b^2 = 9 \)

\( \implies b = \pm 3 \)
When \( b = 3, a = 2 - 3 = -1 \) [Using equation (i)]

\( \implies a = -1 \) when \( b = 3 \).
When \( b = -3, a = 2 - (-3) = 5 \) [Using equation (i)]

\( \implies a = 5 \) when \( b = -3 \).
Case 1: when \( a = -1 \) and \( b = 3 \)
The zeroes of the polynomial are: \( a = -1, a+b = -1+3=2, a+2b = -1+2(3)=5 \).
\( \implies -1, 2 \) and \( 5 \) are the zeroes.
Case 2: when \( a = 5 \) and \( b = -3 \)
The zeroes of the polynomial are: \( a = 5, a+b = 5-3=2, a+2b = 5-2(3)=-1 \).
\( \implies -1, 2 \) and \( 5 \) are the zeroes.
In both the cases, the zeroes of the polynomial are \( -1, 2, 5 \).

Question. Given that \( \sqrt{2} \) is a zero of the cubic polynomial \( 6x^3 + \sqrt{2}x^2 - 10x - 4\sqrt{2} \), find its other two zeroes. 
Answer: Let \( p(x) = 6x^3 + \sqrt{2}x^2 - 10x - 4\sqrt{2} \)
As \( \sqrt{2} \) is one of the zeroes of \( p(x) \).

\( \implies g(x) = (x - \sqrt{2}) \) is one of the factors of \( p(x) \).
\[ \begin{array}{r|l} x - \sqrt{2} & 6x^3 + \sqrt{2}x^2 - 10x - 4\sqrt{2} \\ & 6x^3 - 6\sqrt{2}x^2 \\ \hline & 7\sqrt{2}x^2 - 10x - 4\sqrt{2} \\ & 7\sqrt{2}x^2 - 14x \\ \hline & 4x - 4\sqrt{2} \\ & 4x - 4\sqrt{2} \\ \hline & 0 \end{array} \]
Then, \( 6x^3 + \sqrt{2}x^2 - 10x - 4\sqrt{2} = (x - \sqrt{2}) (6x^2 + 7\sqrt{2}x + 4) \)
\( = (x - \sqrt{2}) \{6x^2 + (3\sqrt{2}x + 4\sqrt{2}x) + 4\} \) (by splitting the middle term)
\( = (x - \sqrt{2}) \{6x^2 + 3\sqrt{2}x + 4\sqrt{2}x + 4\} \)
\( = (x - \sqrt{2}) \{3\sqrt{2}x (\sqrt{2}x + 1) + 4(\sqrt{2}x + 1)\} \)
\( = (x - \sqrt{2}) (\sqrt{2}x + 1) (3\sqrt{2}x + 4) \)

\( \implies x = \sqrt{2}, -\frac{1}{\sqrt{2}} \) or \( -\frac{4}{3\sqrt{2}} \)
Thus, the other two zeroes are \( -\frac{1}{\sqrt{2}} \) or \( -\frac{\sqrt{2}}{2} \) and \( -\frac{4}{3\sqrt{2}} \) or \( -\frac{2\sqrt{2}}{3} \).

Question. Given that \( x - \sqrt{5} \) is a factor of the cubic polynomial \( x^3 - 3\sqrt{5}x^2 + 13x - 3\sqrt{5} \), find all the zeroes of the polynomial. 
Answer: Let \( p(x) = x^3 - 3\sqrt{5}x^2 + 13x - 3\sqrt{5} \)
As \( \sqrt{5} \) is one of the zeroes of \( p(x) \).

\( \implies (x - \sqrt{5}) \) is one of the factors of \( p(x) \).
\[ \begin{array}{r|l} x - \sqrt{5} & x^3 - 3\sqrt{5}x^2 + 13x - 3\sqrt{5} \\ & x^3 - \sqrt{5}x^2 \\ \hline & -2\sqrt{5}x^2 + 13x - 3\sqrt{5} \\ & -2\sqrt{5}x^2 + 10x \\ \hline & 3x - 3\sqrt{5} \\ & 3x - 3\sqrt{5} \\ \hline & 0 \end{array} \]
Now \( p(x) = (x - \sqrt{5})(x^2 - 2\sqrt{5}x + 3) \)
\( = (x - \sqrt{5})(x^2 - \{(\sqrt{5} + \sqrt{2})x + (\sqrt{5} - \sqrt{2})x\} + 3) \) [Note: Error in source image, simpler splitting would be used. Let's provide standard zeroes directly if complex.]
\( = (x - \sqrt{5}) \{x - (\sqrt{5} + \sqrt{2})\} \{x - (\sqrt{5} - \sqrt{2})\} \)
So, all the zeroes of the given polynomial are \( (\sqrt{5} + \sqrt{2}), (\sqrt{5} - \sqrt{2}) \) and \( \sqrt{5} \).

Question. For which values of \( a \) and \( b \) are the zeroes of \( q(x) = x^3 + 2x^2 + a \) also the zeroes of the polynomial \( p(x) = x^5 - x^4 - 4x^3 + 3x^2 + 3x + b \)? 
Answer: Let \( p(x) = x^5 - x^4 - 4x^3 + 3x^2 + 3x + b \) and \( q(x) = x^3 + 2x^2 + a \).
Since, the zeroes of the polynomial \( q(x) \) are also zeroes of \( p(x) \), we can say that \( q(x) \) is a factor of \( p(x) \).
Then, on dividing \( p(x) \) by \( q(x) \):
\[ \begin{array}{r|l} x^3 + 2x^2 + a & x^5 - x^4 - 4x^3 + 3x^2 + 3x + b \\ & x^5 + 2x^4 \quad + ax^2 \\ \hline & -3x^4 - 4x^3 + (3 - a)x^2 + 3x + b \\ & -3x^4 - 6x^3 \quad \quad \quad - 3ax \\ \hline & 2x^3 + (3 - a)x^2 + (3 + 3a)x + b \\ & 2x^3 + 4x^2 \quad \quad + 2a \\ \hline & -(a + 1)x^2 + 3(1 + a)x + b - 2a \end{array} \]
But remainder, \( r(x) = -(a + 1)x^2 + 3(1 + a)x + b - 2a = 0 \)
[since, \( q(x) \) is factor of \( p(x) \)]

\( \implies -(a + 1)x^2 + 3(1 + a)x + b - 2a = 0 \cdot x^2 + 0 \cdot x + 0 \)
On comparing the coefficients of \( x^2 \) and constant term, we get
\( -(a + 1) = 0 \implies a = -1 \)
and \( b - 2a = 0 \implies b = 2a = 2(-1) = -2 \)