CBSE Class 10 Mathematics Polynomials VBQs Set B

Question. The zeroes of the polynomial \( x^2 - 3x - m(m + 3) \) are:
(a) \( m, m + 3 \)
(b) \( -m, m + 3 \)
(c) \( m, -(m + 3) \)
(d) \( -m, -(m + 3) \)

Answer: (b)
Explanation: Given, polynomial can be rewritten as \( x^2 - (m + 3)x + mx - m(m + 3) = x[x - (m + 3)] + m[x - (m + 3)] = [x - (m + 3)] [x + m] \). Hence, the two zeroes are \( m + 3 \) and \( -m \).

Question. If one of the zeroes of the quadratic polynomial \( x^2 + 3x + k \) is 2, then the value of \( k \) is:
(a) 10
(b) -10
(c) -7
(d) -2

Answer: (b)
Explanation: Let, \( p(x) = x^2 + 3x + k \). Since, 2 is one of the zero of \( p(x) \), \( \therefore p(2) = 0 \Rightarrow 2^2 + 3(2) + k = 0 \Rightarrow 4 + 6 + k = 0 \Rightarrow k = -10 \).

Question. The quadratic polynomial, the sum of whose zeroes is -5 and their product is 6, is:
(a) \( x^2 + 5x + 6 \)
(b) \( x^2 - 5x + 6 \)
(c) \( x^2 - 5x - 6 \)
(d) \( -x^2 + 5x + 6 \)

Answer: (a)
Explanation: A polynomial, in which sum of zeroes is -5 and product of zeroes is 6, is: \( x^2 + 5x + 6 \). Since, the quadratic equation is: \( x^2 - (\text{sum of roots})x + \text{product of roots} = 0 \).

Question. If the zeroes of the quadratic polynomial \( x^2 + (a + 1)x + b \) are 2 and -3, then:
(a) \( a = -7, b = -1 \)
(b) \( a = 5, b = -1 \)
(c) \( a = 2, b = -6 \)
(d) \( a = 0, b = -6 \)

Answer: (d)
Explanation: Let \( p(x) = x^2 + (a + 1)x + b \). It is given that 2 and -3 are the zeroes of the given quadratic polynomial. Therefore, \( p(2) = 0 \) and \( p(-3) = 0 \).
\( p(2) = (2)^2 + (a + 1)(2) + b = 0 \Rightarrow 4 + 2a + 2 + b = 0 \Rightarrow 2a + b + 6 = 0 \)...(i)
Also, \( p(-3) = (-3)^2 + (a + 1)(-3) + b = 0 \Rightarrow 9 - 3a - 3 + b = 0 \Rightarrow -3a + b + 6 = 0 \)...(ii)
From (i) and (ii), we get \( 2a + b + 6 = -3a + b + 6 \Rightarrow 5a = 0 \Rightarrow a = 0 \).
Putting the value of 'a' in (i), we have \( 2(0) + b + 6 = 0 \Rightarrow b = -6 \).
Alternate Method: Sum of the zeroes \( = 2 + (-3) = -1 = -(a+1) \Rightarrow a+1=1 \Rightarrow a=0 \). Product of the zeroes \( = 2(-3) = -6 = b \).

Question. The number of polynomials having zeroes as -2 and 5 is:
(a) 1
(b) 2
(c) 3
(d) more than 3

Answer: (d)
Explanation: A quadratic polynomial is given by \( p(x) = k\{x^2 - (\text{sum of the zeroes})x + (\text{product of the zeroes})\} \), where \( k \) is any real number. Sum of the zeroes \( = -2 + 5 = 3 \). Product of the zeroes \( = (-2)5 = -10 \). The polynomial is \( k\{x^2 - 3x - 10\} \). As \( k \) can take any real value, there can be infinite polynomials.

Question. Given that one of the zeroes of the cubic polynomial \( ax^3 + bx^2 + cx + d \) is zero, the product of the other two zeroes is:
(a) \( -\frac{c}{a} \)
(b) \( \frac{c}{a} \)
(c) 0
(d) \( -\frac{b}{a} \)

Answer: (b)
Explanation: Let \( \alpha, \beta \) and \( \gamma \) be the zeroes of the polynomial \( p(x) = ax^3 + bx^2 + cx + d \). Let \( \alpha = 0 \). We know that sum of the product of two zeroes at a time \( = \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} \). Since \( \alpha = 0 \), \( 0 \times \beta + \beta\gamma + \gamma \times 0 = \frac{c}{a} \Rightarrow \beta\gamma = \frac{c}{a} \).

Question. If one of the zeroes of the cubic polynomial \( x^3 + ax^2 + bx + c \) is -1, then the product of the other two zeroes is:
(a) \( b - a + 1 \)
(b) \( b - a - 1 \)
(c) \( a - b + 1 \)
(d) \( a - b - 1 \)

Answer: (a)
Explanation: Let \( \alpha, \beta \) and \( \gamma \) be the zeroes. Let \( \alpha = -1 \). Since -1 is a zero, \( p(-1) = (-1)^3 + a(-1)^2 + b(-1) + c = 0 \Rightarrow -1 + a - b + c = 0 \Rightarrow c = 1 - a + b \). Product of zeroes \( \alpha\beta\gamma = -\frac{\text{constant term}}{\text{coefficient of } x^3} = -c \). So \( (-1)\beta\gamma = -c \Rightarrow \beta\gamma = c \). Substituting \( c \), product \( = 1 - a + b \), which is \( b - a + 1 \).

Question. If \( \alpha, \beta \) are the zeros of the polynomial \( 5x^2 - 7x + 2 \), then the sum of their reciprocal is:
(a) \( \frac{7}{2} \)
(b) \( \frac{7}{5} \)
(c) \( \frac{2}{5} \)
(d) \( \frac{14}{25} \)

Answer: (a)
Explanation: Here, \( \alpha + \beta = -\frac{b}{a} = \frac{-(-7)}{5} = \frac{7}{5} \) and \( \alpha\beta = \frac{c}{a} = \frac{2}{5} \). Sum of reciprocals \( = \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{7/5}{2/5} = \frac{7}{2} \).

Question. The degree of the polynomial \( (x + 1)(x^2 - x + x^4 - 1) \) is:
(a) 2
(b) 3
(c) 4
(d) 5

Answer: (d)
Explanation: Given polynomial can be rewritten as \( x^5 + x^4 + x^3 - 2x - 1 \). This is a polynomial of degree 5.

Question. The zeroes of the quadratic polynomial \( x^2 + 99x + 127 \) are:
(a) both positive
(b) both negative
(c) one positive and one negative
(d) both equal

Answer: (b)
Explanation: For \( p(x) = x^2 + 99x + 127 \), the zeroes are \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-99 \pm \sqrt{(99)^2 - 4(1)(127)}}{2} \). Both values will be negative as \( \sqrt{99^2 - 508} \) is less than 99.

Question. The zeroes of the quadratic polynomial \( x^2 + kx + k \), where \( k \neq 0 \),
(a) cannot both be positive
(b) cannot both be negative
(c) are always unequal
(d) are always equal

Answer: (a)
Explanation: Sum of zeroes \( = -k \) and product of zeroes \( = k \). Case 1: If \( k < 0 \), product is negative, meaning zeroes have opposite signs. Case 2: If \( k > 0 \), product is positive but sum is negative, meaning both zeroes must be negative. In neither case can both be positive.

Question. If the zeroes of the quadratic polynomial \( ax^2 + bx + c \), where \( c \neq 0 \), are equal, then:
(a) \( c \) and \( a \) have opposite signs
(b) \( c \) and \( b \) have opposite signs
(c) \( c \) and \( a \) have the same sign
(d) \( c \) and \( b \) have the same sign

Answer: (c)
Explanation: Zeroes are equal when discriminant \( D = b^2 - 4ac = 0 \Rightarrow 4ac = b^2 \). Since \( b^2 \) is always positive, \( ac \) must be positive, meaning \( a \) and \( c \) must have the same sign.

Question. If one of the zeroes of a quadratic polynomial of the form \( x^2 + ax + b \) is the negative of the other, then it:
(a) has no linear term and the constant term is negative.
(b) has no linear term and the constant term is positive.
(c) can have a linear term but the constant term is negative.
(d) can have a linear term but the constant term is positive.

Answer: (a)
Explanation: Let zeroes be \( \alpha \) and \( -\alpha \). Sum \( = \alpha + (-\alpha) = 0 \Rightarrow -a = 0 \Rightarrow a = 0 \). Product \( = \alpha(-\alpha) = -\alpha^2 = b \). Since \( \alpha^2 > 0 \), \( b \) must be negative. Thus, no linear term (\( a=0 \)) and constant term is negative.

Question. Which of the following is not the graph of a quadratic polynomial?
(a) Parabola opening upwards
(b) Parabola opening downwards intersecting x-axis twice
(c) Parabola touching x-axis at one point
(d) A curve intersecting the x-axis at three points

Answer: (d)
A quadratic polynomial's graph is always a parabola and can intersect the x-axis at most at two points. A curve intersecting the x-axis at three points represents a cubic or higher-degree polynomial.

Question. If the zeroes of the quadratic polynomial \( ax^2 + bx + c \), where \( c \neq 0 \), are equal, then:
(a) \( c \) and \( a \) have opposite signs
(b) \( c \) and \( b \) have opposite signs
(c) \( c \) and \( a \) have the same sign
(d) \( c \) and \( b \) have the same sign

Answer: (c)
Explanation: Given that the zeroes of the quadratic polynomial \( p(x) = ax^2 + bx + c \), where \( c \neq 0 \), are equal. The zeroes of a quadratic polynomial are equal when the discriminant is equal to 0 i.e., \( D = 0 \). \( b^2 - 4ac = 0 \Rightarrow 4ac = b^2 \Rightarrow ac = \frac{b^2}{4} > 0 \). Therefore, for \( ac > 0 \), \( a \) and \( c \) must have the same sign i.e., either \( a > 0 \) and \( c > 0 \) or \( a < 0 \) and \( c < 0 \).

Question. Alternate Method: Given that the zeroes of the quadratic polynomial \( p(x) = ax^2 + bx + c \), where \( c \neq 0 \), are equal. Let \( \alpha \) and \( \beta \) be the zeroes of the polynomial \( p(x) \). If \( \alpha \) and \( \beta \) are equal, these must have the same sign (both positive or both negative).
\( \Rightarrow \alpha\beta > 0 \)
Product of zeroes \( \alpha\beta = \frac{c}{a} \)
\( \Rightarrow \frac{c}{a} > 0 \) [Using \( \alpha\beta > 0 \)]
As \( \frac{c}{a} > 0 \), which is only possible when \( a \) and \( c \) have the same signs, so \( \alpha \) and \( \beta \) have the same sign.

Question. If one of the zeroes of a quadratic polynomial of the form \( x^2 + ax + b \) is the negative of the other, then it:
(a) has no linear term and the constant term is negative.
(b) has no linear term and the constant term is positive.
(c) can have a linear term but the constant term is negative.
(d) can have a linear term but the constant term is positive.

Answer: (a)
Explanation: Let \( p(x) = x^2 + ax + b \). And let \( \alpha \) be one of the zeroes, and \( -\alpha \) is the other zero of the polynomial \( p(x) \). [Given] Product of the zeroes \( = \frac{\text{constant term}}{\text{coefficient of } x^2} \). Product of the zeroes \( = \frac{b}{1} \). \( \alpha(-\alpha) = b \Rightarrow -\alpha^2 = b \) i.e., \( b < 0 \). i.e., the constant term is negative. Sum of the zeroes \( = -\frac{\text{coefficient of } x}{\text{coefficient of } x^2} \). \( \alpha - \alpha = -\frac{a}{1} \Rightarrow 0 = -a \Rightarrow a = 0 \). Hence, it has no linear term and the constant term is negative.

Fill in the Blanks

Question. If one root of the equation \((k - 1)x^2 - 10x + 3 = 0\) is the reciprocal of the other, then the value of \(k\) is .................... .
Answer: 4

Question. The sum and product of the zeroes of a quadratic polynomial are 3 and -10 respectively. The quadratic polynomial is ................... .
Answer: \(x^2 - 3x - 10\)
Explanation: Sum of zeroes = 3
Product of zeroes = -10
Quadratic polynomial: \(x^2 - (\text{sum of zeroes})x + \text{product of zeroes} = x^2 - 3x - 10\)

Question. If two of the zeroes of the cubic polynomial \(ax^3 + bx^2 + cx + d\) are 0, then the third zero is ................... .
Answer: \(-\frac{b}{a}\)
Explanation: Two zeroes of the cubic polynomial are zero. sum of zeroes = \(-\frac{b}{a} \Rightarrow (0 + 0 + x) = -\frac{b}{a}\) (where, \(x\) is the third zero) \(x = -\frac{b}{a}\)

Question. Zeroes of \(p(x) = x^2 - 2x - 3\) are ................... .
Answer: 3 and -1
Explanation: We have, \(x^2 - 2x - 3 = x^2 - 3x + x - 3 \Rightarrow x(x - 3) + 1(x - 3) \Rightarrow (x - 3)(x + 1)\). zeroes of \(p(x)\) are 3 and -1.

Question. If \(x - 2\) is a factor of the polynomial \(x^3 - 6x^2 + ax - 8\), then the value of \(a\) is equal to ................ .
Answer: 12
Explanation: \((x - 2)\) is factor of polynomial \(p(x) = x^3 - 6x^2 + ax - 8\). Therefore, \(x = 2\) is a zero of polynomial \(p(2) = 0 \Rightarrow 2^3 - 6(2)^2 + 2a - 8 = 0 \Rightarrow 8 - 24 + 2a - 8 = 0 \Rightarrow 2a = 24 \Rightarrow a = 12\)

Question. If the sum of the zeroes of the quadratic polynomial \(kx^2 + 2x + 3k\) is equal to the product of its zeroes then \(k = ................... .\)
Answer: \(-\frac{2}{3}\)
Explanation: Given, polynomial \(P(x) = kx^2 + 2x + 3k\). sum of zeroes = \(-\frac{2}{k}\), Product of zeroes = \(\frac{3k}{k} = 3\). According to question, \(-\frac{2}{k} = 3 \Rightarrow k = -\frac{2}{3}\)

Question. If \(\alpha\) and \(\beta\) are the zeroes of the quadratic polynomial \(f(x) = x^2 - x - 4\), find the value of \(\frac{1}{\alpha} + \frac{1}{\beta} - \alpha\beta = ................... .\)
Answer: \(\frac{15}{4}\)
Explanation: \(f(x) = x^2 - x - 4\). Let \(\alpha\) and \(\beta\) are the zeroes of \(f(x)\). \(\alpha + \beta = \frac{-(-1)}{1} = 1\), \(\alpha\beta = \frac{-4}{1} = -4\). So, \(\frac{1}{\alpha} + \frac{1}{\beta} - \alpha\beta = \frac{\alpha + \beta}{\alpha\beta} - \alpha\beta = \frac{1}{-4} - (-4) = -\frac{1}{4} + 4 = \frac{15}{4}\)

Question. A monomial has ................... term/terms.
Answer: One
Explanation: A monomial is number, variable or a product of a number and variable where all exponents are whole numbers. Example: 42, \(5x\), \(2xy\).

Fill in the Blanks

Question. If one root of the equation \( (k - 1)x^2 - 10x + 3 = 0 \) is the reciprocal of the other, then the value of k is .................... .
Answer: 4

Question. The sum and product of the zeroes of a quadratic polynomial are 3 and -10 respectively. The quadratic polynomial is ................... .
Answer: \( x^2 – 3x – 10 \)
Explanation: Sum of zeroes = 3
Product of zeroes = – 10
Quadratic polynomial \( x^2 – (\text{sum of zeroes}) x + \text{product of zeroes} = x^2 – 3x – 10 \)

Question. If two of the zeroes of the cubic polynomial \( ax^3 + bx^2 + cx + d \) are 0, then the third zero is ................... .
Answer: \( \frac{-b}{a} \)
Explanation: Two zeroes of the cubic polynomial are zero. Sum of zeroes = \( \frac{-b}{a} \)
\( \Rightarrow (0 + 0 + x) = \frac{-b}{a} \) (where, x is the third zero)
\( x = \frac{-b}{a} \)

Question. Zeroes of \( p(x) = x^2 – 2x – 3 \) are ................... .
Answer: 3 and – 1
Explanation: We have, \( x^2 – 2x – 3 = x^2 – 3x + x – 3 \)
\( \Rightarrow x (x – 3) + 1 (x – 3) \)
\( \Rightarrow (x – 3) (x + 1) \)
zeroes of \( p(x) \) are 3 and – 1.

Question. If \( x – 2 \) is a factor of the polynomial \( x^3 – 6x^2 + ax – 8 \), then the value of a is equal to ................ .
Answer: 12
Explanation: \( (x – 2) \) is factor of polynomial \( p(x) \)
\( p(x) = x^3 – 6x^2 + ax – 8 \)
Therefore, \( x = 2 \) is a zero of polynomial \( p(2) = 0 \)
\( \Rightarrow 2^3 – 6 (2)^2 + 2a – 8 = 0 \)
\( \Rightarrow 8 – 24 + 2a – 8 = 0 \Rightarrow 2a = 24 \Rightarrow a = 12 \)

Question. The number of zeroes of \( p(x) \) in the given figure is ................... .
Answer: 1
Explanation: The graph \( p(x) \) intersects the x-axis at only one point. So, number of zero is 1.

Question. If the sum of the zeroes of the quadratic polynomial \( kx^2 + 2x + 3k \) is equal to the product of its zeroes then \( k = \)................... .
Answer: \( \frac{-2}{3} \)
Explanation: Given, polynomial \( P(x) = kx^2 + 2x + 3k \)
sum of zeroes = \( \frac{-2}{k} \)
Product of zeroes = \( \frac{3k}{k} = 3 \)
According to question, \( \frac{-2}{k} = 3 \Rightarrow k = \frac{-2}{3} \)

Question. If \( \alpha \) and \( \beta \) are the zeroes of the quadratic polynomial \( f(x) = x^2 – x – 4 \), find the value of \( \frac{1}{\alpha} + \frac{1}{\beta} - \alpha\beta = \) ................... .
Answer: \( \frac{15}{4} \)
Explanation: \( f(x) = x^2 – x – 4 \)
Let \( \alpha \) and \( \beta \) are the zeroes of \( f(x) \)
\( \therefore \alpha + \beta = \frac{-(-1)}{1} = 1 \)
\( \alpha\beta = \frac{-4}{1} = – 4 \)
So, \( \frac{1}{\alpha} + \frac{1}{\beta} - \alpha\beta = \frac{\alpha + \beta}{\alpha\beta} - \alpha\beta \)
\( = \frac{1}{-4} - (- 4) = -\frac{1}{4} + 4 = \frac{15}{4} \)

Question. A monomial has ................... term/terms.
Answer: One
Explanation: A monomial is number, variable or a product of a number and variable where all exponents are whole numbers. Example : 42, 5x, 2xy.

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

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 \( \alpha\cdot\beta = \frac{-3}{4} \).
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,
\( (k – 1)(-3)^2 – k(-3) + 1 = 0 \)
\( \Rightarrow 9k – 9 + 3k + 1 = 0 \)
\( \Rightarrow 12k = 8 \Rightarrow k = \frac{2}{3} \)

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 = \( \frac{\text{Coefficient of } x}{\text{Coefficient of } x^2} \)]
Also, \( \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{0}{-1} = 0 \)
Thus, \( \alpha + \beta = \frac{1}{\alpha} + \frac{1}{\beta} \)