CBSE Class 10 Mathematics Polynomials MCQs Set 13

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) \( -m, m + 3 \)
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) -10
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) \( x^2 + 5x + 6 \)
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) \( a = 0, b = –6 \)
Explanation: Let \( p(x) = x^2 + (a + 1)x + b \).
We know that if \( \alpha \) is one of the zeroes of the quadratic polynomial \( p(x) = ax^2 + bx + c \), then \( p(\alpha) = 0 \).
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
\( 2(0) + b + 6 = 0 \)
\( \Rightarrow b = –6 \)
Hence, if the zeroes of the quadratic polynomial \( x^2 + (a + 1)x + b \) are 2 and –3, then the required values of a and b are \( a = 0 \) and \( b = –6 \).
Alternate Method:
It is given that 2 and –3 are the zeroes of the given quadratic polynomial.
Sum of the zeroes = \( 2 + (–3) = –1 \) ...(i)
Product of the zeroes = \( 2(–3) = –6 \) ...(ii)
The equation of 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 a constant.
Here, \( p(x) = x^2 + (a + 1)x + b \).
Comparing the two equations we get:
Sum of the zeroes = – (coefficient of x) ÷ coefficient of \( x^2 \)
\( \Rightarrow \text{sum of the zeroes} = – (a + 1) \)
\( \Rightarrow –1 = –a – 1 \) [Using (i)]
\( \Rightarrow –1 + 1 = –a \)
\( \Rightarrow –a = 0 \)
\( \Rightarrow a = 0 \)
Product of the zeroes = constant term ÷ coefficient of \( x^2 \)
\( \Rightarrow \text{Product of the zeroes} = b \)
\( –6 = b \) [Using (ii)]
\( \Rightarrow b = –6 \)
Hence, if the zeroes of the quadratic polynomial \( x^2 + (a + 1)x + b \) are 2 and –3, then the required values of a and b are \( a = 0 \) and \( b = –6 \).

Question. The number of polynomials having zeroes as -2 and 5 is:
(a) 1
(b) 2
(c) 3
(d) more than 3
Answer: (d) more than 3
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 a constant.
Sum of the zeroes = – (coefficient of x) ÷ coefficient of \( x^2 \)
and product of the zeroes = constant term ÷ coefficient of \( x^2 \)
Sum of the zeroes = \( – 2 + 5 = 3 \)
and product of the zeroes = \( (–2)5 = –10 \)
A quadratic polynomial is given by
\( = k\{x^2 – (\text{sum of the zeroes})x + (\text{product of the zeroes})\} \)
which becomes,
\( = k\{x^2 – 3x – 10\} \)
where k is any real number.
Thus, we can say that \( kx^2 – 3kx – 10k \) will also have –2 and 5 as their zeroes.
As k can take any real value, there can be infinite polynomials having -2 and 5 as their zeroes.
Hence, the required number of polynomials are infinite i.e., more than 3.

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) \( \frac{c}{a} \)
Explanation: Let \( p(x) = ax^3 + bx^2 + cx + d \).
It is given that one of the zeroes of the cubic polynomial p(x) is zero.
Let \( \alpha, \beta \) and \( \gamma \) be the zeroes of the polynomial \( p(x) = ax^3 + bx^2 + cx + d \)
And let \( \alpha = 0 \) [Given]
We know that:
Sum of the product of two zeroes at a time = coefficient of x ÷ coefficient of \( x^3 \) i.e.,
Sum of the product of two zeroes at a time = \( \frac{c}{a} \)
\( \Rightarrow \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} \)
\( \Rightarrow 0 \times \beta + \beta\gamma + \gamma \times 0 = \frac{c}{a} \) [Using \( \alpha = 0 \)]
\( \Rightarrow 0 + \beta\gamma + 0 = \frac{c}{a} \)
\( \Rightarrow \beta\gamma = \frac{c}{a} \)
Hence, product of the other two zeroes is \( \frac{c}{a} \).

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) \( b – a + 1 \)
Explanation: Let \( p(x) = x^3 + ax^2 + bx + c \)
Let \( \alpha, \beta \) and \( \gamma \) be the zeroes of the polynomial \( p(x) = x^3 + ax^2 + bx + c \)
And \( \alpha = –1 \) [Given]
We know that if \( \alpha \) is one of the zeroes of the polynomial, then \( p(\alpha) = 0 \).
\( \Rightarrow p(\alpha) = p(–1) = 0 \)
\( \Rightarrow (–1)^3 + (–1)^2a + (–1)b + c = 0 \)
\( \Rightarrow –1 + a – b + c = 0 \)
\( \Rightarrow c = 1 – a + b \) ...(i)
We know that:
Product of the zeroes = – constant term ÷ coefficient of \( x^3 \)
i.e., Product of zeroes = \( – \frac{c}{1} \)
\( \Rightarrow \alpha\beta\gamma = –c \)
\( \Rightarrow (–1)\beta\gamma = –c \) [Using \( \alpha = -1 \)]
\( \Rightarrow \beta\gamma = c \)
\( \Rightarrow \beta\gamma = 1 – a + b \) [From equation (i)]
Hence, product of the other two zeroes is \( 1 – a + b \) or \( b – a + 1 \).

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

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) 5
Explanation: Given polynomial can be written 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) both negative
Explanation: Let \( p(x) = x^2 + 99x + 127 \)
Then zeroes of the polynomial are given by
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( x = \frac{-99 \pm \sqrt{(99)^2 - 4(1)(127)}}{2} \)
\( x = \frac{-99 \pm \sqrt{9801 - 508}}{2} \)
\( x = \frac{-99 \pm 96.4}{2} \)
\( x = \left( \frac{-2.6}{2}, \frac{-195.4}{2} \right) \)
\( x = (-1.3, –97.7) \)
Hence both the zeroes are negative.

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) cannot both be positive
Explanation: Let \( p(x) = x^2 + kx + k \) where \( k \neq 0 \).
On comparing p(x) with \( ax^2 + bx + c \), we get
\( a = 1, b = k \) and \( c = k \).
Let \( \alpha \) and \( \beta \) be the zeroes of the polynomial p(x).
We know that:
Sum of the zeroes
\( \alpha + \beta = -\frac{b}{a} \)
\( \Rightarrow \alpha + \beta = -\frac{k}{1} = -k \) ...(i)
And product of the zeroes
\( \alpha\beta = \frac{c}{a} \)
\( \Rightarrow \alpha\beta = \frac{k}{1} = k \) ...(ii)
Case 1: k is negative
If k is negative,
\( \alpha\beta \) [from equation (ii)] is negative.
It means \( \alpha \) and \( \beta \) are of the opposite sign.
\( \Rightarrow \) Both the zeroes are of the opposite signs.
Case 2: k is positive
If k is positive,
\( \alpha\beta \) (from equation (ii)) is positive but \( \alpha + \beta \) is negative.
If the product of the two numbers is positive, then either both are negative or both are positive. But the sum of these numbers is negative, so the numbers must be negative.
\( \Rightarrow \) Both the zeroes are negative.
Hence, in both the cases, both the zeroes cannot be positive.
Alternate Method:
Let \( p(x) = x^2 + kx + k \)
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( x = \frac{-k \pm \sqrt{k^2 - 4k}}{2} \)
\( x = \frac{-k \pm \sqrt{k(k - 4)}}{2} \)
\( \Rightarrow k(k - 4) > 0 \)
\( \Rightarrow k \in (-\infty, 0) \cup (4, \infty) \)
Here, k lies in two intervals; therefore, we need to consider both the intervals separately.
Case 1:
When \( k \in (-\infty, 0) \)
i.e., \( k < 0 \)
We know that in a quadratic equation \( p(x) = ax^2 + bx + c \), if either \( a > 0, c < 0 \) or \( a < 0, c > 0 \), then the zeroes of the polynomial are of the opposite signs.
Here, \( a = 1 > 0, b = k < 0 \) and \( c = k < 0 \).
\( \Rightarrow \) Both the zeroes are of the opposite signs.
Case 2:
When \( k \in (4, \infty) \)
i.e., \( k > 0 \)
We know, in a quadratic polynomial, if the coefficient of the terms are of the same sign, then the zeroes of the polynomial are negative.
i.e., if either \( a > 0, b > 0 \) and \( c > 0 \) or \( a < 0, b < 0 \) and \( c < 0 \), then both the zeroes are negative.
Here, \( a = 1 > 0, b = k > 0 \) and \( c = k > 0 \).
\( \Rightarrow \) Both the zeroes are negative.
Hence, in both cases, both the zeroes cannot 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) c and a have the same sign
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., \( D = 0 \)
\( b^2 – 4ac = 0 \)
\( \Rightarrow 4ac = b^2 \)
\( \Rightarrow ac = \frac{b^2}{4} > 0 \) [square of any positive or negative number is positive]
\( \Rightarrow ac > 0 \)
Therefore, for \( ac > 0 \), a and c must have the same sign
i.e., either \( a > 0 \) and \( c > 0 \) or \( a < 0 \) and \( c < 0 \).
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) has no linear term and the constant term is negative.
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).
Product of the zeroes = constant term ÷ coefficient of \( x^2 \)
Product of the zeroes = \( \frac{b}{1} \)
\( \alpha(-\alpha) = b \)
\( -\alpha^2 = b \) i.e., \( b < 0 \)
i.e., the constant term is negative.
Sum of the zeroes = – (coefficient of x) ÷ coefficient of \( x^2 \)
\( \alpha – \alpha = – \frac{a}{1} \)
\( 0 = – a \)
\( \Rightarrow a = 0 \)
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, \( x^2 – 2x – 3 \)
\( \Rightarrow x^2 – 3x + x – 3 \)
\( \Rightarrow x(x – 3) + 1(x – 3) \)
\( \Rightarrow (x – 3) (x + 1) \)
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. 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, \( 5x \), \( 2xy - 2 \).