Polynomials: 1 mark Questions (CBSE Class 10)
104 different 1 mark questions on Polynomials from CBSE Class 10 Maths board exams 2022–2026, newest first.
The graph of polynomial p(x ) = k is shown here. Number of zeroes of polynomial p(x ) is :
(A) 0(B) 1(C) 2(D) infinitely many
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Answer: (A) 0
The zeroes are the x-coordinates where the graph meets the x-axis. The graph of p(x ) = k is a horizontal line above the x-axis, so it never meets the x-axis. Number of zeroes = 0.
The graph of a polynomial p ( x ) is shown here. The number of zeroes of the polynomial p ( x ) is
(A) 5(B) 1(C) 0(D) 4
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Answer: (D) 4
The number of zeroes equals the number of points where the graph meets the x-axis. The graph cuts the x-axis at 4 points, so p(x) has 4 zeroes.
The value of k for which sum of the zeroes of the polynomial p ( x ) = 3 x 2 − k x + 6 is 2, is
(A) 2(B) − 6 (C) − 2 (D) 6
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Answer: (D) 6
Sum of zeroes = − a b = 3 k 3 k = 2 , so k = 6
If the zeroes of the polynomial p ( x ) = 2 x 2 − 7 x + 6 are α and β , then the value of α 1 + β 1 is
(A) 2 7 (B) 7 6 (C) 6 − 7 (D) 6 7
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Answer: (D) 6 7
α + β = 2 7 , α β = 2 6 = 3 α 1 + β 1 = α β α + β = 3 7/2 = 6 7
The graph of y = f ( x ) is given. The number of zeroes of f ( x ) is :
(A) 0(B) 1(C) 2(D) 4
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Answer: (A) 0
The zeroes of f ( x ) are the x -coordinates of the points where the graph meets the x -axis. The given curve lies wholly above the x -axis and never meets it. So f ( x ) has 0 zeroes.
Assertion (A) : The polynomial p ( y ) = y 2 + 4 y + 3 has two zeroes. Reason (R) : A quadratic polynomial can have at most two zeroes.
(A) Both, Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).(B) Both, Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).(C) Assertion (A) is true, but Reason (R) is false.(D) Assertion (A) is false, but Reason (R) is true.
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Answer: (B) Both, Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
y 2 + 4 y + 3 = ( y + 1 ) ( y + 3 ) , so the zeroes are − 1 and − 3 . A is true.A quadratic polynomial has at most two zeroes, so R is true. R only gives an upper limit; it does not show that this polynomial actually has two zeroes (that comes from the factorisation). So R does not explain A.
The graph of y = f ( x ) is given. The number of distinct zeroes of y = f ( x ) is :
(A) 0(B) 1(C) 2(D) 3
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Answer: (C) 2
The zeroes are the x-coordinates of the points where the graph meets the x-axis. The graph cuts the x-axis at A and touches it at one more point on the right of O. So there are 2 distinct zeroes.
If α and β are two zeroes of a polynomial f ( x ) = p x 2 − 2 x + 3 p and α + β = α β , then value of p is :
(A) − 3 2 (B) 3 2 (C) 3 1 (D) − 3 1
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Answer: (B) 3 2
α + β = p 2 and α β = p 3 p = 3 .p 2 = 3 p = 3 2
How many zeroes does p(x) = (x - 2) (x + 3) have ?
(A) Zero(B) One(C) Two(D) Three
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Answer: (C) Two
p(x) = 0 when x = 2 or x = -3. So p(x) has two zeroes.
The sum and product of zeroes of a quadratic polynomial p(x) are 3 − 1 and 2 respectively. The polynomial p(x) is :
(A) 3 x 2 − x + 6 (B) x 2 + 3 1 x − 2 (C) 3 x 2 − x + 2 (D) − 3 x 2 − x − 6
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Answer: (D) − 3 x 2 − x − 6
p ( x ) = k [ x 2 − ( sum ) x + product ] = k [ x 2 + 3 1 x + 2 ] With k = − 3 : p ( x ) = − 3 x 2 − x − 6 Check: sum = − − 3 − 1 = − 3 1 , product = − 3 − 6 = 2
A polynomial p(x), which has sum of its zeroes equal to their product, is :
(A) 3 x 2 + 2 x + 2 (B) 3 x 2 − 2 x − 3 (C) 3 x 2 − 2 x + 2 (D) x 2 − 3 x + 2
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Answer: (C) 3 x 2 − 2 x + 2
For a x 2 + b x + c : sum = − a b , product = a c ; equal when − b = c (A) sum − 3 2 , product 3 2 ; (B) sum 3 2 , product − 1 ; (D) sum 3, product 2 (C) sum = 3 2 , product = 3 2 : equal
Observe the graph of polynomial p ( x ) . Number of zeroes of p ( x ) is
(A) 5(B) 4(C) 6(D) 3
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Answer: (D) 3
The number of zeroes equals the number of points where the graph meets the x-axis. The curve cuts the x-axis at 3 points, so p ( x ) has 3 zeroes.
Observe the graph of polynomial p ( x ) . The zeroes of the polynomial are
(A) − 2 , 0, 2.5(B) − 2 , 2.5(C) 0, 4(D) − 2 , 0
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Answer: (B) − 2 , 2.5
Zeroes are the x-coordinates of the points where the graph meets the x-axis. The graph touches the x-axis at ( − 2 , 0 ) and cuts it at (2.5, 0). (0, 4) is the y-intercept, not a zero. Zeroes: − 2 , 2.5.
If the zeroes of a polynomial p ( x ) are − 3 and 8, then p ( x ) equals
(A) x 2 + 5 x − 4 (B) ( x + 3 ) ( − x + 8 ) (C) a ( x 2 + 5 x − 24 ) (D) x 2 − 24
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Answer: (B) ( x + 3 ) ( − x + 8 )
A polynomial with zeroes − 3 and 8 is k ( x + 3 ) ( x − 8 ) for some non-zero constant k. With k = − 1 : − ( x + 3 ) ( x − 8 ) = ( x + 3 ) ( − x + 8 ) , whose zeroes are − 3 and 8. Option (C) a ( x 2 + 5 x − 24 ) = a ( x + 8 ) ( x − 3 ) has zeroes − 8 and 3, so it is wrong; (A) and (D) do not have these zeroes either.
If sum and product of zeroes of a polynomial are ( − 3 ) and ( − 2 ) respectively, then a polynomial is
(A) x 2 − 3 x − 2 (B) − x 2 − 3 x + 2 (C) − x 2 + 3 x − 2 (D) x 2 + 3 x + 2
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Answer: (B) − x 2 − 3 x + 2
Required polynomial = k [ x 2 − ( sum ) x + product ] = k ( x 2 + 3 x − 2 ) With k = − 1 : − x 2 − 3 x + 2 Check: sum = − a b = − − 1 − 3 = − 3 , product = a c = − 1 2 = − 2 .
One of the zeroes of the polynomial p ( x ) = k x 2 − 9 x + 3 is ( − 2 3 ) . The value of k is :
(A) 3 22 (B) − 3 14 (C) 3 14 (D) − 3 22
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Answer: (D) − 3 22
p ( − 2 3 ) = 0 k × 4 9 + 2 27 + 3 = 0 4 9 k = − 2 33 k = − 2 33 × 9 4 = − 3 22 .
α , β are zeroes of the polynomial 2 x 2 + 5 x + 1 . The value of ( α 1 + β 1 ) is :
(A) − 4 5 (B) 5(C) 4 5 (D) − 5
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Answer: (D) − 5
α + β = − 2 5 , α β = 2 1 .α 1 + β 1 = α β α + β = 1/2 − 5/2 = − 5 .
In the given figure, graph of p ( x ) is shown. Number of distinct zeroes of p ( x ) is :
(A) 0(B) 1(C) 2(D) many
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Answer: (B) 1
The zeroes are the x -coordinates where the graph meets the x -axis. The graph only touches the x -axis at the origin O, so p ( x ) has 1 distinct zero.
If α , β are zeroes of the polynomial 3 x 2 + 14 x − 5 , then the value of 3 ( α β α + β ) is :
(A) 5 14 (B) 5 42 (C) − 5 14 (D) − 5 42
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Answer: (B) 5 42
α + β = − 3 14 , α β = − 3 5 .α β α + β = − 5/3 − 14/3 = 5 14 .3 × 5 14 = 5 42 .
Observe the given graph of polynomial p(x ). The number of zeroes of p(x ) is
(A) 0(B) 1(C) 3(D) 2
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Answer: (D) 2
The number of zeroes equals the number of points where the graph meets the x -axis. The curve cuts the x -axis at two points, so p(x ) has 2 zeroes.
A quadratic polynomial having only zero (–2) is
(A) ( x − 2 ) 2 (B) x 2 − 2 (C) x 2 + 2 x (D) ( x + 2 ) 2
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Answer: (D) ( x + 2 ) 2
A quadratic with − 2 as its only zero must be k ( x + 2 ) 2 . ( x + 2 ) 2 = 0 gives x = − 2 only.
In the given figure, graph of polynomial p ( x ) is shown. Number of zeroes of p ( x ) is
(A) 3(B) 2(C) 1(D) 4
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Answer: (A) 3
The number of zeroes equals the number of points where the graph meets the x -axis. The graph meets the x -axis at O and at two more points, i.e. 3 points. So p ( x ) has 3 zeroes.
A quadratic polynomial having zeroes 0 and − 2, is
(A) x ( x − 2 ) (B) 4 x ( x + 2 ) (C) x 2 + 2 (D) 2 x 2 + 2 x
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Answer: (B) 4 x ( x + 2 )
A quadratic with zeroes 0 and − 2 is k ⋅ x ( x + 2 ) , k = 0 . 4 x ( x + 2 ) is of this form (k = 4).Check others: x ( x − 2 ) has zeroes 0, 2; x 2 + 2 has no real zeroes; 2 x 2 + 2 x = 2 x ( x + 1 ) has zeroes 0, − 1 .
In the given graph, the polynomial p ( x ) is shown. Number of zeroes of p ( x ) is
(A) 3(B) 2(C) 1(D) 4
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Answer: (B) 2
The graph (a downward parabola) cuts the x -axis at two points, one on each side of O. So p ( x ) has 2 zeroes.
If α and β are the zeroes of polynomial 3 x 2 + 6 x + k such that α + β + α β = − 3 2 , then the value of k is :
(A) − 8 (B) 8 (C) − 4 (D) 4
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Answer: (D) 4
α + β = − 3 6 = − 2 and α β = 3 k .− 2 + 3 k = − 3 2 3 k = 3 4 , so k = 4 .
Two polynomials are shown in the graph below. The number of distinct zeroes of both the polynomials is :
(A) 3 (B) 5 (C) 2 (D) 4
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Answer: (C) 2
Zeroes are the x-coordinates of the points where a graph meets the x-axis. Both parabolas cut the x-axis at the same two points. So the two polynomials together have 2 distinct zeroes.
The sum of the zeroes of the polynomial p ( x ) = 5 x − 7 x 2 + 3 is :
(A) 5 − 7 (B) 5 7 (C) 7 5 (D) 7 − 5
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Answer: (C) 7 5
Write p ( x ) = − 7 x 2 + 5 x + 3 , so a = − 7 , b = 5 . Sum of zeroes = − a b = − − 7 5 = 7 5 .
If − 4 is a zero of the polynomial p ( x ) = x 2 − x − ( 2 + 2 k ) , then the value of k is :
(A) 3(B) 9(C) 6(D) − 9
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Answer: (B) 9
p ( − 4 ) = 16 + 4 − ( 2 + 2 k ) = 18 − 2 k 18 − 2 k = 0 , so k = 9
If one zero of the polynomial q ( x ) = ( p 2 + 4 ) x 2 + 65 x + 4 p is reciprocal of the other, then the value of ‘p ’ is :
(A) − 1 (B) 1(C) − 2 (D) 2
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Answer: (D) 2
If the zeroes are α and α 1 , their product is 1 Product of zeroes = p 2 + 4 4 p = 1 p 2 − 4 p + 4 = 0 , i.e. ( p − 2 ) 2 = 0 , so p = 2
If α and β are the zeroes of the polynomial p ( x ) = x 2 − a x − b , then the value of ( α + β + α β ) is equal to :
(A) a + b (B) − a − b (C) a − b (D) − a + b
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Answer: (C) a − b
α + β = − 1 − a = a .α β = 1 − b = − b .α + β + α β = a − b .
Zeroes of the polynomial p ( x ) = x 2 − 3 2 x + 4 are :
(A) 2 , 2 (B) 2 2 , 2 (C) 4 2 , − 2 (D) 2 , 2
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x 2 − 3 2 x + 4 = x 2 − 2 2 x − 2 x + 4 .= x ( x − 2 2 ) − 2 ( x − 2 2 ) = ( x − 2 2 ) ( x − 2 ) .Zeroes: 2 2 and 2 .
Zeroes of the polynomial p ( y ) = 7 y 2 − 3 11 y − 3 2 are :
(A) − 3 2 , − 7 1 (B) − 7 2 , − 3 1 (C) 3 2 , 7 1 (D) 3 2 , − 7 1
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Answer: (D) 3 2 , − 7 1
p ( y ) = 3 1 ( 21 y 2 − 11 y − 2 ) .21 y 2 − 11 y − 2 = 21 y 2 − 14 y + 3 y − 2 = 7 y ( 3 y − 2 ) + 1 ( 3 y − 2 ) = ( 3 y − 2 ) ( 7 y + 1 ) .Zeroes: y = 3 2 and y = − 7 1 .
If α and β are zeroes of the polynomial p ( x ) = k x 2 − 30 x + 45 k and α + β = α β , then the value of 'k' is :
(A) − 3 2 (B) − 2 3 (C) 2 3 (D) 3 2
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Answer: (D) 3 2
α + β = k 30 , α β = k 45 k = 45 .k 30 = 45 , so k = 45 30 = 3 2 .
Which of the following statements is true for a polynomial p ( x ) of degree 3?
(A) p ( x ) has at most two distinct zeroes.(B) p ( x ) has at least two distinct zeroes.(C) p ( x ) has exactly three distinct zeroes.(D) p ( x ) has at most three distinct zeroes.
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Answer: (D) p ( x ) has at most three distinct zeroes.
A polynomial of degree n has at most n zeroes. A cubic can have 1, 2 or 3 distinct real zeroes (e.g. x 3 has only one), so it has at most three.
If the zeroes of the polynomial a x 2 + b x + b 2 a are reciprocal of each other, then the value of b is
(A) 2(B) 2 1 (C) − 2 (D) − 2 1
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Answer: (A) 2
If the zeroes are α and α 1 , their product is 1. Product of zeroes = a 2 a / b = b 2 b 2 = 1 ⇒ b = 2
If one of the zeroes of the quadratic polynomial ( α − 1 ) x 2 + α x + 1 is − 3 , then the value of α is :
(A) − 3 2 (B) 3 2 (C) 3 4 (D) 4 3
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Answer: (C) 3 4
Put x = − 3 : 9 ( α − 1 ) − 3 α + 1 = 0 . 6 α − 8 = 0 , so α = 3 4 .
For what value of k , the product of zeroes of the polynomial k x 2 − 4 x − 7 is 2 ?
(A) − 14 1 (B) − 2 7 (C) 2 7 (D) − 7 2
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Answer: (B) − 2 7
Product of zeroes = a c = k − 7 . k − 7 = 2 , so k = − 2 7 .
Assertion (A) : Zeroes of a polynomial p ( x ) = x 2 − 2 x − 3 are − 1 and 3. Reason (R) : The graph of polynomial p ( x ) = x 2 − 2 x − 3 intersects x -axis at ( − 1 , 0 ) and ( 3 , 0 ) .
(A) Both, Assertion (A) and Reason (R) are true. Reason (R) explains Assertion (A) completely.(B) Both, Assertion (A) and Reason (R) are true. Reason (R) does not explain Assertion (A).(C) Assertion (A) is true but Reason (R) is false.(D) Assertion (A) is false but Reason (R) is true.
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Answer: (A) Both, Assertion (A) and Reason (R) are true. Reason (R) explains Assertion (A) completely.
x 2 − 2 x − 3 = ( x − 3 ) ( x + 1 ) , so the zeroes are − 1 and 3. A is true.p ( − 1 ) = 0 and p ( 3 ) = 0 , so the graph meets the x-axis at ( − 1 , 0 ) and ( 3 , 0 ) . R is true.The zeroes of a polynomial are exactly the x-coordinates of the points where its graph meets the x-axis, so R explains A.
A quadratic polynomial, the sum of whose zeroes is − 5 and their product is 6, is
(A) x 2 + 5 x + 6 (B) x 2 − 5 x + 6 (C) x 2 − 5 x − 6 (D) − x 2 + 5 x + 6
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Answer: (A) x 2 + 5 x + 6
Required polynomial = x 2 − ( sum ) x + product = x 2 − ( − 5 ) x + 6 = x 2 + 5 x + 6
The zeroes of the polynomial 3 x 2 + 11 x − 4 are :
(A) 3 1 , 4 (B) 3 − 1 , − 4 (C) 3 1 , − 4 (D) 3 − 1 , 4
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Answer: (C) 3 1 , − 4
3 x 2 + 11 x − 4 = 3 x 2 + 12 x − x − 4 = ( 3 x − 1 ) ( x + 4 ) Zeroes: x = 3 1 and x = − 4
The zeroes of the polynomial 3 x 2 − 5 x − 2 , are :
(A) 3 1 , 2 (B) 3 − 1 , 2 (C) 3 − 1 , − 2 (D) 3 1 , − 2
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Answer: (B) 3 − 1 , 2
3 x 2 − 5 x − 2 = 3 x 2 − 6 x + x − 2 = ( 3 x + 1 ) ( x − 2 ) Zeroes: − 3 1 and 2
The zeroes of the polynomial 3 x 2 + 8 x − 3 are :
(A) 3 1 , 3 (B) 3 1 , − 3 (C) 3 − 1 , 3 (D) 3 − 1 , − 3
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Answer: (B) 3 1 , − 3
3 x 2 + 8 x − 3 = 3 x 2 + 9 x − x − 3 = ( 3 x − 1 ) ( x + 3 ) Zeroes: 3 1 and − 3
A quadratic polynomial whose zeroes are 3 and − 2 , is :
(A) x 2 − x − 6 (B) x 2 + x − 6 (C) 2 x 2 − x − 12 (D) x 2 + x + 6
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Answer: (A) x 2 − x − 6
Sum of zeroes = 3 + ( − 2 ) = 1 , product = 3 × ( − 2 ) = − 6 . Polynomial = x 2 − ( sum ) x + product = x 2 − x − 6 .
In the given figure, graph of a polynomial f ( x ) is shown. The number of zeroes of polynomial f ( x ) is :
(A) 3(B) 1(C) 0(D) 2
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Answer: (D) 2
The zeroes are the x-coordinates where the graph meets the x-axis. The curve cuts the x-axis at the origin and at one point on the positive x-axis, so there are 2 zeroes.
A quadratic polynomial whose zeroes are − 8 and 3, is
(A) ( x + 8 ) ( x + 3 ) (B) x 2 + 5 x + 24 (C) ( x − 8 ) ( x − 3 ) (D) x 2 + 5 x − 24
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Answer: (D) x 2 + 5 x − 24
Sum of zeroes = − 8 + 3 = − 5 , product = − 24 . Polynomial = x 2 − ( − 5 ) x + ( − 24 ) = x 2 + 5 x − 24 .
If the two zeroes of a quadratic polynomial are ± 5 , then the quadratic polynomial is :
(A) x 2 + 5 (B) ( x + 5 ) 2 (C) 4 ( x 2 − 5 ) (D) x 2 − 5
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Answer: (C) 4 ( x 2 − 5 )
Sum of zeroes = 0 , product = − 5 , so the polynomial is k ( x 2 − 5 ) for any non-zero constant k . With k = 4 this is 4 ( x 2 − 5 ) , option (C). The other options do not have zeroes ± 5 .
If one zero of a quadratic polynomial k x 2 + 4 x + k is 1, then the value of k is :
(A) 2 (B) − 2 (C) 4 (D) − 4
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Answer: (B) − 2
Put x = 1 : k ( 1 ) 2 + 4 ( 1 ) + k = 0 . 2 k + 4 = 0 ⇒ k = − 2 .
The number of quadratic polynomials having zeroes − 1 and 3 is :
(A) 1(B) 2(C) 3(D) more than 3
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Answer: (D) more than 3
Every polynomial k ( x + 1 ) ( x − 3 ) = k ( x 2 − 2 x − 3 ) with k = 0 has zeroes − 1 and 3. Since k can be any non-zero real number, there are infinitely many such polynomials, i.e. more than 3.
What should be added to the polynomial x 2 − 5 x + 4 , so that 3 is a zero of the resulting polynomial ?
(A) 1(B) 2(C) 4(D) 5
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Answer: (B) 2
Value at x = 3 : 9 − 15 + 4 = − 2 . Adding k gives − 2 + k = 0 , so k = 2 .
If the sum of zeroes of the polynomial p ( x ) = 2 x 2 − k 2 x + 1 is 2 , then value of k is :
(A) 2 (B) 2(C) 2 2 (D) 2 1
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Answer: (B) 2
Sum of zeroes = − a b = 2 k 2 . 2 k 2 = 2 , so k = 2 .
The zeroes of a polynomial x 2 + p x + q are twice the zeroes of the polynomial 4 x 2 − 5 x − 6 . The value of p is :
(A) − 2 5 (B) 2 5 (C) − 5 (D) 10
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Answer: (A) − 2 5
Sum of zeroes of 4 x 2 − 5 x − 6 is 4 5 . Zeroes of x 2 + p x + q are twice these, so their sum is 2 × 4 5 = 2 5 . Sum of zeroes of x 2 + p x + q is − p , so − p = 2 5 and p = − 2 5 .
Assertion (A) : If the graph of a polynomial touches x -axis at only one point, then the polynomial cannot be a quadratic polynomial. Reason (R) : A polynomial of degree n ( n > 1 ) can have at most n zeroes.
(A) Both, Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of Assertion (A).(B) Both, Assertion (A) and Reason (R) are true but Reason (R) is not correct explanation for Assertion (A).(C) Assertion (A) is true but Reason (R) is false.(D) Assertion (A) is false but Reason (R) is true.
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Answer: (D) Assertion (A) is false but Reason (R) is true.
A: The graph of x 2 touches the x -axis only at the origin, and x 2 is quadratic. So A is false. R: A polynomial of degree n has at most n zeroes. R is true.
If the sum and the product of zeroes of a quadratic polynomial are 2 3 and 3 respectively, then a quadratic polynomial is :
(A) x 2 + 2 3 x − 3 (B) ( x − 3 ) 2 (C) x 2 − 2 3 x − 3 (D) x 2 + 2 3 x + 3
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Required polynomial = x 2 − ( sum ) x + product = x 2 − 2 3 x + 3 . x 2 − 2 3 x + 3 = ( x − 3 ) 2 .
If α , β are the zeroes of the polynomial 6 x 2 − 5 x − 4 , then α 1 + β 1 is equal to :
(A) 4 5 (B) − 4 5 (C) 5 4 (D) 24 5
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Answer: (B) − 4 5
α + β = 6 5 , α β = − 6 4 .α 1 + β 1 = α β α + β = − 4/6 5/6 = − 4 5 .
If α and β are zeroes of the polynomial 5 x 2 + 3 x − 7 , the value of α 1 + β 1 is
(A) − 7 3 (B) 5 3 (C) 7 3 (D) − 7 5
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Answer: (C) 7 3
α + β = − 5 3 , α β = − 5 7 α 1 + β 1 = α β α + β = − 7/5 − 3/5 = 7 3
If α and β are zeroes of the polynomial 2 x 2 − 9 x + 5 , then value of α 2 + β 2 is
(A) 4 1 (B) 4 61 (C) 1(D) 4 71
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Answer: (B) 4 61
α + β = 2 9 , α β = 2 5 α 2 + β 2 = ( α + β ) 2 − 2 α β = 4 81 − 5 = 4 61
If α and β ( α > β ) are the zeroes of the polynomial − x 2 + 8 x + 9 , then ( α − β ) is equal to
(A) − 10 (B) 10(C) ± 10 (D) 8
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Answer: (B) 10
− x 2 + 8 x + 9 = − ( x 2 − 8 x − 9 ) = − ( x − 9 ) ( x + 1 ) Zeroes: 9 and − 1 . As α > β , α = 9 , β = − 1 . α − β = 9 − ( − 1 ) = 10
The zeroes of the quadratic polynomial 2 x 2 − 3 x − 9 are :
(A) 3 , 2 − 3 (B) − 3 , 2 − 3 (C) − 3 , 2 3 (D) 3 , 2 3
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Answer: (A) 3 , 2 − 3
2 x 2 − 3 x − 9 = 2 x 2 − 6 x + 3 x − 9 = 2 x ( x − 3 ) + 3 ( x − 3 ) = ( x − 3 ) ( 2 x + 3 ) .Zeroes: x = 3 and x = 2 − 3 .
What should be subtracted from the polynomial x 2 − 16 x + 30 , so that 15 is the zero of the resulting polynomial ?
(A) 30(B) 14(C) 15(D) 16
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Answer: (C) 15
Let k be subtracted: p ( x ) = x 2 − 16 x + 30 − k . 15 is a zero: 225 − 240 + 30 − k = 0 ⇒ 15 − k = 0 . k = 15 .
What should be added from the polynomial x 2 − 5 x + 4 , so that 3 is the zero of the resulting polynomial ?
(A) 1(B) 2(C) 4(D) 5
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Answer: (B) 2
Let k be added: p ( x ) = x 2 − 5 x + 4 + k . 3 is a zero: 9 − 15 + 4 + k = 0 ⇒ − 2 + k = 0 . k = 2 .
If a polynomial p ( x ) is given by p ( x ) = x 2 − 5 x + 6 , then the value of p ( 1 ) + p ( 4 ) is :
(A) 0(B) 4(C) 2(D) − 4
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Answer: (B) 4
p ( 1 ) = 1 − 5 + 6 = 2 .p ( 4 ) = 16 − 20 + 6 = 2 .p ( 1 ) + p ( 4 ) = 4 .
A quadratic polynomial, one of whose zeroes is 2 + 5 and the sum of whose zeroes is 4, is :
(A) x 2 + 4 x − 1 (B) x 2 − 4 x − 1 (C) x 2 − 4 x + 1 (D) x 2 + 4 x + 1
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Answer: (B) x 2 − 4 x − 1
Other zero = 4 − ( 2 + 5 ) = 2 − 5 . Product = ( 2 + 5 ) ( 2 − 5 ) = 4 − 5 = − 1 . Polynomial = x 2 − ( sum ) x + product = x 2 − 4 x − 1 .
The ratio of the sum and product of the roots of the quadratic equation 5 x 2 − 6 x + 21 = 0 is :
(A) 5 : 21(B) 2 : 7(C) 21 : 5(D) 7 : 2
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Answer: (B) 2 : 7
Sum of roots = 5 6 , product of roots = 5 21 . Ratio = 6 : 21 = 2 : 7 .
If α and β are the zeroes of the polynomial p ( x ) = k x 2 − 30 x + 45 k and α + β = α β , then the value of k is :
(A) − 3 2 (B) − 2 3 (C) 2 3 (D) 3 2
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Answer: (D) 3 2
α + β = k 30 and α β = k 45 k = 45 .k 30 = 45 ⇒ k = 3 2 .
Assertion (A) : Degree of a zero polynomial is not defined. Reason (R) : Degree of a non-zero constant polynomial is 0.
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).(C) Assertion (A) is true, but Reason (R) is false.(D) Assertion (A) is false, but Reason (R) is true.
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Answer: (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
A is true: the zero polynomial has no non-zero term, so its degree is not defined. R is true: a non-zero constant c = c x 0 has degree 0. R is about non-zero constants, so it does not explain A.
The graph of y = p ( x ) is shown in the figure for some polynomial p ( x ) . The number of zeroes of p ( x ) is/are :
(A) 0(B) 1(C) 2(D) 3
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Answer: (A) 0
The number of zeroes is the number of points where the graph meets the x-axis. The graph lies entirely above the x-axis and never meets it. So p ( x ) has 0 zeroes.
The sum and the product of zeroes of the polynomial p ( x ) = x 2 + 5 x + 6 are respectively
(A) 5 , − 6 (B) − 5 , 6 (C) 2, 3(D) − 2 , − 3
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Answer: (B) − 5 , 6
For a x 2 + b x + c : sum of zeroes = − a b , product = a c . Sum = − 1 5 = − 5 , product = 1 6 = 6 .
The zeroes of the polynomial p ( x ) = x 2 + 3 x + 2 are given as.
(A) 1, 2(B) 2 , − 1 (C) − 2 , 1 (D) − 2 , − 1
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Answer: (D) − 2 , − 1
x 2 + 3 x + 2 = ( x + 1 ) ( x + 2 ) Zeroes: x = − 1 and x = − 2 .
The graph of y = f ( x ) is shown in the figure for some polynomial f ( x ) . The number of zeroes of f ( x ) is
(A) 5(B) 6(C) 4(D) 8
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Answer: (A) 5
The number of zeroes equals the number of points where the graph meets the x-axis. The wave-shaped graph crosses the x-axis at 5 points. So f ( x ) has 5 zeroes.
The sum and product of zeroes of the polynomial p ( x ) = 3 x 2 − 5 x + 2 are
(A) 3 5 , 3 2 (B) 3 − 5 , 3 2 (C) 1 , 3 2 (D) 3 − 5 , 3 − 2
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Answer: (A) 3 5 , 3 2
For a x 2 + b x + c : sum of zeroes = − a b , product = a c . Here a = 3 , b = − 5 , c = 2 . Sum = 3 5 , product = 3 2 .
Graph of a polynomial p(x) is given in the figure. The number of zeroes of p(x) is :
(A) 2(B) 3(C) 4(D) 5
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Answer: (a) 2
The number of zeroes equals the number of points where the graph meets the x-axis. The graph cuts the x-axis at two points (the dip between the two humps stays above the axis). Number of zeroes = 2
Assertion (A) : Polynomial x 2 + 4 x has two real zeroes. Reason (R) : Zeroes of the polynomial x 2 + a x ( a = 0 ) are 0 and a.
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).(C) Assertion (A) is true, but Reason (R) is false.(D) Assertion (A) is false, but Reason (R) is true.
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Answer: (c) Assertion (A) is true, but Reason (R) is false.
x 2 + 4 x = x ( x + 4 ) , zeroes 0 and − 4 : two real zeroes, so A is true.x 2 + a x = x ( x + a ) , zeroes are 0 and − a , not 0 and a, so R is false.Hence (c).
A quadratic polynomial the sum and product of whose zeroes are − 3 and 2 respectively, is :
(A) x 2 + 3 x + 2 (B) x 2 − 3 x + 2 (C) x 2 − 3 x − 2 (D) x 2 + 3 x − 2
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Answer: (A) x 2 + 3 x + 2
Required polynomial = x 2 − ( sum ) x + product . = x 2 − ( − 3 ) x + 2 = x 2 + 3 x + 2 .
If p ( x ) = x 2 + 5 x + 6 , then p ( − 2 ) is :
(A) 20(B) 0(C) − 8 (D) 8
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Answer: (B) 0
p ( − 2 ) = ( − 2 ) 2 + 5 ( − 2 ) + 6 = 4 − 10 + 6 = 0 .
A quadratic polynomial whose sum and product of zeroes are 2 and − 1 respectively is :
(A) x 2 + 2 x + 1 (B) x 2 − 2 x − 1 (C) x 2 + 2 x − 1 (D) x 2 − 2 x + 1
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Answer: (B) x 2 − 2 x − 1
Required polynomial = x 2 − ( sum ) x + product . = x 2 − 2 x + ( − 1 ) = x 2 − 2 x − 1 .
The number of polynomials having zeroes − 3 and 4 is :
(A) 1(B) 2(C) 3(D) more than 3
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Answer: (D) more than 3
Every polynomial k ( x + 3 ) ( x − 4 ) with k = 0 has zeroes − 3 and 4. So there are infinitely many such polynomials, i.e. more than 3.
If one zero of the quadratic polynomial k x 2 + 3 x + k is 2, then the value of k is :
(A) − 5 6 (B) 5 6 (C) 6 5 (D) − 6 5
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Answer: (A) − 5 6
Put x = 2 : 4 k + 6 + k = 0 . 5 k = − 6 , so k = − 5 6 .
The graph of y = f ( x ) is shown in the figure for some polynomial f ( x ) . The number of zeroes of f ( x ) is
(A) 0(B) 2(C) 3(D) 4
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Answer: (A) 0
The zeroes of f ( x ) are the x-coordinates of the points where the graph meets the x-axis. The graph lies entirely above the x-axis and never meets it, so there are no zeroes.
The zeroes of the quadratic polynomial 16 x 2 − 9 are :
(A) 4 3 , 4 3 (B) − 4 3 , 4 3 (C) 16 9 , 16 9 (D) − 4 3 , − 4 3
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Answer: (B) − 4 3 , 4 3
16 x 2 − 9 = ( 4 x − 3 ) ( 4 x + 3 ) .Zeroes: x = 4 3 and x = − 4 3 .
The zeroes of the polynomial p ( x ) = 25 x 2 − 49 are :
(A) 25 49 , 25 49 (B) − 25 49 , + 25 49 (C) 5 7 , − 5 7 (D) 5 7 , 5 7
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Answer: (C) 5 7 , − 5 7
25 x 2 − 49 = ( 5 x − 7 ) ( 5 x + 7 ) .Zeroes: x = 5 7 and x = − 5 7 .
The zeroes of the polynomial p ( x ) = 2 x 2 − x − 3 are
(A) − 2 3 , 1 (B) 2 3 , 1 (C) − 2 3 , − 1 (D) 2 3 , − 1
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Answer: (D) 2 3 , − 1
2 x 2 − x − 3 = 2 x 2 − 3 x + 2 x − 3 = ( 2 x − 3 ) ( x + 1 ) .Zeroes: x = 2 3 and x = − 1 .
The graph of y = f ( x ) is shown in the figure for some polynomial f ( x ) . The number of zeroes of f ( x ) are
(A) 4(B) 3(C) 2(D) 1
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Answer: (A) 4
The zeroes are the x-coordinates of the points where the graph meets the x-axis. The graph crosses the x-axis at 4 points (one of them the origin), so f ( x ) has 4 zeroes.
The graph of y = p ( x ) is given, for a polynomial p ( x ) . The number of zeroes of p ( x ) from the graph is
(A) 3(B) 1(C) 2(D) 0
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Answer: (B) 1
The zeroes of p ( x ) are the x-coordinates of the points where the graph meets the x-axis. The graph touches the x-axis at exactly one point. So p ( x ) has 1 zero.
If α , β are the zeroes of a polynomial p ( x ) = x 2 + x − 1 , then α 1 + β 1 equals to
(A) 1(B) 2(C) –1(D) 2 − 1
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Answer: (A) 1
α + β = − 1 and α β = − 1 .α 1 + β 1 = α β α + β = − 1 − 1 = 1
Which of the following is a quadratic polynomial having zeroes 3 − 2 and 3 2 ?
(A) 4 x 2 − 9 (B) 9 4 ( 9 x 2 + 4 ) (C) x 2 + 4 9 (D) 5 ( 9 x 2 − 4 )
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Answer: (D) 5 ( 9 x 2 − 4 )
Sum of zeroes = 0, product = − 9 4 . Polynomial: k ( x 2 − 9 4 ) , i.e. a multiple of 9 x 2 − 4 . 5 ( 9 x 2 − 4 ) has zeroes ± 3 2 .
Which of the following is a quadratic polynomial with zeroes 3 5 and 0 ?
(A) 3 x ( 3 x − 5 ) (B) 3 x ( x − 5 ) (C) x 2 − 3 5 (D) 3 5 x 2
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Answer: (A) 3 x ( 3 x − 5 )
3 x ( 3 x − 5 ) = 0 gives x = 0 or x = 3 5 .(B) has zeroes 0 and 5, (C) has zeroes ± 3 5 , (D) has only the zero 0. So the answer is 3 x ( 3 x − 5 ) .
If α , β are zeroes of a polynomial p ( x ) = 2 x 2 − x − 1 then α 2 + β 2 is equal to
(A) 4 − 3 (B) 4 5 (C) 4 1 (D) 4 3
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Answer: (B) 4 5
α + β = 2 1 , α β = − 2 1 .α 2 + β 2 = ( α + β ) 2 − 2 α β = 4 1 + 1 = 4 5
The graph of y = p ( x ) is given in the adjoining figure. Zeroes of the polynomial p ( x ) are
(A) –5, 7(B) 2 − 5 , 2 − 7 (C) –5, 0, 7(D) –5, 2 − 5 , 2 7 , 7
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Answer: (C) –5, 0, 7
Zeroes are the x-coordinates of the points where the graph cuts the x-axis. The graph cuts the x-axis at (–5, 0), (0, 0) and (7, 0). Zeroes: –5, 0, 7. (The points ( − 2 5 , 0 ) and ( 2 7 , 0 ) are only feet of the turning points.)
If one zero of the polynomial 6 x 2 + 37 x − ( k − 2 ) is reciprocal of the other, then what is the value of k ?
(A) − 4 (B) − 6 (C) 6(D) 4
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Answer: (A) − 4
Let the zeroes be α and α 1 ; their product is 1. Product of zeroes = 6 − ( k − 2 ) = 1 − ( k − 2 ) = 6 , so k = − 4
The zeroes of the polynomial p ( x ) = x 2 + 4 x + 3 are given by :
(A) 1, 3(B) − 1 , 3(C) 1, − 3 (D) − 1 , − 3
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Answer: (D) − 1 , − 3
x 2 + 4 x + 3 = ( x + 1 ) ( x + 3 ) Zeroes are − 1 and − 3 .
If α and β are the zeroes of the quadratic polynomial p ( x ) = x 2 − a x − b , then the value of α 2 + β 2 is :
(A) a 2 − 2 b (B) a 2 + 2 b (C) b 2 − 2 a (D) b 2 + 2 a
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Answer: (B) a 2 + 2 b
α + β = a and α β = − b α 2 + β 2 = ( α + β ) 2 − 2 α β = a 2 + 2 b
Assertion (A) : The polynomial p ( x ) = x 2 + 3 x + 3 has two real zeroes. Reason (R): A quadratic polynomial can have at most two real zeroes.
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).(C) Assertion (A) is true, but Reason (R) is false.(D) Assertion (A) is false, but Reason (R) is true.
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Answer: (D) A is false, but R is true.
Discriminant of x 2 + 3 x + 3 is 9 − 12 = − 3 < 0 , so it has no real zeroes. A is false. A quadratic polynomial has at most two zeroes, so R is true.
If one zero of the polynomial x 2 − 3 k x + 4 k be twice the other, then the value of k is :
(A) − 2 (B) 2(C) 2 1 (D) − 2 1
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Answer: (B) 2
Let the zeroes be α and 2 α . Sum: 3 α = 3 k , so α = k . Product: 2 α 2 = 4 k , so 2 k 2 = 4 k . So k = 0 or k = 2 ; k = 0 gives the trivial case x 2 (both zeroes 0) and is not an option, so k = 2 .
If ‘α ’ and ‘β ’ are the zeroes of the polynomial a x 2 − 5 x + c and α + β = α β = 10 , then :
(A) a = 5 , c = 2 1 (B) a = 1 , c = 2 5 (C) a = 2 5 , c = 1 (D) a = 2 1 , c = 5
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Answer: (D) a = 2 1 , c = 5
α + β = a 5 = 10 , so a = 2 1 α β = a c = 10 , so c = 10 × 2 1 = 5
The sum of zeroes of the polynomial 2 x 2 − 17 are given as :
(A) 2 17 2 (B) − 2 17 2 (C) 0(D) 1
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Answer: (C) 0
Here a = 2 , b = 0 , c = − 17 . Sum of zeroes = − a b = 0
If α , β are zeroes of the polynomial x 2 − 1 , then value of ( α + β ) is :
(A) 2(B) 1(C) − 1 (D) 0
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Answer: (D) 0
For x 2 + 0 x − 1 , sum of zeroes = − 1 0 = 0 . (The zeroes are 1 and − 1 .)
If α , β are the zeroes of the polynomial p ( x ) = 4 x 2 − 3 x − 7 , then ( α 1 + β 1 ) is equal to :
(A) 3 7 (B) 3 − 7 (C) 7 3 (D) 7 − 3
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Answer: (D) 7 − 3
α + β = 4 3 , α β = 4 − 7 α 1 + β 1 = α β α + β = − 7/4 3/4 = 7 − 3
The number of polynomials having zeroes − 3 and 5 is :
(A) only one(B) infinite(C) exactly two(D) at most two
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Answer: (B) infinite
Any polynomial k ( x + 3 ) ( x − 5 ) with k = 0 has zeroes − 3 and 5. Higher-degree polynomials such as k ( x + 3 ) ( x − 5 ) ( x − a ) also have these zeroes. Since k can be any non-zero real number, there are infinitely many such polynomials.
The number of polynomials having zeroes − 1 and 2 is :
(A) exactly 2(B) only 1(C) at most 2(D) infinite
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Answer: (D) infinite
Any polynomial k ( x + 1 ) ( x − 2 ) with k = 0 has zeroes − 1 and 2. Higher-degree polynomials with these factors also work. So there are infinitely many such polynomials.
The number of quadratic polynomials having zeroes –5 and –3 is
(A) 1(B) 2(C) 3(D) more than 3
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Answer: (D) more than 3
Any polynomial k ( x + 5 ) ( x + 3 ) = k ( x 2 + 8 x + 15 ) with k = 0 has zeroes –5 and –3. k can take infinitely many values, so there are more than 3 such polynomials.
If α and β are the zeroes of the polynomial x 2 − 1 , then the value of ( α + β ) is
(A) 2(B) 1(C) –1(D) 0
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Answer: (D) 0
For a x 2 + b x + c , sum of zeroes = − a b . Here a = 1 , b = 0 , so α + β = 0 . (The zeroes are 1 and –1.)
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
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Answer: (D) a = 0, b = – 6
Sum of zeroes: 2 + ( − 3 ) = − 1 = − ( a + 1 ) , so a = 0 . Product of zeroes: 2 × ( − 3 ) = − 6 = b .
If one zero of the polynomial x 2 + 3 x + k is 2, then the value of k.
(A) – 10(B) 10(C) 5(D) – 5
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Answer: (A) – 10
Put x = 2: 4 + 6 + k = 0 . k = –10
The zeroes of the polynomial 3 x 2 + 11 x − 4 are :
(A) 3 1 , – 4(B) 3 − 1 , 4(C) 3 1 , 4(D) 3 − 1 , – 4
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Answer: (A) 3 1 , – 4
3 x 2 + 11 x − 4 = 3 x 2 + 12 x − x − 4 = ( 3 x − 1 ) ( x + 4 ) Zeroes: x = 3 1 and x = − 4 .
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