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Real Numbers Polynomials Pair of Linear Equations in Two Variables Quadratic Equations Arithmetic Progressions Triangles Coordinate Geometry Introduction to Trigonometry Some Applications of Trigonometry Circles Constructions Areas related to Circles Surface Areas and Volumes Statistics Probability
If 2 and -3 are the zeroes of the quadratic polynomial x2 + (a + 1) x + b; then find the values of a and b.

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જવાબ :


Obtain all other zeroes of the polynomial x4 + 4x3-2x2-20x -15 if two of its zeroes are √5 and -√5

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જવાબ :


When a polynomial 6x4 + 8x3 + 290x2 + 21x + 7 is divided by another polynomial 3x2 + 4x + 1 the remainder is in the form ax + b. Find a and b.

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જવાબ :


Find the zeroes of the quadratic polynomial 9t2-6t + 1 and verify the relationship between the zeroes and the coefficients.

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જવાબ :


If the product of zeroes of the polynomial ax2 – 6x – 6 is 4, find the value of a. Find the sum of zeroes of the polynomial.

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જવાબ :


Find a quadratic polynomial each with the given numbers as the sum and product of its zeros respectively.
(i) −14,14
(ii) 2–√,13

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જવાબ : Let α, β be the zeros of polynomial.
(i) We have, α + β = −14 and αβ = 14
Thus, polynomial is
p(x) = x2 – (a + B) x + aß


If the product of two zeros of the polynomial p(x) = 2x3 + 6x2 – 4x + 9 is 3, then find its third zero.

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જવાબ : Let α, β, γ be the roots of the given polynomial and αβ = 3
Then αβγ = −92
⇒ 3 x γ = −92 or γ = −32


Given that one of the zeros of the cubic polynomial ax3 + bx2 + cx + d is zero, find the product of the other two zeros.

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જવાબ : Let α, β, γ be the roots of the given polynomial and α = 0.
Then αβ + βγ + γα = c/a ⇒ βγ = c/a


If α and β are zeros of p(x) = x2 + x – 1, then find 1α+1β

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જવાબ : Here, α + β = -1, αβ = -1,
So 1α+1β=β+ααβ=−1−1=1


If α and β are zeros of polynomial p(x) = x2 – 5x + 6, then find the value of α + B – 3aß.

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જવાબ : Here, α + β = 5, αβ = 6
= α + β – 3αβ = 5 – 3 x 6 = -13


If 1 is a zero of the polynomial p(x) = ax2 – 3(a – 1)x -1, then find the value of a.

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જવાબ : Put x = 1 in p(x)
∴ p(1) = a(1)2 – 3(a – 1) x 1 – 1 = 0
⇒ a – 3a + 3 – 1 = 0 ⇒ 2a = -2 ⇒ a = 1


If one of the zeros of the quadratic polynomial (k – 1)x2 + kx + 1 is -3 then find the value of k.

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જવાબ : Since – 3 is a zero of the given polynomial
∴ (k – 1)(-3)2 + k(-3) + 1 = 0 :
⇒ 9k – 9 – 3k + 1 = 0 ⇒ k = 4/3.


Find the zeros of the polynomial p(x) = 4x2 – 12x + 9.

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જવાબ : p(x) = 4x2 – 12x + 9 = (2x – 3)2
For zeros, p(x) = 0
⇒ (2x – 3)(2x – 3) = 0 ⇒ x = 32,32


If one root of the polynomial p(y) = 5y2 + 13y + m is reciprocal of other, then find the value of m.

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જવાબ :


Can x – 2 be the remainder on division of a polynomial p(x) by x + 3?

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જવાબ : No, as degree (x – 2) = degree (x + 3)


Find the quadratic polynomial whose zeros are -3 and 4.
[NCERT Exemplar]

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જવાબ : Sum of zeros = -3 + 4 = 1,
Product of zeros = – 3 x 4 = -12
∴ Required polynomial = x2 – x – 12


If one zero of the quadratic polynomial x2 – 5x – 6 is 6 then find the other zero.

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જવાબ : Let α,6 be the zeros of given polynomial.
Then α + 6 = 5 3 ⇒ α = -1


If both the zeros of the quadratic polynomial ax2 + bx + c are equal and opposite in sign, then find the value of b.

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જવાબ : Let α and -α be the roots of given polynomial.
Then α + (-α) = 0 ⇒ −ba=0 ⇒ b = 0.


What number should be added to the polynomial x2 – 5x + 4, so that 3 is the zero of the polynomial?

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જવાબ : Let f(x) = x2 – 5x + 4
Then f(3) = 32 – 5 x 3 + 4 = -2
For f(3) = 0, 2 must be added to f(x).


Can a quadratic polynomial x2 + kx + k have equal zeros for some odd integer k > 1?

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જવાબ : No, for equal zeros, k = 0,4 ⇒ k should be even.


If the zeros of a quadratic polynomial ax2 + bx + c are both negative, then can we say a, b and c all have the same sign? Justify your answer.

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જવાબ : Yes, because −ba = sum of zeros < 0, so that ba=0 > 0. Also the product of the zeros = ca=0 > 0.


If the graph of a polynomial intersects the x-axis at exactly two points, is it necessarily a quadratic polynomial?

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જવાબ : No, x4 – 1 is a polynomial intersecting the x-axis at exactly two points.


If one of the zeros of the quadratic polynomial f(x) = 4x2 – 8kx – 9 is equal in magnitude but opposite in sign of the other, find the value of k.

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જવાબ : Let one root of the given polynomial be α.
Then the other root = -α
Sum of the roots = (-α) + α = 0
⇒ −ba = 0 or −8k4 = 0 or k = 0


If the graph of a polynomial intersects the x-axis at only one point, can it be a quadratic polynomial?

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જવાબ : Yes, because every quadratic polynomial has at the most two zeros.


If on division of a polynomial p(x) by a polynomial g(x), the quotient is zero, what is the relation between the degrees of p(x) and g(x)?

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જવાબ : Since the quotient is zero, therefore
deg p(x) < deg g(x)


What will the quotient and remainder is on division of ax2 + bx + c by px2 + qx2 + rx + 5, p ≠ 0?

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જવાબ : 0, ax2 + bx + C.


Can (x – 2) be the remainder on division of a polynomial p(x) by (2x + 3)? Justify your answer. (2016 OD)

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જવાબ : In case of division of a polynomial by another polynomial, the degree of the remainder (polynomial) is always less than that of the divisor. (x – 2) cannot be the remainder when p(x) is divided by (2x + 3) as the degree is the same.


Find a quadratic polynomial, the sum and product of whose zeroes are 0 and -√2 respectively. (2015)
 

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જવાબ : Quadratic polynomial is
x2 – (Sum of zeroes) x + (Product of zeroes)
= x2 – (0)x + (-√2)
= x2 – √2


Find the condition that zeroes of polynomial p(x) = ax2 + bx + c are reciprocal of each other. (2017 OD)

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જવાબ : Let α and 1α be the zeroes of P(x).
P(a) = ax2 + bx + c …(given)
Product of zeroes = ca
⇒ α × 1α = ca
⇒ 1 = ca
⇒ a = c (Required condition)
Coefficient of x2 = Constant term


Form a quadratic polynomial whose zeroes are 3 + √2 and 3 – √2. (2012)

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જવાબ : Sum of zeroes,
S = (3 + √2) + (3 – √2) = 6
Product of zeroes,
P = (3 + √2) x (3 – √2) = (3)2 – (√2)2 = 9 – 2 = 7
Quadratic polynomial = x2 – Sx + P = x2 – 6x + 7


If α and β are the zeroes of a polynomial such that α + β = -6 and αβ = 5, then find the polynomial. (2016 D)

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જવાબ : Quadratic polynomial is x2 – Sx + P = 0
⇒ x2 – (-6)x + 5 = 0
⇒ x2 + 6x + 5 = 0


A quadratic polynomial, whose zeroes are -4 and -5, is …. (2016 D)

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જવાબ : x2 + 9x + 20 is the required polynomial.


Find a quadratic polynomial, the stun and product of whose zeroes are √3 and 1√3 respectively. (2014)
 

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જવાબ : Sum of zeroes, (S) = √3
Product of zeroes, (P) = 1√3
Quadratic polynomial = x2 – Sx + P


If the sum of the zeroes of the polynomial p(x) = (k2 – 14) x2 – 2x – 12 is 1, then find the value of k. (2017 D)

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જવાબ : p(x) = (k2 – 14) x2 – 2x – 12
Here a = k2 – 14, b = -2, c = -12
Sum of the zeroes, (α + β) = 1 …[Given]
⇒ −ba = 1
⇒ −(−2)k2−14 = 1
⇒ k2 – 14 = 2
⇒ k2 = 16
⇒ k = ±4


If the sum of zeroes of the quadratic polynomial 3x2 – kx + 6 is 3, then find the value of k. (2012)
 

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જવાબ : Here a = 3, b = -k, c = 6
Sum of the zeroes, (α + β) = −ba = 3 …..(given)
⇒ −(−k)3 = 3
⇒ k = 9


If α and β are the zeroes of the polynomial ax2 + bx + c, find the value of α2 + β2. (2013)

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જવાબ :


If a polynomial x4 – 3x3 – 6x2 + kx – 16 is exactly divisible by x2 – 3x + 2, then find the value of A.

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જવાબ :


Find the zeroes of the quadratic polynomial 3x2  – 2 and verify the relationship between the zeroes and the coefficients.

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જવાબ :


If the zeroes of the polynomial x2 + px + q are double in value to the zeroes of 2x2 – 5x – 3, find the value of p and q. (2012)

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જવાબ : We have, 2x2 – 5x – 3 = 0
= 2x2 – 6x + x – 3
= 2x(x – 3) + 1(x – 3)
= (x – 3) (2x + 1)
Zeroes are:
x – 3 = 0 or 2x + 1 = 0
⇒ x = 3 or x = −12
Since the zeroes of required polynomial is double of given polynomial.
Zeroes of the required polynomial are:
3 × 2, (−12 × 2), i.e., 6, -1
Sum of zeroes, S = 6 + (-1) = 5
Product of zeroes, P = 6 × (-1) = -6
Quadratic polynomial is x2 – Sx + P
⇒ x2 – 5x – 6 … (i)
Comparing (i) with x2 + px + q
p = -5, q = -6


What must be subtracted from the polynomial f(x) = x4 + 2x3 – 13x2 – 12x + 21 so that the resulting polynomial is exactly divisible by x2 – 4x + 3? (2012, 2017 D)

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જવાબ :


Find the zeroes of p(x) = 2x2 – x – 6 and verify the relationship of zeroes with these co-efficients. (2017 OD)

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જવાબ : p(x) = 2x2 – x – 6 …[Given]
= 2x2 – 4x + 3x – 6
= 2x (x – 2) + 3 (x – 2)
= (x – 2) (2x + 3)
Zeroes are:
x – 2 = 0 or 2x + 3 = 0
x = 2 or x = −32
Verification:
Here a = 2, b = -1, c = -6


Find the zeroes of the quadratic polynomial 3x2 – 75 and verify the relationship between the zeroes and the coefficients. (2014)

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જવાબ : We have, 3x2 – 75
= 3(x2 – 25)
= 3(x2 – 52)
= 3(x – 5)(x + 5)
Zeroes are:
x – 5 = 0 or x + 5 = 0
x = 5 or x = -5
Verification:
Here a = 3, b = 0, c = -75
Sum of the zeroes = 5 + (-5) = 0


Find the quadratic polynomial whose zeroes are -2 and -5. Verify the relationship between zeroes and coefficients of the polynomial. (2013)

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જવાબ : Sum of zeroes, S = (-2) + (-5) = -7
Product of zeroes, P = (-2)(-5) = 10
Quadratic polynomial is x2 – Sx + P = 0
= x2 – (-7)x + 10
= x2 + 7x + 10
Verification:
Here a = 1, b = 7, c = 10
Sum of zeroes = (-2) + (-5) = 7


Find a quadratic polynomial whose zeroes are 3+√55 and 3−√55. (2013)

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જવાબ :


Show that 12 and −32 are the zeroes of the polynomial 4x2 + 4x – 3 and verify the relationship between zeroes and co-efficients of polynomial. (2013

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જવાબ :


Verify whether 2, 3 and 12 are the zeroes of the polynomial p(x) = 2x3 – 11x2 + 17x – 6. (2012, 2017 D)

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જવાબ : p(x) = 2x3 – 11x2 + 17x – 6
When x = 2,
p(2) = 2(2)3 – 11(2)2 + 17(2) – 6 = 16 – 44 + 34 – 6 = 0
When x = 3, p(3) = 2(3)3 – 11(3)2 + 17(3) – 6 = 54 – 99 + 51 – 6 = 0

Yes, x = 2, 3 and 12 all are the zeroes of the given polynomial.


Find a quadratic polynomial, the sum and product of whose zeroes are -8 and 12 respectively. Hence find the zeroes. (2014)
 

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જવાબ : Let Sum of zeroes (α + β) = S = -8 …[Given]
Product of zeroes (αβ) = P = 12 …[Given]
Quadratic polynomial is x2 – Sx + P
= x2 – (-8)x + 12
= x2 + 8x + 12
= x2 + 6x + 2x + 12
= x(x + 6) + 2(x + 6)
= (x + 2)(x + 6)
Zeroes are:
x + 2 = 0 or x + 6 = 0
x = -2 or x = -6


Find a quadratic polynomial, the sum and product of whose zeroes are 0 and −35 respectively. Hence find the zeroes. (2015)


 

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જવાબ : Quadratic polynomial = x2 – (Sum)x + Product


Find the zeroes of the quadratic polynomial 6x2 – 3 – 7x and verify the relationship between the zeroes and the coefficients of the polynomial. (2015, 2016 OD)
 

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જવાબ : We have, 6x2 – 3 – 7x
= 6x2 – 7x – 3
= 6x2 – 9x + 2x – 3
= 3x(2x – 3) + 1(2x – 3)
= (2x – 3) (3x + 1)


Find the zeroes of the quadratic polynomial f(x) = x2 – 3x – 28 and verify the relationship between the zeroes and the co-efficient of the polynomial. (2012, 2017 D)

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જવાબ : p(x) = x2 – 3x – 28
= x2 – 7x + 4x – 28
= x(x – 7) + 4(x – 7)
= (x – 7) (x + 4)
Zeroes are:
x – 7 = 0 or x + 4 = 0
x = 7 or x = -4


If α and β are the zeroes of the polynomial 6y2 – 7y + 2, find a quadratic polynomial whose zeroes are 1α and 1β. (2012)

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જવાબ : Given: 6y2 – 7y + 2
Here a = 6, b = -7, c = 2


Divide 3x2 + 5x – 1 by x + 2 and verify the division algorithm. (2013 OD)

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જવાબ :
Quotient = 3x – 1
Remainder = 1
Verification:
Divisor × Quotient + Remainder
= (x + 2) × (3x – 1) + 1
= 3x2 – x + 6x – 2 + 1
= 3x2 + 5x – 1
= Dividend


Check whether polynomial x – 1 is a factor of the polynomial x3 – 8x2 + 19x – 12. Verify by division algorithm. (2014)

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જવાબ : Let P(x) = x3 – 8x2 + 19x – 12
Put x = 1
P(1) = (1)3 – 8(1)2 + 19(1) – 12
= 1 – 8 + 19 – 12
= 20 – 20
= 0
Remainder = 0
(x – 1) is a factor of P(x).
Verification:

 

Since remainder = 0
(x – 1) is a factor of P(x).

Polynomials Class 10 Important Questions Long Answer (4 Marks)


On dividing 3x3 + 4x2 + 5x – 13 by a polynomial g(x) the quotient and remainder were 3x +10 and 16x – 43 respectively. Find the polynomial g(x). (2017 OD)

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જવાબ : Let 3x3 + 4x2 + 5x – 13 = P(x)
q(x) = 3x + 10, r(x) = 16x – 43 …[Given]
As we know, P(x) = g(x) . q(x) + r(x)
3x3 + 4x2 + 5x – 13 = g(x) . (3x + 10) + (16x – 43)
3x3 + 4x2 + 5x – 13 – 16x + 43 = g(x) . (3x + 10)


Given that x – √5 is a factor of the polynomial x3 – 3√5 x2 – 5x + 15√5, find all the zeroes of the polynomial. (2012, 2016)
 

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જવાબ : Let P(x) = x3 – 3√5 x2 – 5x + 15√5
x – √5 is a factor of the given polynomial.
Put x = -√5,

Other zero:
x – 3√5 = 0 ⇒ x = 3√5
All the zeroes of P(x) are -√5, √5 and 3√5.


Divide 4x3 + 2x2 + 5x – 6 by 2x2 + 1 + 3x and verify the division algorithm. (2013)

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જવાબ :

Quotient = 2x – 2
Remainder = 9x – 4
Verification:
Divisor × Quotient + Remainder
= (2x2 + 3x + 1) × (2x – 2) + 9x – 4
= 4x3 – 4x2 + 6x2 – 6x + 2x – 2 + 9x – 4
= 4x3 + 2x2 + 5x – 6
= Dividend


If a polynomial x4 + 5x3 + 4x2 – 10x – 12 has two zeroes as -2 and -3, then find the other zeroes. (2014)
 

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જવાબ : Since two zeroes are -2 and -3.
(x + 2)(x + 3) = x2 + 3x + 2x + 6 = x2 + 5x + 6
Dividing the given equation with x2 + 5x + 6, we get

x4 + 5x3 + 4x2 – 10x – 12
= (x2 + 5x + 6)(x2 – 2)
= (x + 2)(x + 3)(x – √2 )(x + √2 )
Other zeroes are:
x – √2 = 0 or x + √2 = 0
x = √2 or x = -√2

 


Find all the zeroes of the polynomial 8x4 + 8x3 – 18x2 – 20x – 5, if it is given that two of its zeroes are 52−−√ and −52−−√. (2014, 2016 D)

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જવાબ :


If p(x) = x3 – 2x2 + kx + 5 is divided by (x – 2), the remainder is 11. Find k. Hence find all the zeroes of x3 + kx2 + 3x + 1. (2012)

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જવાબ : p(x) = x3 – 2x2 + kx + 5,
When x – 2,
p(2) = (2)3 – 2(2)2 + k(2) + 5
⇒ 11 = 8 – 8 + 2k + 5
⇒ 11 – 5 = 2k
⇒ 6 = 2k
⇒ k = 3
Let q(x) = x3 + kx2 + 3x + 1
= x3 + 3x2 + 3x + 1
= x3 + 1 + 3x2 + 3x
= (x)3 + (1)3 + 3x(x + 1)
= (x + 1)3
= (x + 1) (x + 1) (x + 1) …[∵ a3 + b3 + 3ab (a + b) = (a + b)3]
All zeroes are:
x + 1 = 0 ⇒ x = -1
x + 1 = 0 ⇒ x = -1
x + 1 = 0 ⇒ x = -1
Hence zeroes are -1, -1 and -1.


If α and β are zeroes of p(x) = kx2 + 4x + 4, such that α2 + β2 = 24, find k. (2013)

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જવાબ : We have, p(x) = kx2 + 4x + 4
Here a = k, b = 4, c = 4

⇒ 24k2 = 16 – 8k
⇒ 24k2 + 8k – 16 = 0
⇒ 3k2 + k – 2 = 0 …[Dividing both sides by 8]
⇒ 3k2 + 3k – 2k – 2 = 0
⇒ 3k(k + 1) – 2(k + 1) = 0
⇒ (k + 1)(3k – 2) = 0
⇒ k + 1 = 0 or 3k – 2 = 0
⇒ k = -1 or k = 23


If α and β are the zeroes of the polynomial p(x) = 2x2 + 5x + k, satisfying the relation, α2 + β2 + αβ = 214 then find the value of k. (2017 OD)

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જવાબ : Given polynomial is p(x) = 2x2 + 5x + k
Here a = 2, b = 5, c = k


What must be subtracted from p(x) = 8x4 + 14x3 – 2x2 + 8x – 12 so that 4x2 + 3x – 2 is factor of p(x)? This question was given to group of students for working together. (2015)

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જવાબ :

Polynomial to be subtracted by (15x – 14).


Divide 2x 4– 9x3 + 5x 2 + 3x – 8 bv x 2– 4x+ 1 and verify the division algorithm.

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જવાબ :


Obtain all other zeroes of the polynomial x 4– 3x 3-x 2 + 9x – 6, if two of its zeroes are √3 and -√3.

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જવાબ :


Find other zeroes of the polynomial x 4 – 7x2 + 12 if it is given that two of its zeroes are √3 and -√3.

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જવાબ :


If the polynomial (x4 + 2x3 + 8x2 + 12x + 18) is divided by another polynomial (x2 + 5), the remainder comes out to be (px + q), find the values of p and q.

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જવાબ :
Remainder = 2x + 3
px + q = 2x + 3
p = 2 and q = 3.


If a polynomial 3x4 – 4x3 – 16x2 + 15x + 14 is divided by another polynomial x2 – 4, the remainder comes out to be px + q. Find the value of p and q. (2014)

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જવાબ :


Find the values of a and b so that x4 + x3 + 8x2 +ax – b is divisible by x2 + 1. (2015)

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જવાબ :
If x4 + x3 + 8x2 + ax – b is divisible by x2 + 1
Remainder = 0
(a – 1)x – b – 7 = 0
(a – 1)x + (-b – 7) = 0 . x + 0
a – 1 = 0, -b – 7 = 0
a = 1, b = -7
a = 1, b = -7


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