RBSE Solutions for Class 10 Maths Chapter 3 Polynomials Additional Questions is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 3 Polynomials Additional Questions.

Board |
RBSE |

Te×tbook |
SIERT, Rajasthan |

Class |
Class 10 |

Subject |
Maths |

Chapter |
Chapter 3 |

Chapter Name |
Polynomials |

E×ercise |
Additional Questions |

Number of Questions Solved |
57 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 10 Maths Chapter 3 Polynomials Additional Questions

**Multiple Choice Questions**

Prime factors of 144 are 2x2x2x2, 3×3.

Questions 1.

If 2 is factor of (RBSESolutions.com) polynomial f(x) = x^{4} – x^{3} – 4x^{2} + kx + 10 then find k.

(A) 2

(B) -2

(C) -1

(D) 1

Solution

f(2) = 2^{4} – 2^{3} – 4 × 2^{2} + k × 2 + 10

⇒ 0 = 16 – 8 – 16 + 2k + 10

⇒ 0 = 2k + 2

⇒ 2k = -2

⇒ k = -1

Hence option (C) is correct.

Question 2.

Find that polynomial whose zeros are -5 and 4

(A) x^{2} – x – 20

(B) x^{2} + x – 20

(C) x^{2} – x + 20

(D) x^{2} + x + 20

Solution

Let α = – 5 and β = 4

Then α + β = -5 + 4 = -1 and αβ = -5 x 4 = -20

Quadratic polynomial x^{2} – (α + β) x + αβ = 0

⇒ x^{2} – (-1) x + (-20) = 0

⇒ x^{2} + x – 20 = 0

Hence, option (B) is correct.

Question 3.

Expression (x – 3) will be a (RBSESolutions.com) factor of polynomial f(x) = x^{3} + x^{2} – 17x + 15 if:

(A) f(3) = 0

(B) f(-3) = 0

(C) f(2) = 0

(D) f(-2) = 0

Solution

f(x) = x^{3} + x^{2} – 17x + 15

f(3) = (3)^{3} + (3)^{2} – 17(3) + 15 = 27 + 9 – 51 + 15 = 0

f(3) = 0

Hence, option (A) is correct.

Question 4.

If (x – 5) is a factor of (RBSESolutions.com) polynomial x^{3} – 3x^{2} + kx – 10 then value of k will be:

(A) -8

(B) -7

(C) 5

(D) 8

Solution

Let p(x) = x^{3} – 3x^{2} + kx – 10

if (x – 5), is a factor of p(x)

then p(x) = 0

(5)^{3} – 3(5)^{2} + k(5) – 10 = 0

⇒ 125 – 75 + 5k – 10 = 0

⇒ 40 + 5k = 0

⇒ 5k = -40

⇒ k = -8

Thus k = -8

Hence, option (A) is correct.

Question 5.

Zero of 3y^{3} + 8y^{2} – 1 is

(A) 1

(B) (-1)

(C) 0

(D) None of these

Solution

Let p(y) = 3y^{3} + 8y^{2} – 1

p(1) = 3 × (1)^{3} + 8(1)^{2} – 1 = 3 + 8 – 1 = 10 ≠ 0

p(-1) = 3(-1)^{3} + 8(-1)^{2} – 1 = -3 + 8 – 1 = 4 ≠ 0

and p(0) = 3 × (0)^{3} + 8(0)^{2} – 1 = 0 + 0 – 3 = -3 ≠ 0

Thus by putting y = 1, -1 the expression (RBSESolutions.com) obtained is not equal to zero.

Hence, the option (D) is correct.

Question 6.

If (x – 1) is a factor of polynomial 2x^{2} + kx + √2, then k will be

(A) 2 + √2

(B) 2 – √2

(C) – (2 + √2)

(D) – (2 – √2)

Solution

Let p{x) = 2x^{2} + kx + √2

If (x – 1), is a factor of p(x)

then p(1) = 0

2(1)^{2} + k(1) + √2 = 0

⇒ 2 × 1 + k + √2 = 0

⇒ 2 + k + √2 = 0

⇒ k + ( 2 + √2) = 0

⇒ k = -(2 + √2)

Hence, option (C) is correct.

Question 7.

One zero of p(x) = 2x + 1 will be

(A) \(\frac { 1 }{ 2 }\)

(B) 3

(C) \(\frac { -1 }{ 2 }\)

(D) 1

Solution

p(x) = 2x + 1

For zeros p(x) = 0

0 = 2x + 1

⇒ x = \(\frac { -1 }{ 2 }\)

Hence, option (C) is correct.

**Short Answer Type Questions**

Question 1.

If α, β are zeros of any quadratic (RBSESolutions.com) equation, then write the quadratic equation.

Solution

k[x^{2} – (α + β)x + αβ]

Question 2.

If f(x) are two expressions, then write the relationship between their L.C.M. and H.C.F.

Solution

L.C.M. × H.C.F. = f(x) × g(x)

Question 3.

If α and β are zeros of any quadratic polynomial ax^{2} + bx + c, then write the value of α + β and αβ.

Solution

α + β = \(\frac { -b }{ a }\)

and αβ = \(\frac { c }{ a }\)

Question 4.

What is the zero of (RBSESolutions.com) polynomial?

Solution

The value of x for which polynomial f(x) = 0, is called zero of the polynomial.

Question 5.

If α and β are zeros of polynomial f(x) = x^{2} – 5x + k such that α – β = 1, then find k.

Solution

If α, β are zeros of polynomial x^{2} – 5x + k.

Question 6.

Find the condition for which zeros of (RBSESolutions.com) polynomial, p(x) = ax^{2} + bx + c are inverse of each other. **(CBSE 2012)**

Solution

Let α is first zero of polynomial of p(x) = ax^{2} + bx + c

According to question, second zero of polynomial

Hence, required condition is c = a

Question 7.

If a – b, a + b are zeros of polynomial, x^{3} – 3x^{2} + x + 1, then find a and b.

Solution

Given polynomial = x^{3} – 3x^{2} + x + 1

This degree and leading coefficient calculator finds the degree, leading term, and coefficient associated with it absolutely for free.

Question 8.

Find the zeros of quadratic polynomial x^{2} + x – 2 and (RBSESolutions.com) verify the relationship between zeros and coeffcient **(M.S.B. Raj. 2016)**

Solution

Given quadratic polynomial

f(x) = x^{2} + x – 2 = x^{2} + 2x – x – 2 = x (x + 2) – 1 (x + 2) = (x – 1) (x + 2)

To find zero, f(x) = 0

(x- 1) (x + 2) = 0

x – 1 = 0 or x + 2 = 0

x = 1 or x = -2

Thus 1 and -2 are two zeros of given polynomial

Relation between zeros and coefficient

Sum of zeros = 1 + (- 2) = – 1

and product of zeros = 1 × (-2) = -2

Comparing given polynomial with ax^{2} + bx + c

a = 1, b = 1 and c = -2

Sum of zeros = \(\frac { -b }{ a }\) = \(\frac { -1 }{ 1 }\) = -1

and product of zeros = \(\frac { c }{ a }\) = \(\frac { -2 }{ 1 }\) = -2

Hence, relationship between zeros and coeffi-cient is verified.

Synthetic Division Calculator that can divide polynomials and demonstrate the process in a table format.

Question 9.

Divide x^{3} – 3x^{2} + 3x – 5 by x – 1 – x^{2} and test (RBSESolutions.com) division algorithm.** (M.S.B. Raj. 2013)**

Solution

Question 10.

Find all the zeros of (RBSESolutions.com) polynomial x^{3} – 6x^{2} + 11x – 6 whereas 1 and 2 are its two zeros. **(CBSE 2012)**

Solution

we know that if a is one zero of polynomial f(x) then (x – α) will be a factor of f(x).

According to equation, polynomial f(x) = x^{3} – 6x^{2} + 11x – 6 has two zeros 1 and 2.

Thus (x – 1) (x – 2) = (x^{2} – 3x + 2), will be a factor of f(x).

Now dividing polynomial f(x) by x^{2} – 3x + 2

Using division algorithm for polynomial

Dividend = Divisor × Quotient + Remainder

x^{3} – 6x^{2} + 11x – 6 = (x^{2} – 3x + 2)(x – 3) + 0 = (x – 1)(x – 2)(x – 3)

For zeros of polynomial

f(x) = 0

⇒ (x – 1) (x – 2) (x – 3) = 0

⇒ x – 1 = 0, x – 2 = 0, x – 3 = 0

⇒ x = 1, x = 2, x = 3

Hence, all the zeros of polynomial f(x) are 1, 2, 3.

Question 11.

If α, β are zeros of polynomial p(x) = 2x^{2} + 5x + k which (RBSESolutions.com) satisfies the relation α^{2} + β^{2} + αβ = \(\frac { 21 }{ 4 }\), then find value of k. **(CBSE 2012)**

Solution

**Quadratic Equation**

**Multiple Choice Equations**

Welcome to our step-by-step math roots calculator.

Question 1.

Roots of the equation ax^{2} + bx + c = 0, a ≠ 0 will not be real if

(A) b^{2} < 4ac

(B) b^{2} > 4ac

(C) b^{2} = 4ac

(D) b = 4ac

Solution

Option (A) is correct.

Question 2.

If \(\frac { 1 }{ 2 }\) is one root of (RBSESolutions.com) quadratic equation x^{2} + kx – \(\frac { 5 }{ 4 }\) = 0, then value of k will be [NCERT Exemplar Problem]

(A) 2

(B) -2

(C) \(\frac { 1 }{ 4 }\)

(D) \(\frac { 1 }{ 2 }\)

Solution

Option (A) is correct.

Question 3.

Roots of quadratic equation 2x^{2} – x – 6 = 0 are **[CBSE 2012]**

(A) -2, \(\frac { 3 }{ 2 }\)

(B) 2, \(\frac { -3 }{ 2 }\)

(C) -2, \(\frac { -3 }{ 2 }\)

(D) 2, \(\frac { 3 }{ 2 }\)

Solution

Given quadratic equation is :

2x^{2} – x – 6 = 0

⇒ 2x^{2} – (4 – 3) x – 6 = 0

⇒ 2x^{2} – 4x + 3x – 6 = 0

⇒ (2x^{2} – 4x) + (3x – 6) = 0

⇒ 2x(x – 2) + 3(x – 2) = 0

⇒ (x – 2)(2x + 3) = 0

⇒ x-2 = 0 or 2x + 3 = 0

⇒ x = 2 or x = \(\frac { -3 }{ 2 }\)

Hence, option (B) is correct.

Question 4.

For which value of k, quadratic (RBSESolutions.com) equation 2x^{2} – kx + k = 0 has equal roots** [NCERT Exemplar Problem]**

(A) only 0

(B) only 4

(C) only 8

(D) 0, 8

Solution

option (D) is correct.

Question 5.

Root of equation x^{2} – 9 = 0 are

(A) √3

(B) -√3

(C) 9

(D) ±3

Solution

x^{2} – 9 = 0

⇒ x^{2} = 9

⇒ x = ±√9

⇒ x = ± 3

Hence, option (D) is correct.

Question 6.

Quadratic (RBSESolutions.com) equation 2x^{2} – √5 x + 1 = 0 has** [NCERT Exemplar Problem]**

(A) Two distinct real roots.

(B) Two equal real roots

(C) No real root

(D) More than two real roots

Solution

Option (C) is correct.

Question 7.

Product of root of equation 2x^{2} + x – 6 = 0 will be

(A) -3

(B) -7

(C) 2

(D) 0

Solution

Equation 2x^{2} + x – 6 = 0

Here a = 2, b = 1, c = -6

Product of root = \(\frac { c }{ a }\) = \(\frac { -6 }{ 2 }\) = -3

Hence, option (A) is correct.

Question 8.

Which of the following (RBSESolutions.com) equation has sum of roots 3 ? **[NCERT Exemplar Problem]**

(A) 2x^{2} – 3x + 6 = 0

(B) -x^{2} + 3x – 3 = 0

(C) -√2 x^{2} – \(\frac { 3 }{ \surd 2 }\) x + 1 = 0

(D) 3x^{2} – 3x + 3 = 0

Solution

Sum of roots of option (A) = \(\frac { -b }{ a }\) = \(\frac { 3 }{ 2 }\)

Sum of roots of option (B) = \(\frac { -b }{ a }\) = 3

Hence, option (B) is correct.

Question 9.

Solution of equation x^{2} – 4x = 0 are

(A) 4, 4

(B) 2, 2

(C) 0, 4

(D) 0, 2

Solution

x^{2} – 4x = 0

⇒ x (x – 4) =0

⇒ x = 0 or x – 4 = 0

⇒ x = 0 or x = 4

Thus x = 0, 4

Hence, option (C) is correct.

Solution 10.

Quadratic (RBSESolutions.com) equation px^{2} + qx + r = 0, p ≠ 0 has equal roots if

(A) p^{2} < 4pr

(B) p^{2} > 4qr

(C) q^{2} = 4pr

(D) p^{2} = 4qr

Solution

Option (C) is correct.

Solution 11.

Root of quadratic equation (x^{2} + 1)^{2} – x = 0 are **[NCERT Exemplar Problem]**

(A) Four real roots

(B) Two real roots

(C) One real root

(D) No real root

Solution

Option (D) is correct.

Question 12.

If equation x^{2} + 3ax + k = 0 has x = -a as (RBSESolutions.com) solution, then k will be

(A) 2a^{2}

(B) 0

(C) 2

(D) -2a

Solution

Option (A) is correct.

**Very Short/Short Answer Type Questions**

Question 1.

For quadratic equation ax^{2} + bx + c = 0, a ≠ 0, at which nature of roots depends ?

Solution

Discriminant (D) = b^{2} – 4ac

Question 2.

Describe nature of (RBSESolutions.com) roots of quadratic equation ax^{2} + bx + c = 0, a ≠ 0

Solution

(i) If (b^{2} – 4ac) > 0, then roots will be real and distinct.

(ii) If (b^{2} – 4ac) = 0, then roots will be equal and real.

(iii) If (b^{2} – 4ac) < 0, then roots will be imaginary.

Question 3.

If the sum of two natural numbers is 8 and the product is 15, then find numbers. **(CBSE 2012)**

Solution

Let the first natural number be x.

Sum of two natural numbers is 8 then other natural numbers will be 8 – x.

According to question.

Product of both natural numbers = 15

⇒ x (8 – x) = 15

⇒ 8x – x^{2} = 15

⇒ x^{2} – 8x + 15 = 0

⇒ x^{2} – (5 + 3)x + 15 = 0

⇒ x^{2} – 5x – 3x + 15 = 0

⇒ (x^{2} – 5x) – (3x – 15) = 0

⇒ x (x – 5) – 3 (x – 5) = 0

⇒ (x – 5) (x – 3) = 0

⇒ x – 5 = 0 or x – 3 = 0

⇒ x = 5 or x = 3

Thus, if First natural no. = 5

then Second natural no. = 8

if First natural no. = 3

or Second natural no. = 8

Question 4.

Find the roots of (RBSESolutions.com) quadratic equation √2 x^{2} + 7x + 5√2 = 0. **(CBSE 2013)**

Solution

Given quadratic equation is

Question 5.

Solve for x

Solution

Given the quadratic equation is

Question 6.

Find k for which quadratic (RBSESolutions.com) equation (k – 12) x^{2} + 2(k – 12) x + 2 = 0 has equal and real roots.

Solution

Given

(k – 12) x^{2} + 2(k – 12)x + 2 = 0

Comparing it with quadratic equation ax^{2} + bx + c = 0

a = (k – 12),b = 2(k – 12), c = 2

Discriminant (D) = b^{2} – 4ac

= 4(k – 12)^{2} – 4 × (k – 12) × 2

= 4(k – 12) [k – 12 – 2]

= 4(k – 12)(k – 14)

Given equation will have real and equal roots, if discriminant = 0.

D = 0

⇒ 4(k – 12)(k – 14) =0

⇒ k – 12 = 0 or k – 14 = 0

⇒ k = 12 or k = 14

Question 7.

A field is in the shape of a right-angled (RBSESolutions.com) triangle. Its hypotenuse is 1 m larger than twice the smallest side. If its third side is 7 m larger than the smallest side, then find sides of the field.

Solution

Let length of smallest side = x m

hypotenuse = (2x + 1) m and third side = (x + 7) m

By pythagorus theorem,

(hypotenuse)^{2} = sum of square of other both

⇒ (2x + 1)^{2} = x^{2} + (x + 7)^{2}

⇒ 4x^{2} + 4x + 1 = 2x^{2} + 14x + 49

⇒ 2x^{2} – 10x – 48 = 0

⇒ x^{2} – 5x – 24 = 0

⇒ x^{2} – 8x + 3x – 24 = 0

⇒ x(x – 8) + 3(x – 8) = 0

⇒ (x – 8) (x + 3) =0

⇒ x = 8, – 3

⇒ x = 8 [∵ x = -3 not possible]

hypotenuse = 2 × 8 + 1 = 17 m

and third side = 8 + 7 = 15 m.

Hence, the length of sides of the field is 8m, 17 m, and 15 m.

Question 8.

Solve for x:

x^{2} – 4ax – b^{2} + 4a^{2} = 0.** [CBSE 2012]**

Solution

Given quadratic equation x^{2} – 4ax – b^{2} + 4a^{2} = 0

Comparing it by quadratic (RBSESolutions.com) standard equation Ax^{2} + Bx + C = 0

A = 1, B = -4a and C = -(b^{2} – 4a^{2})

By Shridhar Acharya formula

Question 9.

Solve for x: √3 x^{2} – 2√2 x – 2√3 = 0. **[CBSE 2015]**

Solution

Question 10.

If 12 is added to a natural number, (RBSESolutions.com) then it becomes 160 times of its reciprocal. Find that number. **[NCERT Exemplar Problem]**

Solution

Let x be a natural number. According to the question, when 12 is added to a natural number.

It becomes 160 times its reciprocal

x + 12 = 160 × \(\frac { 1 }{ x }\)

⇒ x^{2} + 12x = 160

⇒ x^{2} + 12x – 160 = 0

⇒ x^{2} + (20 – 8) x – 160 = 0

⇒ x^{2} + 20x – 8x – 160 = 0

⇒ x (x + 20) – 8 (x + 20) = 0

⇒ (x + 20) (x – 8) = 0

if x + 20 = 0, then x = -20

or x – 8 = 0, then x = 8

x is a natural number, so it cannot be negative

x = 8

Hence, number is 8.

Question 11.

If a two digit number is 4 times the sum (RBSESolutions.com) of its digit and three times the product of it digit, then find the number.

Solution

Let digit of ten and unit place are x and y respectively.

number = 10x + y

According to first condition

number = 4 × sum of digit

⇒ 10x + y = 4 × (x + y)

⇒ 10x + y = 4x + 4y

⇒ 10x – 4x = 4y – y

⇒ 6x = 3y

⇒ 2x = y

⇒ y = 2x …(i)

According to second condition

number = 3 × Product of digit

10x + y = 3 × x × y

⇒ 10x + y = 3xy …(ii)

From (RBSESolutions.com) equation (i) putting y = 2x in equation

10x + 2x = 3x × 2x

⇒ 12x = 6x^{2}

⇒ 6x^{2} – 12x = 0

⇒ 6x (x – 2) = 0

when 6x = 0, then x = 0

when x – 2 = 0, then x = 2

Hence, given number is two digit number,

so x ≠ 0 and hence x = 2

⇒ y = 2 × 2 = 4 [∵ y = 2x]

Hence, required number = 10x + y = 10 × 2 + 4 = 24

**Long Answer Type Questions**

Question 1.

A pillar has to be fitted at the point of the (RBSESolutions.com) boundary of a circular park of diameter 13 m. Such that two gates A and B situated at both ends of diameter having a difference in distance of 7 m from this pillar. Is this possible? If yes, find the distance of the pillar from both gates. **(M.S.B. Raj 2013)**

Solution

Let pillar is fitted at point C and distance from gate B to point C = x m.

According to the question, the difference (RBSESolutions.com) in the distance from gate A and B to the pillar is 7 m.

AC = (x + 7) m

AB is diameter

∠ACB = 90° (∵ angle in semi circle is right angle)

Now in right angled triangle ACB

AB^{2} = AC^{2} + BC^{2} (By pythagorus Theorem)

⇒ 132 = (x + 7)^{2} + x^{2}

⇒ 169 = x^{2} + 49 + 14x + x^{2}

⇒ 169 = 2x^{2} + 14x + 49

⇒ 2x^{2} + 14x + 49 – 169 = 0

⇒ 2x^{2} + 14x – 120 = 0

⇒ x^{2} + 7x – 60 = 0

Now b^{2} – 4ac = (7)^{2} – 4 × 1 × (-60) = 49 + 240 = 289 > 0

Thus, given quadratic equation has two (RBSESolutions.com) real roots and so pillar can be fitted at boundary of park.

Now x^{2} + 7x – 60 = 0

⇒ x^{2} + (12 – 5)x – 60 = 0

⇒ x^{2} + 12x – 5x – 60 = 0

⇒ (x^{2} + 12x) – (5x + 60) = 0

⇒ x (x + 12) – 5(x + 12) = 0

⇒ (x + 12) (x – 5) = 0

⇒ x + 12 = 0 and x – 5 = 0

⇒ x = – 12 and x = 5

x is distance between pillar and gate B so Ignore x = -12

x = 5 m and x + 7 = 5 + 7 = 12 m.

Hence distance from pillar to gate A = 12 m.

and from pillar to gate B = 5 m.

Question 2.

The difference of square of two (RBSESolutions.com) numbers is 180. Square of the smaller number is 8 times the larger number. Find two numbers.

Solution

Let larger number = x

Smaller number = y

According to first condition of question

x^{2} – y^{2} = 180 …(i)

According to second condition of question

y^{2} = 8x …(ii)

Putting value of y^{2} from equation (ii) in equation (i)

x^{2} – 8x = 180

⇒ x^{2} – 8x – 180 = 0

Comparing it by ax^{2} + bx + c = 0

a = 1, b = -8, c = -180

Then by (RBSESolutions.com) quadratic formula

Thus x = 18 and -10

When x = 18, then from equation (ii)

y^{2} = 8 × 18 = 144

⇒ y = ±√144

⇒ y = ± 12

When x = -10, then (RBSESolutions.com) from equation (ii)

y^{2} = 8 × (-10)

⇒ y^{2} = -80 (not possible)

∴ y = ± 12

y = + 12 and – 12

Hence, required numberes will be 18 and 12 or 18, -12

Question 3.

The sum of areas of two squares is 468 sq.m. If the difference between their perimeter is 24 m, then find sides of both squares.

Solution

Let the side of a square is x m.

Perimeter of that square = 4x m

Difference in peimeter is 24 m

Perimeter of second square = 4x + 24 m

Then, side of (RBSESolutions.com) second square = \(\frac { 4x+24 }{ 4 }\) = \(\frac { 4(x+6) }{ 4 }\) = (x + 6) m

Area of first square = x^{2} sq m

Area of second square = (x + 6)^{2} sq m = x^{2} + 12x + 36 sq m

Sum of areas of both squares = 468 sq m

x^{2} + (x^{2} + 12x + 36) = 468

⇒ 2x^{2} + 12x + 36 – 468 = 0

⇒ 2x^{2}+ 12x – 432 =0

⇒ 2(x^{2} + 6x – 216) = 0

⇒ x^{2} + 6x – 216 = 0

⇒ x^{2} + 18x – 12x – 216 = 0

⇒ x(x + 18) – 12(x + 18) = 0

⇒ (x + 18)(x – 12) = 0

when x + 18 = 0, then x = -18 (not possible)

or x – 12 = 0, then x = 12

∴ x = 12

Side of smaller square = 12 m

and sideof larger square = x + 6 = 12 + 6 = 18 m

Thus sides of both (RBSESolutions.com) squares are 12 m. and 18 m respatively

**H.C.F. and L.C.M.**

**Multiple Choice Questions**

Question 1.

H.C.F. of 4x^{2}y and x^{3}y^{2} will be

(A) x^{2}y

(B) x^{2}y^{2}

(C) 4x^{3}y^{2}

(D) 4x^{2}y^{2}

Solution

Option (C) is correct.

Question 2.

H.C.F. of x^{2} – 4 and x^{2} + 4x + 4 will be

(A) (x – 2)

(B) (x – 4)

(C) (x + 2)

(D) (x + 4)

Solution

Option (C) is (RBSESolutions.com) correct.

Question 3.

H.C.F. of 36a^{5}b^{2} and 90a^{3}b^{4} will be

(A) 36a^{3}b^{2}

(B) 18a^{3}b^{2}

(C) 90a^{3}b^{4}

(D) 180a^{5}b^{4}

Solution

36a^{5}b^{2} = 3 × 3 × 2 × 2 × a^{5} × b^{2}

90a^{3}b^{4} = 3 × 3 × 2 × 5 × a^{3} × b^{4} = 3 × 3 × 2 × a^{3} × b^{2}

H.C.F. = 18a^{3}b^{2}

Hence, Option (B) is correct.

Question 4.

L.C.M. of x^{2} – 1 and x^{2} – x – 2 will be

(A) (x^{2} – 1)(x – 2)

(B) (x^{2} – 1)(x + 2)

(C) (x – 1)^{2 }(x + 2)

(D) (x + 1)^{2 }(x – 2)

Solution

x^{2} – 1 = (x – 1)(x + 1) …(i)

x^{2} – x – 2 = x^{2} – 2x + x – 2 = x (x – 2) + 1 (x – 2) = (x + 1)(x – 2) …(ii)

from eqn (i) and (ii)

(x + 1)(x – 1 )(x – 2)

L.C.M. = (x^{2} – 1)(x – 2)

Hence, Option (A) is correct.

Question 5.

If 5p^{2}q and 15pq^{2}r^{2} are two (RBSESolutions.com) expression 5pq, then L.C.M. will be

(A) 75p^{2}q^{3}r^{2}

(B) 5p^{2}q^{2}r^{2}

(C) 15p^{2}q^{2}r^{2}

(D) 15p^{3}q^{2}r^{2}

Solution

Hence, option (C) is correct.

Question 6.

H.C.F. of expression x^{2} – 1 and x + 1 will be

(A) x – 1

(B) x + 1

(C) (x^{2} – 1)(x + 1)

(D) (x – 1)(x + 1)

Solution

Option (B) is correct.

Question 7.

If (2 + p) and 100 – 25p^{2} are (RBSESolutions.com) two expression, then their L.C.M. will be

(A) 100 – 25p^{2}

(B) 2 + p

(C) 98 – 25p^{2}

(D) (100 – 25p^{2}) (2 + p)

Solution

First expression = (2 + p)

and other expression = 100 – 25p^{2}

= (10 – 5p) (10 + 5p)

= 5(2 – p) × 5(2 + p)

= 25(2 – p) (2 + p)

Thus L.C.M. of two expression = 100 – p^{2}

Hence, Option (A) is correct.

**Very Short/Short Answer Type Questions**

Question 1.

If one expression is u(x) and (RBSESolutions.com) other is v(x) their H.C.F. is r(x), then find L.C.M.

Solution

L.C.M. = \(\frac { u(x)\times v(x) }{ r(x) }\)

Question 2.

Find the H.C.F. of the following :

(i) x^{2} – 4 and x^{2} + 4x + 4

(ii) 4x^{4} – 16x^{3} + 12x^{2} and 6x^{3} + 6x^{2} – 72x

Solution

(i) x^{2} – 4 = (x + 2)(x – 2)

x^{2} + 4x + 4 = (x + 2)^{2}

H.C.F. of coefficient = 1

H.C.F. of other factors = (x + 2)^{1} = x + 2

H.C.F. = x + 2

(ii) 4x^{4} – 16x^{3} + 12x^{2} = 4x^{2}(x^{2} – 4x + 3) = 4x^{2}(x – 1)(x – 3)

6x^{3} + 6x^{2} – 72x = 6x(x^{2} + x – 12) = 6x(x + 4)(x – 3) = 2x(x – 3) [because H.C.F. of coefficients is 2]

= 2x^{2} – 6x

Question 3.

Find the L.C.M. of the (RBSESolutions.com) following

(i) x^{2} – 1 and x^{4} – 1

(ii) (x + 1)^{2} (x + 5)^{3} and x^{2} + 10x + 25

(iii) 6(x^{2} – 3x + 2) and 18(x^{2} – 4x + 3)

Solution

Question 4.

L.C.M. of two quadratic (RBSESolutions.com) equations is (x^{2} – y^{2}) (x^{2} + xy + y^{2}) and H.C.F is (x – y). Find expressions

Solution

L.C.M. = (x^{2} – y^{2}) (x^{2} + xy + y^{2}) = (x – y)(x + y)(x^{2} + xy + y^{2})

and H.C.F. = (x – y)

In L.C.M. and H.C.F. common factor is (x – y)

First expression = (x – y) (x + y) = x^{2} – y^{2}

and second egression = (x – y) (x^{2} + xy + y^{2})

= x^{3} – x^{2}y + x^{2}y – xy^{2} + xy^{2} – y^{3}

= (x^{3} – y^{3})

Hence, two expression are (x^{2} – y^{2}) and (x^{3} – y^{3})

Question 5.

Least common multiples of two (RBSESolutions.com) polynomials is x^{3} – 3x^{2} + 3x – 2 and the highest common factor is x – 2. If one polynomial is x^{2} – x + 1, find the other.

Solution

Question 6.

Find L.C.M and H.C.F. of the following (RBSESolutions.com) expression

x^{2} + 6x + 9, x^{2} – x – 12, x^{3} + 4x^{2} + 4x + 3

Solution

Expression x^{2} + 6x + 9 = (x)^{2} + 2 × x × 3 + (3)^{2}

x^{3} + 4x^{2} + 4x + 3 = (x + 3)(x^{2} + x + 1) …(iii)

common (RBSESolutions.com) factor of eqn. (i), (ii) and (iii) = (x + 3)

Thus, required H.C.F. = (x + 3)

in eqns. (i), (ii) (iii)

= (x + 3)^{2} (x – 4) (x^{2} + x + 1)

= (x^{2} + 6x + 9) (x^{3} – x^{2} + x – 4x^{2} – 4x – 4)

Hence, required L.C.M = (x^{2} + 6x + 9) (x^{3} – 5x^{2} – 3x – 4)

We hope the given RBSE Solutions for Class 10 Maths Chapter 3 Polynomials Additional Questions will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Maths Chapter 3 Polynomials Additional Questions, drop a comment below and we will get back to you at the earliest.

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