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) = x4 – x3 – 4x2 + kx + 10 then find k.
(A) 2
(B) -2
(C) -1
(D) 1
Solution
f(2) = 24 – 23 – 4 × 22 + 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) x2 – x – 20
(B) x2 + x – 20
(C) x2 – x + 20
(D) x2 + x + 20
Solution
Let α = – 5 and β = 4
Then α + β = -5 + 4 = -1 and αβ = -5 x 4 = -20
Quadratic polynomial x2 – (α + β) x + αβ = 0
⇒ x2 – (-1) x + (-20) = 0
⇒ x2 + x – 20 = 0
Hence, option (B) is correct.
Question 3.
Expression (x – 3) will be a (RBSESolutions.com) factor of polynomial f(x) = x3 + x2 – 17x + 15 if:
(A) f(3) = 0
(B) f(-3) = 0
(C) f(2) = 0
(D) f(-2) = 0
Solution
f(x) = x3 + x2 – 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 x3 – 3x2 + kx – 10 then value of k will be:
(A) -8
(B) -7
(C) 5
(D) 8
Solution
Let p(x) = x3 – 3x2 + 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 3y3 + 8y2 – 1 is
(A) 1
(B) (-1)
(C) 0
(D) None of these
Solution
Let p(y) = 3y3 + 8y2 – 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 2x2 + kx + √2, then k will be
(A) 2 + √2
(B) 2 – √2
(C) – (2 + √2)
(D) – (2 – √2)
Solution
Let p{x) = 2x2 + 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[x2 – (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 ax2 + 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) = x2 – 5x + k such that α – β = 1, then find k.
Solution
If α, β are zeros of polynomial x2 – 5x + k.
Question 6.
Find the condition for which zeros of (RBSESolutions.com) polynomial, p(x) = ax2 + bx + c are inverse of each other. (CBSE 2012)
Solution
Let α is first zero of polynomial of p(x) = ax2 + 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, x3 – 3x2 + x + 1, then find a and b.
Solution
Given polynomial = x3 – 3x2 + 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 x2 + x – 2 and (RBSESolutions.com) verify the relationship between zeros and coeffcient (M.S.B. Raj. 2016)
Solution
Given quadratic polynomial
f(x) = x2 + x – 2 = x2 + 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 ax2 + 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 x3 – 3x2 + 3x – 5 by x – 1 – x2 and test (RBSESolutions.com) division algorithm. (M.S.B. Raj. 2013)
Solution
Question 10.
Find all the zeros of (RBSESolutions.com) polynomial x3 – 6x2 + 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) = x3 – 6x2 + 11x – 6 has two zeros 1 and 2.
Thus (x – 1) (x – 2) = (x2 – 3x + 2), will be a factor of f(x).
Now dividing polynomial f(x) by x2 – 3x + 2
Using division algorithm for polynomial
Dividend = Divisor × Quotient + Remainder
x3 – 6x2 + 11x – 6 = (x2 – 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) = 2x2 + 5x + k which (RBSESolutions.com) satisfies the relation α2 + β2 + 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 ax2 + bx + c = 0, a ≠ 0 will not be real if
(A) b2 < 4ac
(B) b2 > 4ac
(C) b2 = 4ac
(D) b = 4ac
Solution
Option (A) is correct.
Question 2.
If \(\frac { 1 }{ 2 }\) is one root of (RBSESolutions.com) quadratic equation x2 + 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 2x2 – 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 :
2x2 – x – 6 = 0
⇒ 2x2 – (4 – 3) x – 6 = 0
⇒ 2x2 – 4x + 3x – 6 = 0
⇒ (2x2 – 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 2x2 – 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 x2 – 9 = 0 are
(A) √3
(B) -√3
(C) 9
(D) ±3
Solution
x2 – 9 = 0
⇒ x2 = 9
⇒ x = ±√9
⇒ x = ± 3
Hence, option (D) is correct.
Question 6.
Quadratic (RBSESolutions.com) equation 2x2 – √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 2x2 + x – 6 = 0 will be
(A) -3
(B) -7
(C) 2
(D) 0
Solution
Equation 2x2 + 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) 2x2 – 3x + 6 = 0
(B) -x2 + 3x – 3 = 0
(C) -√2 x2 – \(\frac { 3 }{ \surd 2 }\) x + 1 = 0
(D) 3x2 – 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 x2 – 4x = 0 are
(A) 4, 4
(B) 2, 2
(C) 0, 4
(D) 0, 2
Solution
x2 – 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 px2 + qx + r = 0, p ≠ 0 has equal roots if
(A) p2 < 4pr
(B) p2 > 4qr
(C) q2 = 4pr
(D) p2 = 4qr
Solution
Option (C) is correct.
Solution 11.
Root of quadratic equation (x2 + 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 x2 + 3ax + k = 0 has x = -a as (RBSESolutions.com) solution, then k will be
(A) 2a2
(B) 0
(C) 2
(D) -2a
Solution
Option (A) is correct.
Very Short/Short Answer Type Questions
Question 1.
For quadratic equation ax2 + bx + c = 0, a ≠ 0, at which nature of roots depends ?
Solution
Discriminant (D) = b2 – 4ac
Question 2.
Describe nature of (RBSESolutions.com) roots of quadratic equation ax2 + bx + c = 0, a ≠ 0
Solution
(i) If (b2 – 4ac) > 0, then roots will be real and distinct.
(ii) If (b2 – 4ac) = 0, then roots will be equal and real.
(iii) If (b2 – 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 – x2 = 15
⇒ x2 – 8x + 15 = 0
⇒ x2 – (5 + 3)x + 15 = 0
⇒ x2 – 5x – 3x + 15 = 0
⇒ (x2 – 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 x2 + 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) x2 + 2(k – 12) x + 2 = 0 has equal and real roots.
Solution
Given
(k – 12) x2 + 2(k – 12)x + 2 = 0
Comparing it with quadratic equation ax2 + bx + c = 0
a = (k – 12),b = 2(k – 12), c = 2
Discriminant (D) = b2 – 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 = x2 + (x + 7)2
⇒ 4x2 + 4x + 1 = 2x2 + 14x + 49
⇒ 2x2 – 10x – 48 = 0
⇒ x2 – 5x – 24 = 0
⇒ x2 – 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:
x2 – 4ax – b2 + 4a2 = 0. [CBSE 2012]
Solution
Given quadratic equation x2 – 4ax – b2 + 4a2 = 0
Comparing it by quadratic (RBSESolutions.com) standard equation Ax2 + Bx + C = 0
A = 1, B = -4a and C = -(b2 – 4a2)
By Shridhar Acharya formula
Question 9.
Solve for x: √3 x2 – 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 × \(\frac { 1 }{ x }\)
⇒ x2 + 12x = 160
⇒ x2 + 12x – 160 = 0
⇒ x2 + (20 – 8) x – 160 = 0
⇒ x2 + 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 = 6x2
⇒ 6x2 – 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
AB2 = AC2 + BC2 (By pythagorus Theorem)
⇒ 132 = (x + 7)2 + x2
⇒ 169 = x2 + 49 + 14x + x2
⇒ 169 = 2x2 + 14x + 49
⇒ 2x2 + 14x + 49 – 169 = 0
⇒ 2x2 + 14x – 120 = 0
⇒ x2 + 7x – 60 = 0
Now b2 – 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 x2 + 7x – 60 = 0
⇒ x2 + (12 – 5)x – 60 = 0
⇒ x2 + 12x – 5x – 60 = 0
⇒ (x2 + 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
x2 – y2 = 180 …(i)
According to second condition of question
y2 = 8x …(ii)
Putting value of y2 from equation (ii) in equation (i)
x2 – 8x = 180
⇒ x2 – 8x – 180 = 0
Comparing it by ax2 + 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)
y2 = 8 × 18 = 144
⇒ y = ±√144
⇒ y = ± 12
When x = -10, then (RBSESolutions.com) from equation (ii)
y2 = 8 × (-10)
⇒ y2 = -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 = x2 sq m
Area of second square = (x + 6)2 sq m = x2 + 12x + 36 sq m
Sum of areas of both squares = 468 sq m
x2 + (x2 + 12x + 36) = 468
⇒ 2x2 + 12x + 36 – 468 = 0
⇒ 2x2+ 12x – 432 =0
⇒ 2(x2 + 6x – 216) = 0
⇒ x2 + 6x – 216 = 0
⇒ x2 + 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 4x2y and x3y2 will be
(A) x2y
(B) x2y2
(C) 4x3y2
(D) 4x2y2
Solution
Option (C) is correct.
Question 2.
H.C.F. of x2 – 4 and x2 + 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 36a5b2 and 90a3b4 will be
(A) 36a3b2
(B) 18a3b2
(C) 90a3b4
(D) 180a5b4
Solution
36a5b2 = 3 × 3 × 2 × 2 × a5 × b2
90a3b4 = 3 × 3 × 2 × 5 × a3 × b4 = 3 × 3 × 2 × a3 × b2
H.C.F. = 18a3b2
Hence, Option (B) is correct.
Question 4.
L.C.M. of x2 – 1 and x2 – x – 2 will be
(A) (x2 – 1)(x – 2)
(B) (x2 – 1)(x + 2)
(C) (x – 1)2 (x + 2)
(D) (x + 1)2 (x – 2)
Solution
x2 – 1 = (x – 1)(x + 1) …(i)
x2 – x – 2 = x2 – 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. = (x2 – 1)(x – 2)
Hence, Option (A) is correct.
Question 5.
If 5p2q and 15pq2r2 are two (RBSESolutions.com) expression 5pq, then L.C.M. will be
(A) 75p2q3r2
(B) 5p2q2r2
(C) 15p2q2r2
(D) 15p3q2r2
Solution
Hence, option (C) is correct.
Question 6.
H.C.F. of expression x2 – 1 and x + 1 will be
(A) x – 1
(B) x + 1
(C) (x2 – 1)(x + 1)
(D) (x – 1)(x + 1)
Solution
Option (B) is correct.
Question 7.
If (2 + p) and 100 – 25p2 are (RBSESolutions.com) two expression, then their L.C.M. will be
(A) 100 – 25p2
(B) 2 + p
(C) 98 – 25p2
(D) (100 – 25p2) (2 + p)
Solution
First expression = (2 + p)
and other expression = 100 – 25p2
= (10 – 5p) (10 + 5p)
= 5(2 – p) × 5(2 + p)
= 25(2 – p) (2 + p)
Thus L.C.M. of two expression = 100 – p2
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) x2 – 4 and x2 + 4x + 4
(ii) 4x4 – 16x3 + 12x2 and 6x3 + 6x2 – 72x
Solution
(i) x2 – 4 = (x + 2)(x – 2)
x2 + 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) 4x4 – 16x3 + 12x2 = 4x2(x2 – 4x + 3) = 4x2(x – 1)(x – 3)
6x3 + 6x2 – 72x = 6x(x2 + x – 12) = 6x(x + 4)(x – 3) = 2x(x – 3) [because H.C.F. of coefficients is 2]
= 2x2 – 6x
Question 3.
Find the L.C.M. of the (RBSESolutions.com) following
(i) x2 – 1 and x4 – 1
(ii) (x + 1)2 (x + 5)3 and x2 + 10x + 25
(iii) 6(x2 – 3x + 2) and 18(x2 – 4x + 3)
Solution
Question 4.
L.C.M. of two quadratic (RBSESolutions.com) equations is (x2 – y2) (x2 + xy + y2) and H.C.F is (x – y). Find expressions
Solution
L.C.M. = (x2 – y2) (x2 + xy + y2) = (x – y)(x + y)(x2 + xy + y2)
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) = x2 – y2
and second egression = (x – y) (x2 + xy + y2)
= x3 – x2y + x2y – xy2 + xy2 – y3
= (x3 – y3)
Hence, two expression are (x2 – y2) and (x3 – y3)
Question 5.
Least common multiples of two (RBSESolutions.com) polynomials is x3 – 3x2 + 3x – 2 and the highest common factor is x – 2. If one polynomial is x2 – x + 1, find the other.
Solution
Question 6.
Find L.C.M and H.C.F. of the following (RBSESolutions.com) expression
x2 + 6x + 9, x2 – x – 12, x3 + 4x2 + 4x + 3
Solution
Expression x2 + 6x + 9 = (x)2 + 2 × x × 3 + (3)2
x3 + 4x2 + 4x + 3 = (x + 3)(x2 + 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) (x2 + x + 1)
= (x2 + 6x + 9) (x3 – x2 + x – 4x2 – 4x – 4)
Hence, required L.C.M = (x2 + 6x + 9) (x3 – 5x2 – 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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