RBSE Class 10 Maths Important Questions Chapter 2 Polynomials

Rajasthan Board RBSE Class 10 Maths Important Questions Chapter 2 Polynomials Important Questions and Answers.

RBSE Class 10 Maths Chapter 2 Important Questions Polynomials

Objective Type Questions

Question 1.
The zero of the polynomial p(x), where p(x) = ax + 1, a ≠ 0, is—
(A) 1
(B) – a
(C) 0
(D) \(-\frac{1}{a}\)
Answer:
(D) \(-\frac{1}{a}\)

Question 2.
The maximum number of the zeroes of a cubic polynomial can be—
(A) 1
(B) 2
(C) 3
(D) 4
Answer:
(C) 3

Question 3.
If α, β and γ and the zeroes of the cubic polynomial ax³ + bx² + cx + d = 0, then the value of α + β + γ will be—
(A) a
(B) b
(C) c
(D) \(-\frac{b}{a}\)
Answer:
(D) \(-\frac{b}{a}\)

Question 4.
The quotient and the remainder on dividing x⁴ – 3x² + 2x + 5 by x – 1 using division algorithm will be—
(A) quotient = x⁴ – 3x² + 2x + 5,
remainder = 6
(B) quotient = x³ – 3x + 2,
remainder = – 5
(C) quotient = x³ + x² – 2x,
remainder = 5
(D) quotient = 3x² + 2x + 5,
remainder = 3
Answer:
(C) quotient = x³ + x² – 2x,
remainder = 5

Question 5.
The expression (x – 3) will be a factor of the polynomial p(x) = x³ + x² – 17x + 5 if—
(A) p(3) = 0
(B) p(- 3) = 0
(C) p(-3) = 3
(D) p(- 3) = – 3
Answer:
(A) p(3) = 0

Question 6.
If α, β and γ are the zeroes of the cubic polynomial ax³ + bx² + cx + d, then the value of αβ + βγ + γα will be—
(A) c
(B) \(\frac{c}{a}\)
(C) a
(D) ac
Answer:
\(\frac{c}{a}\)

Question 7.
If x³ + a³ is divided by (x + a) then the remainder will be—
(A) x² + a²
(B) x + a
(C) 0
(D) ²
Answer:
(C) 0

Question 8.
If the zeroes of the cubic polynomial ax³ + bx² + cx + d be α, β and γ, then the value of αβγ will be—
(A) \(-\frac{d}{a}\)
(B) d
(C) a
(D) ad
Answer:
\(-\frac{d}{a}\)

Very Short Answer Type Questions

Question 1.
What is a zero of a polynomial?
Answer:
A real number a will be called a zero of any polynomial p(x) if p(a) = 0, i.e., that value of the variable for which the value of the polynomial becomes zero is called a zero of the polynomial.

Question 2.
Given an example of a linear polynomial.
Answer:
ax + b, where a, b ∈ R and a ≠ 0

Question 3.
If the zeroes of the quadratic polynomial ax² + bx + c be α and β then write the value of α + β and αβ.
Answer:
α + β = \(-\frac{b}{a}\) and αβ = \(\frac{c}{a}\)

Question 4.
Write division algorithm.
Answer:
For given polynomial p(x) and non-zero polynomial g(x), there exist two polynomials q{x) and r(x) such that p(x) = g(x) q(x) + r(x)
Where r(x) = 0 or deg r(x) < deg g(x)

Question 5.
What is called a polynomial?
Answer:
Algebraic expressions consisting of one or more than one terms are called polynomials if in its terms the power of any variable is not negative.

Question 6.
Given an example of a quadratic polynomial.
Answer:
y² – 2

Question 7.
Write the definition of a cubic polynomial with example.
Answer:
A polynomial in which the maximum number of powers is three is called a cubic polynomial.
For example 3x³ – 2x² + x – 1

Question 8.
What is the value of a polynomial?
Answer:
If p(x) is any polynomial in the variable x and k is any real number, then the real number that is obtained by replacing, x by k in p(x) is called the value of p(x) at x = k and it is devoted by p(k).

Question 9.
If one factor of the expression x³ – 2x + 1 is (x – 1) then write the remainder.
Answer:
Let (x – 1) be a factor of the expression f(x) = x³ – 2x + 1.
Then, f(1) = (1)³ – 2(1) + 1 = 1 – 2 + 1 = 0
∴ Remainder = 0

Question 10.
Find the zeroes of the quadratic polynomial x² + 7x + 10.
Solution:
x² + 7x + 10
= x² + 5x + 2x + 10
= x(x + 5) + 2(x + 5)
= (x + 2) (x + 5)
∴ The value of x² + 7x + 10 is zero when x + 5 = 0 or x + 2 = 0
i.e., When x = – 5 orx = -2
∴ Zeroes of x² + 7x + 10 are – 5 and – 2.

Question 11.
If 2 is a factor of the polynomial f(x) = x⁴ – x³ – 4x² + kx + 10, then write the value of k.
Solution:
f(2) = 2⁴ – 2³ – 4 × 2² + k × 2 + 10
⇒ 0 = 16 – 8 – 16 + 2k + 10
⇒ 0 = 2k + 2
⇒ k = – 1

Question 12.
If (x – 2) is a factor of the expression x² + 2x – a then write the value of a.
Answer:
Putting x – 2 = 0 or x = 2 in the expression, its value must be zero. Therefore
(2)² + 2 × 2 – a = 0
⇒ 4 + 4 – a = 0
⇒ a = 8

Question 13.
If x³ + y³ + z³ – 3xyz = (x + y + z) (x² + y² + z² + k), then write the value of k.
Answer:
k = – xy – yz. – zx

Question 14.
Find a quadratic polynomial whose sum and product of zeroes are – 4 and 3 respectively.
Solution:
x² – (sum of the zeroes) x + product of the zeroes
= x² – (- 4)x + 3
= x² + 4x + 3

Question 15.
Write the maximum number of zeroes of a polynomial of degree n.
Answer:
n zeroes.

Question 16.
What type is the shape of the graph of a quadratic polynomial ax² + bx + c?
Answer:
Shape of a parabola.

Question 17.
Find a polynomial whose zeroes are – 5 and 4.
Solution:
[x – (- 5)] (x – 4)
⇒ (x + 5) (x – 4)
⇒ x² – 4x + 5x – 20
⇒ x² + x – 20

Question 18.
If zeroes α and β of the polynomial f(x) = x² – 5x + k are such that α – β = 1, then write the value of k.
Solution:
Given that α and β are the zeroes of the polynomial x² – 5x + k
α + β = \(-\left(\frac{-5}{1}\right)\) = 5
αβ = \(\frac{k}{1}\) = k
α – β = 1 or (α – β )² = 1
or (α + β )² – 4αβ = 1
or 25 – 4k = 1
4k = 24 ∴ k= \(\frac{24}{4}\) = 6

Question 19.
Find the zeroes of the polynomial x² – 9.
Solution:
x² – 9 = (x)² – (3)²
= (x + 3) (x – 3)
For zeroes of the polynomial
x + 3 = 0 ⇒ x = – 3
and x – 3 = 0 ⇒ x = 3
Therefore zeroes of the polynomial x² – 9 = ± 3

Question 20.
If the product of the zeroes of the polynomial ax² – 6x – 6 be 6, then find the value of a.
Solution:
The zeroes of the given polynomial are α and β . Then, from product of the zero

Short Answer Type Questions

Question 1.
Without applying the process of division prove that
(i) x² + (a – 3) x – 3a is exactly divisible by (x + a)
(ii) 3x³ + 11x² + x – 15 is exactly divisible by (x – 1)
Solution:
(i) If the given expression is exactly divisible by (x + a), then putting the value x = – a, the value of the expression will become zero. Therefore putting the value x = – a in the expression
(- a)² + (a – 3) (- a) – 3a
= a² – a² + 3a – 3a = 0
Hence the expression x² + (a – 3) x – 3a is exactly divisible by (x + a).

(ii) Putting x = 1 in the expression
3(1)³ + 11(1)² + 1 – 15
= 3 + 11 + 1 – 15 = 0
Hence the expression 3x³ + 11x² + x – 15 is exactly divisible by the expression (x – 1).

Question 2.
For what value of a is the polynomial x³ + 2x² – 3ax – 8 completely divisible by the expression (x – 4).
Solution:
We know that on dividing by (x – 4), the remainder f(4) will be obtained
f(x) = x³ + 2x² – 3ax – 8
f(4) = (4)³ + 2(4)² – 3a × 4 – 8
= 64 + 32 – 12a – 8 = 0
0 = 88 – 12a
or 12a = 88
a = \(\frac{88}{12}\) = \(\frac{22}{3}\)

Question 3.
Prove that the zeroes of the polynomial x² + 2x + 3 do not exist.
Solution:
Suppose that
f(x) = x² + 2x + 3
f(x)= {x² + 2 × x+ 1} + 2
= (-x + 1)² + 2
Here for real values of x, the value of (x + 1)² will not be negative. Therefore (x + 1)² will always be greater than zero. Hence the value of f(x) will always be greater than 2.

Question 4.
For what value of p will the polynomial px³ + 9x² + 6x – 1 be completely divisible by (3x + 2)?
Solution:
It is given that (3x + 2) is a factor of the given expression.
∴ 3x + 2 = 0
∴ 3x= – 2 ∴ x = \(\frac{-2}{3}\)
Putting the value x = \(\frac{-2}{3}\) in the expression f(x) = px³ + 9x²+ 6x – 1

Question 5.
Find a quadratic polynomial the sum and product of whose zeroes are 4 and 1 respectively.
Solution:
Given that
α + β = 4 and
αβ = 1
Where the zeroes of the quadratic polynomial are α and β.
∴ Quadratic polynomial = (x – α) (x – β)
= x² – (α + β ) x + αβ
= x² – 4x + 1
Hence the required polynomial = x² – 4x + 1

Question 6.
Find the sum and the product of the zeroes of the quadratic polynomial 2x² – 6x + 4.
Solution:
The given polynomial is 2x² – 6x + 4
Therefore,

Question 7.
If (x – 1) and (x + 2) are the factors of the polynomial 2x³ + mx² + nx – 14 then find the values of m and rt.
Solution:
Let f(x) = 2x³ + mx² + nx – 14
Then if (x – 1) and (x + 2) are the factors of f(x). Then
f(1) = 0 and f(- 2) = 0
Putting x = 1 in the polynomial
2x³ + mx² + nx – 14
f(1) = m²(1)³ + m(1)² + n( 1) – 14 = 0
f(1) = 2 + m + n – 14
⇒ m + n — 12 = 0 …..(1)
Putting x = – 2 in the polynomial
2x³ + mx² + nx – 14
f(- 2) = 2(- 2)³ + m(- 2)² + n(- 2) – 14 = 0
⇒ f(-2) = – 16 + 4m – 2n – 14 = 0
⇒ 4m – 2n – 30= 0 …..(2)
From equation (1) and (2)
m + n – 12 ……(3)
2m – n = 15 ……(4)
Adding
3m = 27
⇒ m = 9
From equation (3), n = 3
Therefore m = 9 and n = 3

Question 8.
If the polynomial x³ + lx + m is divided by (x – 1) or (x + 1), then the remainder 7 is left. Find the values of l and m.
Solution:
f(x) = x³ + lx + m and (x – 1) or (x + 1) are its factors when we subtract 7 from it. Hence
f(1)= 1 + l + m – 7 = 0
⇒ l + m – 6 = 0 ……..(1)
and f(- 1) = – 1 – l + m – 7 = 0
⇒ – l + m – 8=0 ……(2)
Solving equations (1) and (2), l = – 1 and m = 7

Question 9.
Find the zeroes of the quadratic polynomial x² + 7x + 10 and verify the relationship between the zeroes and the coefficients.
Solution:
According to the question
x² + 7x + 10
= (x + 2) (x + 5)
Therefore the value of x² + 7x + 10 is zero when x + 2 = 0 or x + 5 = 0, i.e., x = – 2 or x = – 5. Therefore, the zeroes of x² + 7x + 10 are – 2 and – 5. Now,
Sum of the zeroes = – 2 + (- 5)
= (-7)

Therefore the sum and product of zeroes – 2 and – 5 are the same as there are in it.
∴ The relationship between the zeroes and the coefficient of the polynomial is verified.

Question 10.
Find a quadratic polynomial the sum and product of whose zeroes are 0 and \(\sqrt{5}\) respectively.
Solution:
Let the zeroes of the quadratic polynomial be α and β,
Given that, α + β = 0
α β = \(\sqrt{5}\)
∵ α and β are the zeroes
∴ Quadratic polynomial = (x – α) (x – β)
= x² – (α + β) x + αβ
= x² – 0x + \(\sqrt{5}\)
= x² + 75
Hence the required polynomial = x² + \(\sqrt{5}\)

Question 11.
If the sum of the zeroes of the quadratic polynomial kx² + 5x + 3k is equal to their product, then find the value of k.
Solution:
According to the given polynomial kx² + 5x + 3k,
Sum of the zeroes (α + β) = \(-\frac{5}{k}\)
and product of the zeroes (αβ) = \(\frac{3k}{k}\) = 3
Now according to the question
\(-\frac{5}{k}\) = \(\frac{3}{1}\)
⇒ 3k = 5
⇒ k = \(-\frac{5}{3}\)

Question 12.
Divide 3x³ + x² + 2x + 5 by 1 + 2x + x².
Solution:
Here It is given that
f(x) = 3x³ + x² + 2x + 5
and g(x) = 1 + 2x + x² = x² + 2x + 1

Quotient = 3x – 5 and Remainder = 9x + 10

Long Answer Type Questions

Question 1.
Find the factors of the following polynomials using the remainder theorem :
(i) x³ – 2x² – 5x + 6
(ii) 2x³ – x² – 13x – 6
(iii) y³ – 7y + 6
(iv) x³ – 6x² + 3x + 10
Solution:
(i) f(x) = x³ – 2x² – 5x + 6
Putting x = 1
f(1)= 1³ – 2.1² – 5.1 + 6
= 1 – 2 – 5 + 6 = 0
∵ (x – 1) is a factor of f(x).
Hence expression = x³ – x² – x² + x – 6x + 6 [We arrange first two terms such that taking common (x – 1) is left, also onwards we make such pairs of two terms that this process continues.]
= x²(x – 1) – x (x – 1) – 6 (x – 1)
= (x – 1) (x² – x – 6)
= (x – 1) (x + 2) (x – 3)

Second Method—
∵ (x – 1) is one factor of f(x)
∴ by division process

So x³ – 2x² – 5x + 6 = (x – 1) (x² – x – 6)
= (x – 1) (x – 3) (x + 2)

(ii) f(x) = 2x³ – x² – 13x – 6
Putting the value x = – 2
2(- 2)³ – (-2)² – 13(- 2) – 6
= -16 – 4 + 26 – 6 = 0
Therefore (x + 2) will be a factor of f(x)
So, 2x³ + 4x² – 5x² – 10x – 3x – 6
= 2x²(x + 2) – 5x(x + 2) – 3(x + 2)
= (x + 2) (2x² – 5x – 3)
= (x + 2) {2x² – 6x + x – 3}
= (x + 2) {2x(x – 3) + 1(x – 3)}
= (x + 2) (x – 3) (2x + 1)

(iii) f(y) = y³ – 7y + 6
In the given expression. Putting y = 1
f(1) = (1)²³ – 7(1) + 6 = 1 – 7 + 6 = 0
Therefore (y – 1) will be a factor of the given expression.
Now we group in twos the terms of the given expression such that (y – 1) may become a common factor in these.
y³ – 7y + 6
= y³ – y² + y² – 7y + 6 (∵ y² – y² = 0)
= y²(y – 1) + y² – y + y – 7y + 6 (∵ y – y = 0)
= y²(y – 1) + y(y – 1) – 6y + 6
= y²(y – 1) + y(y – 1) – 6(y – 1)
In the whole expression taking (y – 1) common
(y – 1) (y² + y – 6)
= (y – 1) {y² + 3y – 2y – 6}
= (y – i) {y(y + 3) – 2(y + 3)}
= (y – 1) {(y – 2) (y + 3)}
= (y – 1) (y – 2) (y + 3)

(iv) f(x) = x³ – 6x² + 3x + 10
Putting x = – 1 in the given expression
(-1)³ – 6(-1)² + 3(-1) + 10
= -1 – 6 – 3 + 10
= – 10 + 10 = 0
Therefore (x + 1) will be a factor of the given expression.
Now we group in twos the given terms of the given expression such that (x + 1) may become a common factor in these. x³ – 6x² + 3x + 10
= x³ + x² – x² – 6x² + 3x + 10 (∵ x² – x² = 0)
= x² (x + 1) – 7x² – 7x + 7x + 3x + 10 (∵ 7x – 7x = 0)
= x²(x + 1) – 7x(x + 1) + 10x + 10
= x²(x + 1) – 7x(x + 1) + 10(x + 1)
Taking (x + 1) common in the whole expression
= (x + 1) (x² – 7x + 10)
= (x + 1) (x² – 5x – 2x + 10)
= (x + 1) {x(x – 5) – 2(x – 5)}
= (x + I) {(x – 2) (x – 5)}
= (x + 1) (x – 2) (x – 5)

Question 2.
If (x + 1) and (x – 2) are the factors of the polynomial x³ + kx² + hx + 6 then find the value of h and k.
Solution:
Let f(x) = x³ + kx² + hx + 6
If (x + 1) and (x – 2) are the factors of f(x), then
f(-1) = 0 and f(2) = 0
∴ f(-1) = (- 1)³ + k(-1)² + h(- 1) + 6 = 0
⇒ -1 + k – h + 6 = 0
⇒ k – h + 5=0 ….(1)
and f(2) = (2)³ + k(2)² + h(2) + 6 = 0
= 8 + 4k + 2h + 6 = 0
⇒ 4k + 2h + 14 = 0 …….(2)
Solving equation (1) and (2)
Multiplying equation (1) by 2 and adding to equation (2)

Putting the value of ft in equation (1)
k – h = – 5
⇒ – 4 – h = – 5
⇒ -h = -5 + 4 = -1
⇒ h = 1
Hence we get h = 1 and k = – 4.

Question 3.
One factor of the polynomial f(x) = 3x³ + ax² + 4x + b is (x + 2). If it is divided by (x – 3) then a remainder of – 5 is left. Find the values of a and b.
Solution:
f(x) = 3x³ + ax² + 4x + b
Its factor is (x + 2). Therefore,
f(- 2) = 3(- 2)³ + a(-2)² + 4(-2) + ² = 0
⇒ – 24 + 4a – 8 + b
⇒ 4a + b – 32 = 0 …..(1)
and adding + 5 the factor is (x – 3). Hence
f(x) = 3x³ + ax² + 4x + b
f(3) + 5 = 3(3)³ + a(3)² + 4 × 3 + b + 5 = 0
⇒ 81 + 9a + 12 + b + 5 = 0
⇒ 9a + b + 98 = 0 ……(2)
Solving equations (1) and (2)

Putting the value of a in equation (1)
4a + b = 32
4(- 26) + b = 32
– 104 + b = 32
b = 104 + 32 = 136
Therefore a = – 26, b = 136

Question 4.
Verify that 3, -1 and \(-\frac{1}{3}\) are the zeroes of the cubic polynomial p(x) = 3x³ – 5x² – 11x – 3, and then verify the relationship between the zeroes and the coefficients.
Solution:
Comparing the given polynomial with ax³ + bx² + cx + d, we get a = 3, b = -5, c = -11, d = -3.
Again,
p(3)= 3 × 3³ – (5 × 3²) – (11 × 3) – 3
= 81 – 45 – 33 – 3 = 0
p(- 1) = 3 × (- 1)³ – 5 × (- 1)² – 11 × (-1) – 3
= -3 – 5 + 11 – 3 = 0

Hence, 3, -1 and \(-\frac{1}{3}\) are the zeroes of 3x³ – 5x² – 11x – 3.
So, we take α = 3, β = – 1 and γ = \(-\frac{1}{3}\).
Now,

Hence the sum and product by the zeroes 3, – 1 and \(-\frac{1}{3}\) are the same as there are in it.
Therefore the relationship between the zeroes and the coefficients of the polynomial is verified.

Question 5.
Divide 3x² – x³ – 3x + 5 by x – 1 – x², and verify the division algorithm.
Solution:
The given polynomials are not in standard form. To carry out division, we first write both. The dividend and divisor in decreasing order of their degrees.
So, dividend = – x³ + 3x² – 3x + 5 and divisor = – x² + x – 1.
Division process is shown below :

We stop here since degree 0 of 3 is less than degree 2 of -x² + x – 1.
So by carrying out the division process, the remainder obtained is 3 and the quotient obtained is x – 2.
Now, divisor x quotient + remainder
= (- x² + x – 1) (x – 2) + 3
= -x³ + x² – x + 2x² – 2x + 2 + 3
= -x³ + 3x² – 3x + 5
= dividend
Hence, the division algorithm is verified.

Question 6.
Find all the zeroes of 2x⁴ – 3x³ – 3x² + 6x – 2, if you know that two of
its zeroes are \(\sqrt{2}\) and \(-\sqrt{2}\).
Solution:
Since two zeroes are \(\sqrt{2}\) and \(-\sqrt{2}\). Therefore (x – \(\sqrt{2}\) ) (x + \(\sqrt{2}\)) = x² – 2 is a factor of the given polynomial.
Therefore,

Now, by splitting – 3x, we factorise 2x² – 3x + 1 as (2x – 1) (x – 1). Therefore, its zeroes will be given by x = \(\frac{1}{2}\) and x = 1.
Therefore, the zeroes of the given polynomial are \(\sqrt{2}\), \(-\sqrt{2}\), \(\frac{1}{2}\) and 1.

Question 7.
Divide x³ – 3x² + 3x – 5 by x – 1 – x² and verify the division algorithm.
Solution:
Here f(x) = x³ – 3x² + 3x – 5
and g(x) = -x² + x – 1
Now

We find that
quotient q(x) = -x + 2
and remainder r(x) = -3
Now,
quotient × divisor + remainder
= (-x + 2) (-x² + x – 1) – 3
= x³ – x² + x – 2x² + 2x – 2 – 3
= x³ – 3x² + 3x – 5
= dividend
Hence, the division algorithm is verified.