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All questions of Polynomials for Class 9 Exam
Can you explain the answer of this question below: If α and β are the zeroes of the polynomial 5x2 – 7x + 2, then sum of their reciprocals is:- A:14/25
- B:7/5
- C:2/5
- D:7/2
The answer is d.
If α and β are the zeroes of the polynomial 5x2 – 7x + 2, then sum of their reciprocals is:
A:
14/25
B:
7/5
C:
2/5
D:
7/2
|
Anshu Shah answered |
We have 2 find (1/α + 1/β)
now 1/α + 1/β = (α + β)/ α β (taking LCM)
now by the given poly. we get
(α + β) = -b/a = 7/5
α β = c/a = 2/5
so, (α + β)/ α β = (7/5) / (2/5)
= 7/2
So, 1/α + 1/β = (α + β)/ α β = 7/2
Hence, 1/α + 1/β = 7/2
If one of the factors of x2 + x – 20 is (x + 5), then other factor is- a)(x – 4)
- b)(x – 5)
- c)(x – 6)
- d)(x – 7)
Correct answer is 'A'. Can you explain this answer?
If one of the factors of x2 + x – 20 is (x + 5), then other factor is
a)
(x – 4)
b)
(x – 5)
c)
(x – 6)
d)
(x – 7)
|
Arun Sharma answered |
Using mid-term splitting,
x2+x-20=x2+5x-4x-20=x(x+5)-4(x+5)
Taking common x+5
(x+5)(x-4) , so the other factor is x-4
x2+x-20=x2+5x-4x-20=x(x+5)-4(x+5)
Taking common x+5
(x+5)(x-4) , so the other factor is x-4
If the graph of a polynomial intersects the x – axis at three points, then the number of zeroes =- a)3
- b)0
- c)at most three
- d)at least three
Correct answer is option 'A'. Can you explain this answer?
If the graph of a polynomial intersects the x – axis at three points, then the number of zeroes =
a)
3
b)
0
c)
at most three
d)
at least three
|
Megha Roy answered |
If the graph of a polynomial intersects the x-axis at three points, then the number of zeroes are 3 because number of zeroes of the polynomial are the number of the coordinates of the points where its graph intersects the x-axis.
If p and q are the zeroes of the polynomial x2- 5x + k. Such that p - q = 1, find the value of K
- a)6
- b)7
- c)8
- d)9
Correct answer is option 'A'. Can you explain this answer?
If p and q are the zeroes of the polynomial x2- 5x + k. Such that p - q = 1, find the value of K
a)
6
b)
7
c)
8
d)
9
|
Zachary Foster answered |
Given α and β are the zeroes of the polynomial x2 − 5x + k
Also given that α − β = 1 → (1)
Recall that sum of roots (α + β) = −(b/a)
∴ α + β = 5 → (2)
Add (1) and (2), we get
α − β = 1
α + β = 5
2α = 6
∴ α = 3
Put α = 3 in α + β = 5
3 + β = 5
∴ β = 2
Hence 3 and 2 are zeroes of the given polynomial
Put x = 2 in the given polynomial to find the value of k ( Since 2 is a zero of the polynomial, f(2) will be 0 )
x2 − 5x + k = 0
⇒ 22 − 5(2) + k = 0
⇒ 4 − 10 + k = 0
⇒ − 6 + k = 0
∴ k = 6
Also given that α − β = 1 → (1)
Recall that sum of roots (α + β) = −(b/a)
∴ α + β = 5 → (2)
Add (1) and (2), we get
α − β = 1
α + β = 5
2α = 6
∴ α = 3
Put α = 3 in α + β = 5
3 + β = 5
∴ β = 2
Hence 3 and 2 are zeroes of the given polynomial
Put x = 2 in the given polynomial to find the value of k ( Since 2 is a zero of the polynomial, f(2) will be 0 )
x2 − 5x + k = 0
⇒ 22 − 5(2) + k = 0
⇒ 4 − 10 + k = 0
⇒ − 6 + k = 0
∴ k = 6
The number of zeroes of a cubic polynomial is- a)3
- b)at least 3
- c)2
- d)at most 3
Correct answer is option 'D'. Can you explain this answer?
The number of zeroes of a cubic polynomial is
a)
3
b)
at least 3
c)
2
d)
at most 3
|
Nilima singh answered |
The number of zeroes of a cubic polynomial is at most 3 because the highest power of the variable in cubic polynomial is 3, i.e. ax3+bx2+cx+d
If 5 is a zero of the quadratic polynomial, (x^2) - kx - 15 then the value of k isa) 2b)-2c) 4d)- 4Correct answer is option 'A'. Can you explain this answer?- 4
If 5 is a zero of the quadratic polynomial, (x^2) - kx - 15 then the value of k is
a) 2
b)-2
c) 4
d)- 4
Correct answer is option 'A'. Can you explain this answer?
|
Rohit Sharma answered |
p(x) = x2- kx - 15
Given: p(5) = 0
⇒ (5)2- k(5) - 15 = 0
⇒ 25 - 5k - 15 = 0
⇒ 5k = 10
⇒ k = 10/5 = 2
⇒ (5)2- k(5) - 15 = 0
⇒ 25 - 5k - 15 = 0
⇒ 5k = 10
⇒ k = 10/5 = 2
Thus, Value of k is 2
The three zeroes of the polynomial 2x3 + 5x2 – 28x – 15 _____- a)All three are not real numbers
- b)Are all Natural numbers
- c)Are all rational numbers
- d)Are all Integers
Correct answer is option 'C'. Can you explain this answer?
The three zeroes of the polynomial 2x3 + 5x2 – 28x – 15 _____
a)
All three are not real numbers
b)
Are all Natural numbers
c)
Are all rational numbers
d)
Are all Integers
|
Gaurav Kumar answered |
Since the solution is not a complex number, it is a real number. Since rational numbers include all integers, natural numbers and fractions , So the numbers are rational numbers.
If one of the zeroes of the cubic polynomial x3−7x+6 is 2, then the product of the other two zeroes is- a)2
- b)3
- c)– 3
- d)– 2
Correct answer is 'C'. Can you explain this answer?
If one of the zeroes of the cubic polynomial x3−7x+6 is 2, then the product of the other two zeroes is
a)
2
b)
3
c)
– 3
d)
– 2
|
Navya Patel answered |
Let α,β,γ are the zeroes of the given polynomial. Given: α = 2
Since αβγ = -d/a
Since αβγ = -d/a
If α,β be the zeros of the quadratic polynomial 2 – 3x – x2, then α + β =- a)2
- b)3
- c)1
- d)None of these
Correct answer is option 'D'. Can you explain this answer?
If α,β be the zeros of the quadratic polynomial 2 – 3x – x2, then α + β =
a)
2
b)
3
c)
1
d)
None of these
|
Pooja Shah answered |
If α and β are the zeros of the polynomial then
(x−α)(x−β) are the factors of the polynomial
Thus, (x−α)(x−β) is the polynomial.
So, the polynomial =x2 − αx − βx + αβ
=x2 − (α + β)x + αβ....(i)
Now,the quadratic polynomial is
2 − 3x − x2 = x2 + 3x − 2....(ii)
(x−α)(x−β) are the factors of the polynomial
Thus, (x−α)(x−β) is the polynomial.
So, the polynomial =x2 − αx − βx + αβ
=x2 − (α + β)x + αβ....(i)
Now,the quadratic polynomial is
2 − 3x − x2 = x2 + 3x − 2....(ii)
Now, comparing equation (i) and (ii),we get,
−(α + β) = 3
α + β = −3
−(α + β) = 3
α + β = −3
A polynomial of degree ____ is called a quadratic polynomial.- a)0
- b)2
- c)1
- d)3
Correct answer is option 'B'. Can you explain this answer?
A polynomial of degree ____ is called a quadratic polynomial.
a)
0
b)
2
c)
1
d)
3
|
Abhiram Gupta answered |
A polynomial of degree 2 is called a quadratic polynomial.
What value/s can x take in the expression k(x – 10) (x + 10) =0 where k is any real number.- a)100, -100
- b)Infinitely many
- c)Depends on value of k
- d)10, -10
Correct answer is option 'D'. Can you explain this answer?
What value/s can x take in the expression k(x – 10) (x + 10) =0 where k is any real number.
a)
100, -100
b)
Infinitely many
c)
Depends on value of k
d)
10, -10
|
|
Avinash Patel answered |
k(x – 10) (x + 10) =0
⇒ either k=0
Or x-10=0
Or x+10=0
Since we don’t know the value of k
So either x-10=0
x=10
Or x+10=0
x=-10
So values of x can be 10,-10
⇒ either k=0
Or x-10=0
Or x+10=0
Since we don’t know the value of k
So either x-10=0
x=10
Or x+10=0
x=-10
So values of x can be 10,-10
If “1” is a zero of the polynomial P(a) = x2a2 – 2xa + 3x – 2 , then x = ______
- a)2
- b)-2, 1
- c)+2, -2
- d)-2, 0
Correct answer is option 'B'. Can you explain this answer?
If “1” is a zero of the polynomial P(a) = x2a2 – 2xa + 3x – 2 , then x = ______
a)
2
b)
-2, 1
c)
+2, -2
d)
-2, 0
|
|
Pooja Shah answered |
The correct solution of this question is given below:
Here, P(a) = x2a2 - 2xa + 3x - 2
1 is a zero of P(a), so P(1) = 0
Therefore, x212 - 2x.1 + 3x - 2 = 0
x2 + x - 2 = 0
(x + 2)(x - 1) = 0
x = -2, 1
A polynomial of degree three is called ……- a)cubic polynomial
- b)quadratic polynomial
- c)linear polynomial
- d)zero polynomial
Correct answer is option 'A'. Can you explain this answer?
A polynomial of degree three is called ……
a)
cubic polynomial
b)
quadratic polynomial
c)
linear polynomial
d)
zero polynomial
|
Maitri Kulkarni answered |
A cubic polynomial is a polynomial of the form 
. A cubic polynomial is a polynomial of degree 3. A univariate cubic polynomial has the form . An equation involving a cubic polynomial is called a cubic equation. A closed-form solution known as the cubic formula exists for the solutions of an arbitrary cubic equation.
. A cubic polynomial is a polynomial of degree 3. A univariate cubic polynomial has the form . An equation involving a cubic polynomial is called a cubic equation. A closed-form solution known as the cubic formula exists for the solutions of an arbitrary cubic equation.
If one root of the polynomial p(y) = 5y2 + 13y + m is reciprocal of other, then the value of m is- a)6
- b)0
- c)5
- d)1/5
Correct answer is option 'C'. Can you explain this answer?
If one root of the polynomial p(y) = 5y2 + 13y + m is reciprocal of other, then the value of m is
a)
6
b)
0
c)
5
d)
1/5
|
Vivek Rana answered |
Let the roots be α and 1/α.
Product of roots = α(1/α) = 1
⇒ m/5 = 1
⇒ m = 5
⇒ m/5 = 1
⇒ m = 5
Quadratic polynomial having sum of it's zeros 5 and product of it's zeros – 14 is –- a)x2 – 5x – 14
- b)x2 – 10x – 14
- c)x2 – 5x + 14
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
Quadratic polynomial having sum of it's zeros 5 and product of it's zeros – 14 is –
a)
x2 – 5x – 14
b)
x2 – 10x – 14
c)
x2 – 5x + 14
d)
None of these
|
|
Amit Kumar answered |
The quadratic equation is of the form x2 - (sum of zeros) x + (product of zeros)
=x2 - 5x - 14
=x2 - 5x - 14
If (x + 1) is a factor of x2 – 3ax + 3a – 7, then the value of a is- a)-2
- b)0
- c)1
- d)-1
Correct answer is option 'C'. Can you explain this answer?
If (x + 1) is a factor of x2 – 3ax + 3a – 7, then the value of a is
a)
-2
b)
0
c)
1
d)
-1
|
|
Anjana Khatri answered |
Since (X+1) is a factor of x^2 - 3ax + 3a - 7
Therefore , x + 1 = 0
or, x=-1
Putting the value of x in the the p(x) = x^2 - 3ax + 3a - 7
or, (-1)^2 - 3(a)(-1) + 3a - 7 = 0
1+3a+3a-7=0
6a=6
or, a=1
Therefore, the value of a = 1.
Direction: Read the following text and answer the following questions on the basis of the same:The below picture are few natural examples of parabolic shape which is represented by a quadratic polynomial. A parabolic arch is an arch in the shape of a parabola. In structures, their curve represents an efficient method of load, and so can be found in bridges and in architecture in a variety of forms.
If α are 1/α the zeroes of the quadratic polynomial 2x2 – x + 8k, then k is- a)4
- b)1/4
- c)-1/4
- d)2
Correct answer is option 'B'. Can you explain this answer?
Direction: Read the following text and answer the following questions on the basis of the same:
The below picture are few natural examples of parabolic shape which is represented by a quadratic polynomial. A parabolic arch is an arch in the shape of a parabola. In structures, their curve represents an efficient method of load, and so can be found in bridges and in architecture in a variety of forms.
If α are 1/α the zeroes of the quadratic polynomial 2x2 – x + 8k, then k is
a)
4
b)
1/4
c)
-1/4
d)
2
|
|
Vikram Kapoor answered |
Given equation, 2x2 – x + 8k
Product of zeroes = c/a = 8k/2
So, 8k/2 = 1
k = 2/8
k = 1/4
Can you explain the answer of this question below:If (x + 1) is a factor of 2x3 + ax2 + 2bx + 1, then find the values of a and b given that 2a – 3b = 4.
- A:
a = 2, b = 2
- B:
a = 5, b = 5
- C:
a = 2, b = 5
- D:
a = 5, b = 2
The answer is d.
If (x + 1) is a factor of 2x3 + ax2 + 2bx + 1, then find the values of a and b given that 2a – 3b = 4.
a = 2, b = 2
a = 5, b = 5
a = 2, b = 5
a = 5, b = 2
|
|
Vikram Kapoor answered |
Given is the value of a and b.
given that (x+1) is a factor of p(x)
therefore, -1 is a zero of given p(x)
p(x) = 2x3 +ax2 +2bx +1
substituting the value of -1 in the given p(x), we get
p(x)=2*(-1)3 +a *(-1)2 +2*b*(-1)+ 1
= -2 + a -2b + 1
= -1 + a - 2b
or,a - 2b = 1
also given that 2a - 3b = 4
so we got two equations;
a - 2b = 7 ...(1)
2a - 3b = 4 ...(2)
(1)* (2)* (4)- 2 = 2a - 4b = (3)
(2)* 1 = 2a - 3b = 4 (4)
(4) - (3) = [2a - 2a ] + [-3b - (-4b)] = 4 - 2
-3b + 4b = 2
therefore b = 2
substituting the value of bin (3)
2a - 4b = 2
2a - (4*2) = 2
2a - 8 = 2
2a = 2 + 8
2 = 10
a = 10 / 2
therefore a = 5
so we get the value of a and b
that is ; a = 5 and b = 2
given that (x+1) is a factor of p(x)
therefore, -1 is a zero of given p(x)
p(x) = 2x3 +ax2 +2bx +1
substituting the value of -1 in the given p(x), we get
p(x)=2*(-1)3 +a *(-1)2 +2*b*(-1)+ 1
= -2 + a -2b + 1
= -1 + a - 2b
or,a - 2b = 1
also given that 2a - 3b = 4
so we got two equations;
a - 2b = 7 ...(1)
2a - 3b = 4 ...(2)
(1)* (2)* (4)- 2 = 2a - 4b = (3)
(2)* 1 = 2a - 3b = 4 (4)
(4) - (3) = [2a - 2a ] + [-3b - (-4b)] = 4 - 2
-3b + 4b = 2
therefore b = 2
substituting the value of bin (3)
2a - 4b = 2
2a - (4*2) = 2
2a - 8 = 2
2a = 2 + 8
2 = 10
a = 10 / 2
therefore a = 5
so we get the value of a and b
that is ; a = 5 and b = 2
If the zeroes of the quadratic polynomial x2 + (a + 1) x + b are 2 and -3, the- a)a = -7, b = -1
- b)a = 5, b = -1
- c)a = 2, b = -6
- d)a = 0, b = -6
Correct answer is option 'D'. Can you explain this answer?
If the zeroes of the quadratic polynomial x2 + (a + 1) x + b are 2 and -3, the
a)
a = -7, b = -1
b)
a = 5, b = -1
c)
a = 2, b = -6
d)
a = 0, b = -6
|
|
Amit Sharma answered |
p(x) = x2 + (a + 1)x + b
Given: zeros of polynomial is 2, -3
For x = 2
(2)2 + (a + 1)2 + b = 0
⇒ 4 + 2a + 2 + b = 0
⇒ 2a + b = -6 ...(i)
Given: zeros of polynomial is 2, -3
For x = 2
(2)2 + (a + 1)2 + b = 0
⇒ 4 + 2a + 2 + b = 0
⇒ 2a + b = -6 ...(i)
For x = -3
(-3)2 + (a + 1) (-3) + b = 0
⇒ 9 - 3a - 3 + b = 0
⇒ -3a + b = -6 ..(ii)
(-3)2 + (a + 1) (-3) + b = 0
⇒ 9 - 3a - 3 + b = 0
⇒ -3a + b = -6 ..(ii)
Solving (i) and (ii), we get 5a = 0
⇒ a = 0 and b = -6
⇒ a = 0 and b = -6
Alternative method:
p(x) = x2 + (a + 1)x + b
Given: zeros of polynomial is 2, -3
Given: zeros of polynomial is 2, -3
Sum of zeros: -(a+1) = 2-3
⇒ a = 0
Product of zeros: b = 2.(-3) = -6
⇒ a = 0
Product of zeros: b = 2.(-3) = -6
Which of the following is not the graph of a quadratic polynomial ?- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
Which of the following is not the graph of a quadratic polynomial ?
a)
b)
c)
d)
|
Dr Manju Sen answered |
For any quadratic polynomial ax2 + bx + c, a≠0, graph for the corresponding equation:
- Has one of the two shapes either open upwards like ∪ or open downwards like ∩ depending on whether a > 0 or a < 0. These curves are called parabolas.
- The curve of a quadratic polynomial crosses the X-axis on at most two points.
Thus, option A is correct.
If x = 2 and x = 3 are zeros of the quadratic polynomial x2 + ax + b, the values of a and b respectively are :- a)5, 6
- b)- 5, - 6
- c)- 5, 6
- d)5, - 6
Correct answer is option 'C'. Can you explain this answer?
If x = 2 and x = 3 are zeros of the quadratic polynomial x2 + ax + b, the values of a and b respectively are :
a)
5, 6
b)
- 5, - 6
c)
- 5, 6
d)
5, - 6
|
|
Gaurav Kumar answered |
Zeros of the polynomials are the values which gives zero when their value is substituted in the polynomial
When x=2,
x2+ax+b =(2)2+a*2+b=0
4+2a+b=0
b=-4-2a ….1
When x=3,
(3)2+ 3a + b=0
9 + 3a + b=0
Substituting
9 + 3a - 4 - 2a =0
5 + a =0
a = -5
b = 6
When x=2,
x2+ax+b =(2)2+a*2+b=0
4+2a+b=0
b=-4-2a ….1
When x=3,
(3)2+ 3a + b=0
9 + 3a + b=0
Substituting
9 + 3a - 4 - 2a =0
5 + a =0
a = -5
b = 6
The number of polynomials having zeros as - 2 and 5 is- a)1
- b)2
- c)3
- d)more than 3
Correct answer is option 'D'. Can you explain this answer?
The number of polynomials having zeros as - 2 and 5 is
a)
1
b)
2
c)
3
d)
more than 3
|
Shubham Mukherjee answered |
let - 2 and 5 are the zeroes of the polynomials of the formp(x) = ax^2 + bx + c.
sum of the zeroes = - (coefficient of x) � coefficient of x^2 i.e.
Sum of the zeroes = - b/a
- 2 + 5 = - b/a
3 = - b/a
⇒ b = - 3 and a = 1
Product of the zeroes = constant term � coefficient of x^2 i.e.
Product of zeroes = c/a
(- 2)5 = c/a
- 10 = c
Put the values of a, b and c in the polynomial p(x) = ax^2 + bx + c.
∴ The equation is x^2 - 3x - 10
OR
The equation of a quadratic polynomial is given by x^2 - (sum of the zeroes)x + (product of the zeroes) where,
Sum of the zeroes = - (coefficient of x) � coefficient of x^2 and
Product of the zeroes = constant term � coefficient of x^2
∴ Sum of the zeroes = - 2 + 5 and product of the zeroes = (- 2)5
= 3 = - 10
∴ the equation is x^2 - 3x - 10
We know, the zeroes do not change if the polynomial is divided or multiplied by a constant
⇒ kx^2 - 3kx - 10k will also have - 2 and 5 as their zeroes.
As k can take any real value, there can be many polynomials having - 2 and 5 as their zeroes.
Find the quadratic polynomial whose zeros are 2 and -6
- a)x2 + 4x + 12
- b)x2 – 4x – 12
- c)x2 + 4x – 12
- d)x2 – 4x + 12
Correct answer is option 'C'. Can you explain this answer?
Find the quadratic polynomial whose zeros are 2 and -6
a)
x2 + 4x + 12
b)
x2 – 4x – 12
c)
x2 + 4x – 12
d)
x2 – 4x + 12
|
|
Naina Sharma answered |
We know that quadratic equation is of the form x2-(sum of zeros)x+product of zeros
Sum of zeros=2-6=-4
Product of zeros=2*(-6)=-12
x2-(sum of roots )x + product of roots
x2-(-4)x + 12
x2+4x-12
So the equation is x2+4x-12
Sum of zeros=2-6=-4
Product of zeros=2*(-6)=-12
x2-(sum of roots )x + product of roots
x2-(-4)x + 12
x2+4x-12
So the equation is x2+4x-12
The value of 155 mod 9 is- a)0
- b)1
- c)2
- d)3
Correct answer is option 'C'. Can you explain this answer?
The value of 155 mod 9 is
a)
0
b)
1
c)
2
d)
3
|
|
Raghav Bansal answered |
By the Division algorithm 155 = 9(17) + 2. Where remainder is 155 mod 9.
If two zeroes of a polynomial 4x4 -20x3 +23 x2 + 5x – 6 are ½ and – ½ , the how many more zeroes does it have?- a)3
- b)None
- c)1
- d)2
Correct answer is option 'D'. Can you explain this answer?
If two zeroes of a polynomial 4x4 -20x3 +23 x2 + 5x – 6 are ½ and – ½ , the how many more zeroes does it have?
a)
3
b)
None
c)
1
d)
2
|
Krishna Iyer answered |
The polynomial with degree 4 is a biquadratic equation. And a polynomial has no. of zeros equal to the degree of the polynomial. So total no. of zeros are 4. 2 zeros are already provided to us , so it has 2 more zeros.
Which of the given is the set of zeroes of the polynomial p(x) = 2x3 + x2 – 5x + 2- a)1/2 , -1 , -2
- b)1/2 , 1 , -2
- c)-1/2 , -1 , -2
- d)-1/2 , 1 , -2
Correct answer is option 'B'. Can you explain this answer?
Which of the given is the set of zeroes of the polynomial p(x) = 2x3 + x2 – 5x + 2
a)
1/2 , -1 , -2
b)
1/2 , 1 , -2
c)
-1/2 , -1 , -2
d)
-1/2 , 1 , -2
|
|
Krishna Iyer answered |
We have the polynomial 2x3+x2-5x+2
By Hit and trial method, x=1 is the solution.
Dividing the polynomial by x-1
We get quotient as = 2x2+3x-2
By Hit and trial method, x=1 is the solution.
Dividing the polynomial by x-1
We get quotient as = 2x2+3x-2
Can you explain the answer of this question below: When x2 – 2x + k divides the polynomial x4 – 6x3 + 16x2 – 25x + 10, the remainder is (x + a) . The value of a is _________
- A:
-3
- B:
3
- C:
-5
- D:
5
The answer is c.
When x2 – 2x + k divides the polynomial x4 – 6x3 + 16x2 – 25x + 10, the remainder is (x + a) . The value of a is _________
-3
3
-5
5
|
|
Pooja Shah answered |
Given that the remainder is (x + a)
⇒ (4k – 25 + 16 – 2k)x + [10 – k(8 – k) ] = x + a
⇒ (2k – 9)x + [10 – 8k + k2 ] = x + a
On comparing both the sides, we get
2k – 9 = 1
⇒ 2k = 10
∴ k = 5
Also 10 – 8k + k2 = a
⇒ 10 – 8(5) + 52 = a
⇒ 10 – 40 + 25 = a
∴ a = – 5
⇒ (4k – 25 + 16 – 2k)x + [10 – k(8 – k) ] = x + a
⇒ (2k – 9)x + [10 – 8k + k2 ] = x + a
On comparing both the sides, we get
2k – 9 = 1
⇒ 2k = 10
∴ k = 5
Also 10 – 8k + k2 = a
⇒ 10 – 8(5) + 52 = a
⇒ 10 – 40 + 25 = a
∴ a = – 5
The value of p when x3 + 9x2 + px – 10 is exactly divisible by (x+ 2 ) is ____- a)1
- b)9
- c)3
- d)6
Correct answer is option 'B'. Can you explain this answer?
The value of p when x3 + 9x2 + px – 10 is exactly divisible by (x+ 2 ) is ____
a)
1
b)
9
c)
3
d)
6
|
Trisha das answered |
Solution:
To find the value of p that makes x^3 + 9x^2 + px + 10 exactly divisible by (x + 2), we can use the remainder theorem.
The remainder theorem states that if a polynomial f(x) is divided by (x - a), then the remainder is f(a).
Therefore, if we divide x^3 + 9x^2 + px + 10 by (x + 2), the remainder will be equal to f(-2), where f(x) = x^3 + 9x^2 + px + 10.
We can find f(-2) by substituting -2 for x in the expression for f(x):
f(-2) = (-2)^3 + 9(-2)^2 + p(-2) + 10
= -8 + 36 - 2p + 10
= 18 - 2p
Since we want the remainder to be zero (i.e. we want (x + 2) to be a factor of x^3 + 9x^2 + px + 10), we must have:
18 - 2p = 0
Solving for p, we get:
p = 9
Therefore, the value of p that makes x^3 + 9x^2 + px + 10 exactly divisible by (x + 2) is 9.
To find the value of p that makes x^3 + 9x^2 + px + 10 exactly divisible by (x + 2), we can use the remainder theorem.
The remainder theorem states that if a polynomial f(x) is divided by (x - a), then the remainder is f(a).
Therefore, if we divide x^3 + 9x^2 + px + 10 by (x + 2), the remainder will be equal to f(-2), where f(x) = x^3 + 9x^2 + px + 10.
We can find f(-2) by substituting -2 for x in the expression for f(x):
f(-2) = (-2)^3 + 9(-2)^2 + p(-2) + 10
= -8 + 36 - 2p + 10
= 18 - 2p
Since we want the remainder to be zero (i.e. we want (x + 2) to be a factor of x^3 + 9x^2 + px + 10), we must have:
18 - 2p = 0
Solving for p, we get:
p = 9
Therefore, the value of p that makes x^3 + 9x^2 + px + 10 exactly divisible by (x + 2) is 9.
Find the sum and the product of the zeroes of the polynomial: x2-3x-10- a)3, 10
- b)-3, -10
- c)3,-10
- d)-3, 10
Correct answer is option 'C'. Can you explain this answer?
Find the sum and the product of the zeroes of the polynomial: x2-3x-10
a)
3, 10
b)
-3, -10
c)
3,-10
d)
-3, 10
|
|
Raghav Bansal answered |
X²-3x-10
x² -(5x-2x)-10
x² - 5x+2x-10
x(x-5)+2(x-5)
(x-5)(x+2)
x=5
x=-2
Sum of zeroes = α+β = 5-2 = 3
α+β = -b/a = -(-3)/1 = 3
Product of zeroes = αβ = 5*-2 = -10
αβ = c/a = -10/1 = -10
x² -(5x-2x)-10
x² - 5x+2x-10
x(x-5)+2(x-5)
(x-5)(x+2)
x=5
x=-2
Sum of zeroes = α+β = 5-2 = 3
α+β = -b/a = -(-3)/1 = 3
Product of zeroes = αβ = 5*-2 = -10
αβ = c/a = -10/1 = -10
If the degree of the dividend is 5 and the degree of the divisor is 3, then the degree of the quotient will be- a)2
- b)1
- c)-2
- d)0
Correct answer is option 'A'. Can you explain this answer?
If the degree of the dividend is 5 and the degree of the divisor is 3, then the degree of the quotient will be
a)
2
b)
1
c)
-2
d)
0
|
|
Vikas Kumar answered |
The degree of the dividend is 5 and that of the divisor is 3;Therefore, if expression with degree 5 is divided by expression with degree 3, then the degree of the quotient is 2.
Can you explain the answer of this question below:The sum of two zeroes of the polynomial f(x) = 2x2+(p+3) x+5 is zero, then the value of ‘p’ is
- A:
4
- B:
– 4
- C:
3
- D:
– 3
The answer is d.
The sum of two zeroes of the polynomial f(x) = 2x2+(p+3) x+5 is zero, then the value of ‘p’ is
4
– 4
3
– 3
|
|
Shubham Mukherjee answered |
Let one zeroes of the given polynomial be α and β. According to the question,
The value of quadratic polynomial f (x)=2x2– 3x- 2 at x =-2 is ……- a)12
- b)15
- c)-12
- d)16
Correct answer is option 'A'. Can you explain this answer?
The value of quadratic polynomial f (x)=2x2– 3x- 2 at x =-2 is ……
a)
12
b)
15
c)
-12
d)
16
|
|
Amit Kumar answered |
We put the value x=-2 in the polynomial 2x2-3x-2
f(-2)=2(-2)2-3(-2)-2=2*4+6-2=12
f(-2)=2(-2)2-3(-2)-2=2*4+6-2=12
The number of polynomials having zeroes -2 and 5 is:- a)1
- b)3
- c)2
- d)more than 3
Correct answer is option 'D'. Can you explain this answer?
The number of polynomials having zeroes -2 and 5 is:
a)
1
b)
3
c)
2
d)
more than 3
|
|
Amit Kumar answered |
Since the question doesn’t say that there are only 2 zeros of the polynomial we , there can be n number of polynomials which have two of its zeros as -2 and 5 .So the correct answer is more than 3.
If p(x) is a polynomial of at least degree one and p(k) = 0, then k is known as- a)value of p(x)
- b)zero of p(x)
- c)constant term of p(x)
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
If p(x) is a polynomial of at least degree one and p(k) = 0, then k is known as
a)
value of p(x)
b)
zero of p(x)
c)
constant term of p(x)
d)
none of these
|
Ishan Choudhury answered |
Let p(x) = ax + b
p(k) = ak + b = 0
∴ k is zero of p(x)
p(k) = ak + b = 0
∴ k is zero of p(x)
If one zero of the quadratic polynomial x2+ 3x + k is 2, then the value of ‘k’ is- a)– 10
- b)5
- c)– 5
- d)10
Correct answer is option 'A'. Can you explain this answer?
If one zero of the quadratic polynomial x2+ 3x + k is 2, then the value of ‘k’ is
a)
– 10
b)
5
c)
– 5
d)
10
|
|
Rajiv Gupta answered |
Let the given quadratic polynomial be p(x) = x^2 + 3x + k
Given one of the zero of the quadratic polynomial is 2.
Hence p(2) = 0
Put x = 2 in p(x), we get
p(2) = 2^2 + 3(2) + k
⇒ 0 = 4 + 6 + k
⇒ 0 = 10 + k
∴ k = – 10
The number of polynomials having zeroes as -2 and 5 is:- a)1
- b)2
- c)3
- d)More than 3
Correct answer is option 'D'. Can you explain this answer?
The number of polynomials having zeroes as -2 and 5 is:
a)
1
b)
2
c)
3
d)
More than 3
|
Kamna Science Academy answered |
The polynomials x2-3x-10, 2x2-6x-20, (1/2)x2-(3/2)x-5, 3x2-9x-30, have zeroes as -2 and 5.
The expression that should be subtracted from the polynomial f(x) = x4 + 2x3-13x2 – 12x + 21 so that the resulting polynomial is exactly divisible by g(x) = x2 – 4x + 3 is- a)x2 – 3
- b)2x – 3
- c)x – 4
- d)x2 + 4
Correct answer is option 'B'. Can you explain this answer?
The expression that should be subtracted from the polynomial f(x) = x4 + 2x3-13x2 – 12x + 21 so that the resulting polynomial is exactly divisible by g(x) = x2 – 4x + 3 is
a)
x2 – 3
b)
2x – 3
c)
x – 4
d)
x2 + 4
|
|
Rahul Kapoor answered |
Any division statement can be rewritten in the form of
Dividend = Divisor x Quotient + Remainder.
When the remainder is zero, the dividend is said to be the multiple of the divisor.
When the remainder is not equal to zero, and if it is subtracted from the dividend, then the dividend would become a multiple of divisor.
Dividend = x^4+2x^3-13x^2-12x+21
Divisor = x^2-4x+3
Step 1: Divide the dividend by divisor
Quotient = x^2+6x+8
Remainder = 2x – 3
Step 2: Hence, on subtracting 2x – 3 from the dividend, it will be exactly divisible by the given divisor.
Verification:
On subtracting 2x – 3 from the dividend, we get
x^4+2x^3-13x^2-12x+21-(2x-3) = x^4+2x^3-13x^2-14x+24
Dividing x^4+2x^3-13x^2-14x+24 by x^2-4x+3
Quotient = x^2+6x+8
Remainder = 0
If sum of the zeroes of the polynomial is 4 and their product is 4, then the quadratic polynomial is- a)x2 – 2x + 2
- b)x2 – 4x + 4
- c)x2 + 4x + 4
- d)x2 + 2x + 2
Correct answer is option 'B'. Can you explain this answer?
If sum of the zeroes of the polynomial is 4 and their product is 4, then the quadratic polynomial is
a)
x2 – 2x + 2
b)
x2 – 4x + 4
c)
x2 + 4x + 4
d)
x2 + 2x + 2
|
|
Naina Sharma answered |
Sum of zeros= α+β =4
Product of zeros=αβ=4
We have Quadratic equation as
x2-(Sum of zeros)x+Product of zeros
=x2-4x+4
Product of zeros=αβ=4
We have Quadratic equation as
x2-(Sum of zeros)x+Product of zeros
=x2-4x+4
If the point (5,0) , (0-2) and (3,6) lie on the graph of a polynomial . Then which of the following is a zero of the polynomial?- a)5
- b)6
- c)not defined
- d)-2
Correct answer is option 'A'. Can you explain this answer?
If the point (5,0) , (0-2) and (3,6) lie on the graph of a polynomial . Then which of the following is a zero of the polynomial?
a)
5
b)
6
c)
not defined
d)
-2
|
Sanaya sharma answered |
Explanation:
Since, a zero of a polynomial always lies on the x-axis as a point. In the given question the point (5,0) is a point that lies on x-axis, so its corresponding x value represents zero of the polynomial.
Therefore, 5 is the zero of the polynomial.
Hence, the correct option is A.
You can know more about Geometrical Meaning of the Zeroes of a Polynomial through the document:
The quadratic polynomial whose sum of zeroes is 3 and product of zeroes is –2 is :- a) x2 + 3x – 2
- b)x2 – 2x + 3
- c)x2 – 3x + 2
- d)x2 – 3x – 2
Correct answer is 'C'. Can you explain this answer?
The quadratic polynomial whose sum of zeroes is 3 and product of zeroes is –2 is :
a)
x2 + 3x – 2
b)
x2 – 2x + 3
c)
x2 – 3x + 2
d)
x2 – 3x – 2
|
|
Pooja Shah answered |
Sum of zeros = 3/1
-b/a = 3/1
Product of zeros = 2/1
c/a = 2/1
This gives
a = 1
b = -3
c = -2,
The required quadratic equation is
ax2+bx+c
So, x2-3x+2
-b/a = 3/1
Product of zeros = 2/1
c/a = 2/1
This gives
a = 1
b = -3
c = -2,
The required quadratic equation is
ax2+bx+c
So, x2-3x+2
If sum of the squares of zeros of the quadratic polynomial f(x) = x2 – 8x + k is 40, find the value of k.- a)12
- b)-12
- c)14
- d)-14
Correct answer is option 'A'. Can you explain this answer?
If sum of the squares of zeros of the quadratic polynomial f(x) = x2 – 8x + k is 40, find the value of k.
a)
12
b)
-12
c)
14
d)
-14
|
|
Shantala nair answered |
p (x)= x^2-8x+k
p (x)=Ax^2+Bx+C(The equation is in this form)
p (x)=Ax^2+Bx+C(The equation is in this form)
Let the zeroes be 'a' and 'b'
It is given that
=> a^2+b^2=40
=> (a+b)^2 - 2ab = 40 ----1
=> a^2+b^2=40
=> (a+b)^2 - 2ab = 40 ----1
sum of zeroes
=> a+b = -B/A = -(-8)/1 = 8
=> a+b = -B/A = -(-8)/1 = 8
Product of zeroes
=> a*b = C/A = k
=> a*b = C/A = k
Substitute these values in the equation1 we get
=> 8^2 - 2k = 40
=> 64 - 2k = 40
=> 2k = 24
=> k = 12
=> 8^2 - 2k = 40
=> 64 - 2k = 40
=> 2k = 24
=> k = 12
Therefore the value of k is 12
The sum and product of the zeroes of the polynomial f(x) = 4x2−27x+3k2 are equal, then the value of ‘k’ is- a)±3
- b)0
- c)±2
- d)±1
Correct answer is option 'A'. Can you explain this answer?
The sum and product of the zeroes of the polynomial f(x) = 4x2−27x+3k2 are equal, then the value of ‘k’ is
a)
±3
b)
0
c)
±2
d)
±1
|
|
Vikram Kapoor answered |
Let α,β are the zeroes of the given polynomial.
Given:
⇒ −(−27) = 3k2 ⇒ k2 = 0 ⇒ k= ±3
Given:
⇒ −(−27) = 3k2 ⇒ k2 = 0 ⇒ k= ±3
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