1. Explain whether the expression is a univariate polynomial? Legal evidence.
4x2–3x+7
1. Explain whether the expression is a univariate polynomial? Legal evidence.
4x2–3x+7
Answer:
We can write the equation 4x2–3x+7 as 4x2–3x1+7x0 Because x is the variable of the above formula and x to the power is 0 ,1,2, the expression is a univariate polynomial
2. Show that the expression is a univariate polynomial? Legal evidence.
y2+√2
2. Show that the expression is a univariate polynomial? Legal evidence.
y2+√2
Answer:
We can write the equation
y2+√2 as y2+√2y0
In the above formula, y is variable and x to the power is 0.2, yes since the expression a is the
difference polynomial.
3. Are there more than one verb in a single variable? Legal evidence.
3√t+t√2
3. Are there more than one verb in a single variable? Legal evidence.
3√t+t√2
Answer:
We can write the expression 3√t+t√2 as 3t1/2+√2t
Since t is a variable in the above formula, x to the power is 1/2, So the expression is a univariate polynomial
4. Show whether the expression is a univariate polynomial? Legal evidence.
y + 2/y
4. Show whether the expression is a univariate polynomial? Legal evidence.
y + 2/y
Answer:
y + 2/y can be written as
y+2y-1
Since y is a variable in the above formula, the power of x is -1. It is not a number, that is, the expression is not a differential polynomial
5. The differential of the expression Specify whether it is a polynomial or not.
x10+y3+t50
5. The differential of the expression Specify whether it is a polynomial or not.
x10+y3+t50
Answer:
Here, since the even powers of
10, 3 and 50 are numbers in the x10+y3+t50 equation, 3 variables are used in the
x10 expression. So it is not a univariate polynomial
6 . Enter the coefficient of x2 in the equation below
2+x2+x
6 . Enter the coefficient of x2 in the equation below
2+x2+x
Answer:
We can write the equation
2+x2+x as 2 + (1) x2 + x
. The coefficient is the number given by the variable
. Here, the number given by the variable x2 is 1
. That is, the coefficient of x2 in 2 + x2 + x is 1.
7. Check the coefficient of x2 below
2–x2+x3
7. Check the coefficient of x2 below
2–x2+x3
Answer:
We can Write the equation 2–x2+x3 as 2+(–1)x2+x3
Number is equal. variable number
Here, the number given by the variable x2 is -1.That is, x2 -1 in the coefficient 2–x2+x3.
8. Express the x2 coefficient in the equation below
(π/2)x2+x
8. Express the x2 coefficient in the equation below
(π/2)x2+x
Answer:
We can write the equation
(π/2)x2+x as (π/2)x2+x.
The coefficient is the number to be divided by the difference
Here the number to be divided by the difference x2 is π/2.
Therefore, the coefficient of x2 in (π/2)x2 + x is π/2.
9. Check the coefficient of x2 in the equation below
√2x-1
9. Check the coefficient of x2 in the equation below
√2x-1
Answer:
We can write the equation √2x1 as 0x2+√2x-1. [because 0x2 is related to 0] 44 coefficient difference
Number of Competitions Here x2 the number to be divided by the difference is 0
So the coefficient of x2 in √2x-1 is 0.
10.Give an example of degree 35 Binomial and degree 100 monomial.
Answer:
Degree 35 Binomial: A binomial polynomial up to 35 degree was known as the degree 35 binomial
Example: 3x35+5
100 degree monomial: A polynomial of one term with a maximum of 100 degrees is called a 100 degree monomial.
Example: 4x100
11. Give the polynomial degree for the below expression:
5x3+4x2+7x
11. Give the polynomial degree for the below expression:
5x3+4x2+7x
Answer:
In polygamy, variable highest power is the polynomial degree.
In this case 5x3+4x2+7x = 5x3+4x2+7x1
to the power of variable x: 3, 2, 1
5x3+4x2+7x has a degree of 3 because 3 is the highest equation of x.
12. Write the degree of the polynomial:
4–y2
12. Write the degree of the polynomial:
4–y2
Answer:
In polygamy, variable highest power is the polynomial degree
Here in 4–y2, the power of the variable
4–y2 of 2 because 2, is the highest power for y in the equation.
13. Write degree for the below polynomial:
5t–√7
13. Write degree for the below polynomial:
5t–√7
Answer:
Polynomial variable to highest power was polynomial degree
the power of the variable t is 5t–√7 :
5t -√7 has the degree 1 because 1 is highest power of y in the equation.
14. Write the degree of the polynomial:
3
14. Write the degree of the polynomial:
3
Answer:
The greatest power difference in polygamy is a polynomial degree
3 = 3(1) = 3(0)
The difference in the power is 0
3 has a degree of 0.
15. Determine whether the polynomial is linear, quadratic or cubic.
x2+x
15. Determine whether the polynomial is linear, quadratic or cubic.
x2+x
Answer:
x2+x has the highest power of 2
It has a degree of 2
So x2+x was a quadratic polygamy
16. Determine whether the polynomial is linear, quadratic or cubic.
x–x3
16. Determine whether the polynomial is linear, quadratic or cubic.
x–x3
Answer:
x–x3 has the highest power of 3
it has a degree of 3
So x–x3 was a cubic polygamy
17. Show whether a given polynomial is linear, quadruple or cubic
y+y2+4
17. Show whether a given polynomial is linear, quadruple or cubic
y+y2+4
Answer:
y+y2+4 has the highest power of 2
it has a degree of 2
So y+y2+4 was a quadratic polygamy
18. Is a given polynomial linear, quadratic or square Classify it.
1 + x
18. Is a given polynomial linear, quadratic or square Classify it.
1 + x
Answer:
1 + x to the greatest power of 1
it has the degree of one
So 1+x is a linear polygamy
19. Divide whether a given polynomial is linear, quadratic, or cubic.
3t
19. Divide whether a given polynomial is linear, quadratic, or cubic.
3t
Answer:
3t has the highest power of 1.
It has the degree of 1
Therefore 3t was a linear polygamy
20. Divide whether a given polynomial is linear, quadratic, or cubic.
r2
20. Divide whether a given polynomial is linear, quadratic, or cubic.
r2
Answer:
r2 has the highest power of 2
It has the degree of 2
therefore, r2 was a quadratic polygamy
21. Divide whether the given polynomial is linear, quadratic or cubic
7x3
21. Divide whether the given polynomial is linear, quadratic or cubic
7x3
Answer:
7x3 has highest power of 3
it has the degree 3
therefore 7x3 was the cubic polynomial
22. Given the polygamy value (x) = 5x−4x2+3
(i) x = 0
(ii) x = – 1
(iii) x = 2
22. Given the polygamy value (x) = 5x−4x2+3
(i) x = 0
(ii) x = – 1
(iii) x = 2
Answer:
f(x) = let 5x−4x2+3
(i) x = 0
f (0) = 5(0) -4(0) 2 +3
= 3
(ii) x = -1
f(x) = 5x−4x2+3
f(−1) = 5(−1)−4(−1) 2 +3
= −5–4+3
= −6
(iii) x = 2
f(x) = 5x−4x2+3
f(2) = 5(2)−4(2) 2 +3
= 10–16+3
= -3
23. Solve the p(0),p(1),p(2) of the below polynomial p(y) = y2−y+1
23. Solve the p(0),p(1),p(2) of the below polynomial p(y) = y2−y+1
Answer:
p(y) = y2−y+1
∴ p(0) = (0)2−(0) + 1 = 1
p(1) = (1)2–(1)+1 = 1
p(2) = (2)2–(2)+1 = 3
24. Solve the p(0),p(1),p(2) of the below polynomials:
p(t) = 2+t+2t2−t3
24. Solve the p(0),p(1),p(2) of the below polynomials:
p(t) = 2+t+2t2−t3
Answer:
p(t) = 2+t+2t2−t3
∴ p(0) = 2 + 0 + 2 ( 0)2–(0)3 = 2
p(1) = 2+1+2(1)2–(1)3 =2+1+2–1 = 4
p(2) = 2+2 + 2(2)2–(2)3 =2+2+8–8 = 4
25. Find p(0), p(1) and p(2) for the following polynomial:
p(x) = x3
25. Find p(0), p(1) and p(2) for the following polynomial:
p(x) = x3
Answer:
p(x) = x3
∴ p(0) = (0)3 = 0
p(1) = (1)3 = 1
p(2) = (2)3 = 8
26. Find p(0), p(1), and p(2) for the following polynomial:
P(x)=(x-1)(x+1)
26. Find p(0), p(1), and p(2) for the following polynomial:
P(x)=(x-1)(x+1)
Answer:
p(x)=(x-1)(x+1)
P(0)=(0-1)(0+1)=(-1)(1)=-1
P(1)=(1-1)(1+1)=(0)(2)=0
P(2)=(2-1)(2+1)=(1)(3)=3
27. Express the below polynomials as zeros. p(x) = 3x+1, x = -1/3
27. Express the below polynomials as zeros. p(x) = 3x+1, x = -1/3
Answer:
For, x = -1/3, p(x) = 3x+1
∴ p(−1/3) = 3(-1/3)+ 1 = −1 + 1 = 0
-1/3 is the zero point of above polynomial.
28. Verify that the following polynomials are zero. p(x) = 5x–π, x = 4/5
28. Verify that the following polynomials are zero. p(x) = 5x–π, x = 4/5
Answer:
For, x = 4/5, p(x) = 5x–π
∴ p(4/5) = 5(4/5) - π = 4-π
4/5 is not the zero point of above polynomial
29. Show that the following polynomial is zero. p(x) = x2−1, x = 1, −1
29. Show that the following polynomial is zero. p(x) = x2−1, x = 1, −1
Answer:
For, x = 1, −1; p(x) = x2−1
p(1)= 12−1=1−1 = 0
p(−1)= (-1)2−1 = 1−1 = 0
1,-1 is the zero point of p(x ).
30. Express the following polynomials as zeros.
p(x) = (x+1)(x–2), x=−1, 2
30. Express the following polynomials as zeros.
p(x) = (x+1)(x–2), x=−1, 2
Answer:
For, x = −1.2; p(x) = (x+1)(x–2)
∴ p (−1) = (−1+1)(−1–2)= (0)(−3) = 0
p(2) = (2 +1)(2–2) = (3)(0) = 0
-1,,2 are the zeros of the above polynomial.
31. Extend the following polynomial to zero..p(x) = x2, x = 0
31. Extend the following polynomial to zero..p(x) = x2, x = 0
Answer:
For , x = 0 p(x) = x2
p(0) = 02 = 0
0 was the zero of the above polynomial
32. Express the following polynomials as zeros.
p(x) = lx+m, x = -m/l
32. Express the following polynomials as zeros.
p(x) = lx+m, x = -m/l
Answer:
For x = -m/l ; p(x) = lx+m
∴ p(-m/l) = l(-m/l) + m = −m + m = 0
-m/l was a zero of the above polynomial
33. Show that the following polynomial is zero.
p(x) = 3x2−1, x = -1/√3, 2/√3
33. Show that the following polynomial is zero.
p(x) = 3x2−1, x = -1/√3, 2/√3
Answer:
For, x = -1/√3 , 2/√3 ; p(x) = 3x2−1
∴p( -1/√3) = 3(-1/√3)2 -1 = 3(1/3)-1 = 1-1 = 0
∴p(2/ √3) = 3(2/√3)2 1 = 3(4/3)-1 = 4−1 = 3 ≠ 0
∴ -1/√3 is the zero of above polynomial. And 2/ √ 3 was not a zero of above polynomial.
34. Extend the following polynomial to zero.
p(x) = 2 x + 1, x = 1/2
34. Extend the following polynomial to zero.
p(x) = 2 x + 1, x = 1/2
Answer:
For, x = 1/2 p(x) = 2 x + 1
∴ p(1/2) = 2 (1/2)+1 = 1+1 = 2≠0
∴ 1/2 is not the zero of p(x).
35. Find the zero point of the polynomial:
p(x) = x + 5
35. Find the zero point of the polynomial:
p(x) = x + 5
Answer:
p(x) = x + 5⇒ x + 5 = 0
∴ -5
36. Find the zero point of the polynomial
p( x) = x-5
36. Find the zero point of the polynomial
p( x) = x-5
Answer:
p(x) = x- 5 = 0
x = 5 ⇒ 5 is a zero polynomial
37. Find the zero point of the polynomial
p(x) = 2 x + 5
37. Find the zero point of the polynomial
p(x) = 2 x + 5
Answer:
p(x) = 2 x + 5
⇒ 2x = 5 ⇒ x = - 5/2
∴ x = -5/2 is the zero polynomial of p(x).
38. Find the zero point of the polynomial
P(x) = 3x − 2
38. Find the zero point of the polynomial
P(x) = 3x − 2
Answer:
P(x) = 3x − 2
⇒ x ⇒ 3 x − 2 = 0
⇒ 3x = 2
⇒ x = 2/3
2/3 is the zero of polynomial p (x).
39. Find the zero of the following polynomial:
p(x) = 3x
39. Find the zero of the following polynomial:
p(x) = 3x
Answer:
p(x) = 3x ⇒ 3x = 0 ⇒ x = 0
0 is the zero polynomial of the polynomial p(x).
40.find zero of the below polynomial
p (x) = cx + d, c ≠ 0, c, d they are real numbers.
40.find zero of the below polynomial
p (x) = cx + d, c ≠ 0, c, d they are real numbers.
Answer:
p(x) = cx + d
⇒ cx + d = 0
⇒ x = -d/c
∴ x = -d/c is zero of the polynomial
Also Read: Polynomials Class 9 Extra Questions
