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(x + 1) is a factor of the polynomial
- a)x4+3x3+3x2+x+1
- b)x3+x2−x+1
- c)x4+x3+x2+1
- d)x3+x2+x+1
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
(x + 1) is a factor of the polynomiala)x4+3x3+3x2+x+1b)x3+x2−x+1...
Option a
f(x) = x4 + 3x3 + 3x2 + x + 1
Substitute x = -1:
( -1 )4 + 3( -1 )3 + 3( -1 )2 + ( -1 ) + 1 = 1 - 3 + 3 - 1 + 1 = 1 ≠ 0
So, (x + 1) is not a factor of this polynomial.
f(x) = x4 + 3x3 + 3x2 + x + 1
Substitute x = -1:
( -1 )4 + 3( -1 )3 + 3( -1 )2 + ( -1 ) + 1 = 1 - 3 + 3 - 1 + 1 = 1 ≠ 0
So, (x + 1) is not a factor of this polynomial.
Option b
f(x) = x3 + x2 - x + 1
Substitute x = -1:
( -1 )3 + ( -1 )2 - ( -1 ) + 1 = -1 + 1 + 1 + 1 = 2 ≠ 0
So, (x + 1) is not a factor of this polynomial.
f(x) = x3 + x2 - x + 1
Substitute x = -1:
( -1 )3 + ( -1 )2 - ( -1 ) + 1 = -1 + 1 + 1 + 1 = 2 ≠ 0
So, (x + 1) is not a factor of this polynomial.
Option c
f(x) = x4 + x3 + x2 + 1
Substitute x = -1:
( -1 )4 + ( -1 )3 + ( -1 )2 + 1 = 1 - 1 + 1 + 1 = 2 ≠ 0
So, (x + 1) is not a factor of this polynomial.
f(x) = x4 + x3 + x2 + 1
Substitute x = -1:
( -1 )4 + ( -1 )3 + ( -1 )2 + 1 = 1 - 1 + 1 + 1 = 2 ≠ 0
So, (x + 1) is not a factor of this polynomial.
Option d
f(x) = x3 + x2 + x + 1
Substitute x = -1:
( -1 )3 + ( -1 )2 + ( -1 ) + 1 = -1 + 1 - 1 + 1 = 0
Since the result is zero, (x + 1) is a factor of this polynomial.
f(x) = x3 + x2 + x + 1
Substitute x = -1:
( -1 )3 + ( -1 )2 + ( -1 ) + 1 = -1 + 1 - 1 + 1 = 0
Since the result is zero, (x + 1) is a factor of this polynomial.
Correct answer: d) x3 + x2 + x + 1
Most Upvoted Answer
(x + 1) is a factor of the polynomiala)x4+3x3+3x2+x+1b)x3+x2−x+1...
Option d is the best answer in india reason is given in r d sharma and r s aggrawal or Google search
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Community Answer
(x + 1) is a factor of the polynomiala)x4+3x3+3x2+x+1b)x3+x2−x+1...
A) To check if (x+1) is a factor, we can use synthetic division:
-1 | 1 3 3 1 1
|__ -1 -2 -1 0
| 1 2 1 0 1
Since the last number in the result is not zero, (x+1) is not a factor of the polynomial.
b) To check if (x+1) is a factor, we can use synthetic division:
-1 | 1 0 1 0
|__ -1 1 -1
| 1 -1 2 -1
Since the last number in the result is not zero, (x+1) is not a factor of the polynomial.
-1 | 1 3 3 1 1
|__ -1 -2 -1 0
| 1 2 1 0 1
Since the last number in the result is not zero, (x+1) is not a factor of the polynomial.
b) To check if (x+1) is a factor, we can use synthetic division:
-1 | 1 0 1 0
|__ -1 1 -1
| 1 -1 2 -1
Since the last number in the result is not zero, (x+1) is not a factor of the polynomial.
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