Given f(x)=2x3−5x2−68x+35
, apply the Factor Theorem and determine which of the following is a factor of f(x)
.(1 point)
Responses
(x+5)
left parenthesis x plus 5 right parenthesis
(x−3)
left parenthesis x minus 3 right parenthesis
(2x+1)
left parenthesis 2 x plus 1 right parenthesis
(x+7)
To determine which of the given factors is a factor of f(x), we need to check if f(x) is equal to zero when we plug in the value of that factor.
1. Let's check if f(x) is equal to zero when x = -5:
f(-5) = 2(-5)^3 - 5(-5)^2 - 68(-5) + 35
= 2(-125) - 5(25) + 340 + 35
= -250 - 125 + 340 + 35
= 0
Since f(-5) is equal to zero, (x+5) is a factor of f(x).
2. Let's check if f(x) is equal to zero when x = 3:
f(3) = 2(3)^3 - 5(3)^2 - 68(3) + 35
= 2(27) - 5(9) - 204 + 35
= 54 - 45 - 204 + 35
= -160
Since f(3) is not equal to zero, (x-3) is not a factor of f(x).
3. Let's check if f(x) is equal to zero when x = -1/2:
f(-1/2) = 2(-1/2)^3 - 5(-1/2)^2 - 68(-1/2) + 35
= 2(-1/8) - 5(1/4) + 34 + 35
= -1/4 - 5/4 + 34 + 35
= 68 - 6
= 62
Since f(-1/2) is not equal to zero, (2x+1) is not a factor of f(x).
4. Let's check if f(x) is equal to zero when x = -7:
f(-7) = 2(-7)^3 - 5(-7)^2 - 68(-7) + 35
= 2(-343) - 5(49) + 476 + 35
= -686 - 245 + 476 + 35
= -420 + 511
= 91
Since f(-7) is not equal to zero, (x+7) is not a factor of f(x).
Therefore, the only factor of f(x) among the given options is (x+5).
To determine if a given polynomial is a factor of f(x), you can use the Factor Theorem. According to the Factor Theorem, if the polynomial is a factor, then when you substitute the value for x into f(x), the result should be zero.
Let's check each option:
1. (x+5): Substitute x = -5 into f(x):
f(-5) = 2(-5)^3 - 5(-5)^2 - 68(-5) + 35
= 2(-125) - 5(25) + 340 + 35
= -250 - 125 + 375 + 35
= -250 + 375 + 35 - 125
= 135
Since the result is not zero, (x+5) is not a factor of f(x).
2. (x-3): Substitute x = 3 into f(x):
f(3) = 2(3)^3 - 5(3)^2 - 68(3) + 35
= 2(27) - 5(9) - 204 + 35
= 54 - 45 - 204 + 35
= 89 - 249
= -160
Since the result is not zero, (x-3) is not a factor of f(x).
3. (2x+1): Substitute x = -1/2 into f(x):
f(-1/2) = 2(-1/2)^3 - 5(-1/2)^2 - 68(-1/2) + 35
= 2(-1/8) - 5(1/4) + 34 + 35
= -1/4 - 5/4 + 34 + 35
= 33 - 6/4 + 35
= 33 - 3/2 + 35
= 68 - 3/2
= 67.5 - 1.5
= 66
Since the result is not zero, (2x+1) is not a factor of f(x).
4. (x+7): Substitute x = -7 into f(x):
f(-7) = 2(-7)^3 - 5(-7)^2 - 68(-7) + 35
= 2(-343) - 5(49) + 476 + 35
= -686 - 245 + 476 + 35
= -686 - 210 + 511
= -896 + 511
= -385
Since the result is not zero, (x+7) is not a factor of f(x).
Therefore, none of the given options is a factor of f(x).
To apply the Factor Theorem and determine which of the given options is a factor of f(x), we need to check if the option will make f(x) equal to zero.
First, let's substitute the given options into f(x) and see if any of them result in f(x) being equal to zero.
1. For (x+5):
Substituting x+5 into f(x) gives us f(x) = 2(x+5)^3 - 5(x+5)^2 - 68(x+5) + 35.
Expanding and simplifying, f(x) = 2(x^3 + 15x^2 + 75x + 125) - 5(x^2 + 10x + 25) - 68x - 340 + 35.
Simplifying further, f(x) = 2x^3 + 30x^2 + 150x + 250 - 5x^2 - 50x - 125 - 68x - 340 + 35.
Combining like terms, f(x) = 2x^3 + 25x^2 + 32x - 180.
Since f(x) ≠ 0, (x+5) is not a factor of f(x).
2. For (x-3):
Substituting x-3 into f(x) gives us f(x) = 2(x-3)^3 - 5(x-3)^2 - 68(x-3) + 35.
Expanding and simplifying, f(x) = 2(x^3 - 9x^2 + 27x - 27) - 5(x^2 - 6x + 9) - 68x + 204 + 35.
Simplifying further, f(x) = 2x^3 - 18x^2 + 54x - 54 - 5x^2 + 30x - 45 - 68x + 239.
Combining like terms, f(x) = 2x^3 - 23x^2 + 16x + 140.
Since f(x) ≠ 0, (x-3) is not a factor of f(x).
3. For (2x+1):
Substituting 2x+1 into f(x) gives us f(x) = 2(2x+1)^3 - 5(2x+1)^2 - 68(2x+1) + 35.
Expanding and simplifying, f(x) = 2(8x^3 + 12x^2 + 6x + 1) - 5(4x^2 + 4x + 1) - 68(2x) + 35.
Simplifying further, f(x) = 16x^3 + 24x^2 + 12x + 2 - 20x^2 - 20x - 5 - 136x + 35.
Combining like terms, f(x) = 16x^3 + 4x^2 - 144x + 32.
Since f(x) ≠ 0, (2x+1) is not a factor of f(x).
4. For (x+7):
Substituting x+7 into f(x) gives us f(x) = 2(x+7)^3 - 5(x+7)^2 - 68(x+7) + 35.
Expanding and simplifying, f(x) = 2(x^3 + 21x^2 + 147x + 343) - 5(x^2 + 14x + 49) - 68x - 476 + 35.
Simplifying further, f(x) = 2x^3 + 42x^2 + 294x + 686 - 5x^2 - 70x - 245 - 68x - 476 + 35.
Combining like terms, f(x) = 2x^3 + 37x^2 + 156x - 6.
Since f(x) ≠ 0, (x+7) is not a factor of f(x).
After substituting each option and verifying that none of them resulted in f(x) being equal to zero, we can conclude that none of the given options (x+5), (x-3), (2x+1), and (x+7) are factors of f(x).