Answer:
Let, 2x^3-5x^2-28x+15= f (x)
where f (x) is the function of variable x.
If (x-5) is factor of f (x), let's consider that (x-5) is just equal to zero.
Therefore, x-5=0
x=5
Put value of x in f (x).
Therefore,
f (x) = [2. (5)^3] - [5. (5)^2] - [28. (5)] + 15
f (x) = 250 - 125 - 140 +15
f (x) = 0
As f (x) = 0, no one could ever stop me from writing that (x-5) is a factor of f (x) in this case.
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Answers & Comments
Answer:
Let, 2x^3-5x^2-28x+15= f (x)
where f (x) is the function of variable x.
If (x-5) is factor of f (x), let's consider that (x-5) is just equal to zero.
Therefore, x-5=0
x=5
Put value of x in f (x).
Therefore,
f (x) = [2. (5)^3] - [5. (5)^2] - [28. (5)] + 15
f (x) = 250 - 125 - 140 +15
f (x) = 0
As f (x) = 0, no one could ever stop me from writing that (x-5) is a factor of f (x) in this case.