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Given f(x)=2ˣ³+5ˣ²-28x-15 and 2x+1 is a facter of f(x). Find all the noots of f(x).

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Final answer:

The roots of the equation f(x) = 2x³ + 5x² - 28x - 15, given that 2x+1 is a factor, are x = -1/2, x = -5, and x = 3.

Step-by-step explanation:

To find the roots of the equation f(x) = 2x³ + 5x² - 28x - 15 given that 2x+1 is a factor, we need to set f(x) equal to zero and solve for x.

By setting f(x) = 0 and factoring out 2x+1, we get (2x+1)(x²+4x-15) = 0.

Therefore, the roots of f(x) are x = -1/2 (from 2x+1 = 0) and x = -5 or x = 3 (from x²+4x-15 = 0).

corrected question:

Given f(x)=2ˣ³+5ˣ²-28x-15 and 2x+1 is a facter of f(x). Find all the roots of f(x)

User Simon Hazelton
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