What are all the roots of the function f(x) = 5x^3 + 5x^2 + 170x + 280 given that one factor is x – 7 using the Remainder Theorem?

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To find all the roots of the function f(x) = 5x3 + 5x2 + 170x + 280 given that x – 7 is a factor, we will use the Remainder Theorem and polynomial division.

According to the Remainder Theorem, if (x – c) is a factor of f(x), then f(c) = 0. Since we know that x – 7 is a factor, we can substitute x = 7 into f(x):

f(7) = 5(7)3 + 5(7)2 + 170(7) + 280 = 0

Next, we can perform polynomial long division to divide f(x) by (x – 7). This will give us the other factors of the polynomial.

1. Polynomial Division:

We divide: f(x) ÷ (x – 7). This yields:

               5x2 + 40x + 280
          _______________________
 (x - 7) | 5x3 + 5x2 + 170x + 280
   - (5x3 - 35x2)
   ------------------------
                   40x + 280
                   - (40x - 280)
   ------------------------
                       0

The quotient is 5x2 + 40x + 280. Now, we need to find the roots of this quadratic equation using the quadratic formula:

x = (-b ± √(b2 – 4ac)) / 2a

In this case, a = 5, b = 40, c = 280:

Discriminant = b2 – 4ac = 402 – 4(5)(280) = 1600 – 5600 = -4000

Since the discriminant is negative, there are no real roots for the quadratic polynomial. Instead, the roots will be complex. We apply the quadratic formula:

x = (-40 ± √(-4000)) / (2 * 5) = (-40 ± 20i√10) / 10

This results in:

x = -4 ± 2i√10

Thus, the complete set of roots for the function f(x) is:

1. x = 7 (from the factor x – 7)
2. x = -4 + 2i√10 (complex root)
3. x = -4 – 2i√10 (complex root)

In summary, the roots of the polynomial function f(x) = 5x3 + 5x2 + 170x + 280 are:
7, -4 + 2i√10, -4 – 2i√10.

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