Given that one root of f(x) = x³ + 4x² + 20x + 48 is x = 6, what are all the factors of this function using the Remainder Theorem?

Understanding the Problem:

We start with the polynomial function f(x) = x³ + 4x² + 20x + 48. We know that one of its roots is x = 6. This means that when we substitute 6 into the polynomial, f(6) should equal 0.

Using the Remainder Theorem:

The Remainder Theorem states that if a polynomial f(x) is divided by x – r, the remainder of this division is f(r). Since we already know that f(6) = 0, we can conclude that x – 6 is a factor of f(x).

Performing Polynomial Long Division:

Next, we can perform polynomial long division to divide f(x) by x – 6. Here’s how it works step by step:

  • Divide the leading term of by the leading term of x to get .
  • Multiply the entire divisor (x – 6) by and subtract from f(x).
  • This yields a new polynomial to simplify.
  • Repeat the process until you cannot divide anymore.

Finding All Factors:

After performing the division, suppose you arrive at:

  • Once the polynomial division is complete, let’s say we obtained x² + 10x + 8.
  • Now, we need to factor x² + 10x + 8.
  • This factors further to (x + 2)(x + 8).

Final Factors:

Now, we can compile all the factors of the original polynomial:

  • f(x) = (x – 6)(x + 2)(x + 8)

Summary:

In conclusion, the factors of the polynomial function f(x) = x³ + 4x² + 20x + 48 are (x – 6), (x + 2), and (x + 8). Thank you for exploring polynomial factoring using the Remainder Theorem!

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