Factors Of X^3-4x^2-20x+48 Revealed!

Nov 12, 2025 by Andrew McMorgan 37 views

Factors of $f(x)=x^3-4 x^2-20 x+48$ Revealed!

Hey there, math whizzes and future mathematicians! Today, we're diving deep into the fascinating world of polynomial functions. We've got a juicy problem on our hands: a cubic function, $f(x)=x^3-4 x^2-20 x+48$ , and we're told that one of its roots is $x=6$ . Our mission, should we choose to accept it, is to find all the factors of this function. And the special tool we're going to use? The Remainder Theorem, a neat little trick that simplifies finding factors and roots. So, grab your calculators, your notebooks, and let's break this down together. We'll explore why knowing one root is a game-changer and how the Remainder Theorem guides us to the complete factorization. Get ready to flex those brain muscles, guys, because this is where the magic happens!

Understanding the Power of a Known Root and the Remainder Theorem

Alright, so we're given this function: $f(x)=x^3-4 x^2-20 x+48$ . The key piece of information here is that $x=6$ is a root. What does that actually mean in the grand scheme of things? It means that when you plug $x=6$ into the function, the result is zero. That is, $f(6) = 0$ . This is super important because it tells us that $(x-6)$ is a factor of our polynomial. This is directly related to the Factor Theorem, which is a special case of the Remainder Theorem. The Remainder Theorem states that when a polynomial $f(x)$ is divided by a linear factor $(x-c)$ , the remainder is $f(c)$ . If $f(c)=0$ , then $(x-c)$ is a factor, and $c$ is a root. Since we know $x=6$ is a root, we know for a fact that $(x-6)$ must be one of the factors of $f(x)$ . Now, our job is to find the other factors. Since $f(x)$ is a cubic polynomial (meaning it has a degree of 3), it will have a total of three roots (counting multiplicity) and therefore three linear factors (or one linear and one irreducible quadratic factor, but let's stick to linear for now as the options suggest). We've found one factor, $(x-6)$ , so we're looking for the remaining two linear factors. To find these, we need to divide our original polynomial $f(x)$ by the factor we already know, $(x-6)$ . This division will give us a quadratic polynomial, and then we can factor that quadratic to find the remaining roots and factors. This is where the Remainder Theorem, or more specifically, polynomial division, comes into play. We can use either synthetic division or long division to perform this task. Synthetic division is usually quicker and less prone to errors for linear divisors like $(x-6)$ . So, let's get our tools ready and perform this division. The result will be a much simpler polynomial that we can then factorize easily, revealing all the secrets hidden within $f(x)=x^3-4 x^2-20 x+48$ . It's like having a map and finding one landmark; the rest of the journey becomes so much clearer!

Performing the Division: Unveiling the Quadratic Factor

Okay, guys, we know that $(x-6)$ is a factor of $f(x)=x^3-4 x^2-20 x+48$ . The next logical step is to divide $f(x)$ by $(x-6)$ to see what polynomial is left. This remaining polynomial will be a quadratic (degree 2) because we're dividing a cubic (degree 3) by a linear factor (degree 1), and $3-1=2$ . We can use synthetic division, which is a super efficient method for dividing by linear factors of the form $(x-c)$ . In our case, $c=6$ . So, we set up our synthetic division like this:

Write down the coefficients of $f(x)$ : 1, -4, -20, and 48. Write down the root, 6, to the left.

6 | 1  -4  -20   48
  |_________________

Now, bring down the first coefficient (1):

6 | 1  -4  -20   48
  |_________________
    1

Multiply the number you just brought down (1) by the root (6), and write the result (6) under the next coefficient (-4):

6 | 1  -4  -20   48
  |    6
  |_________________
    1

Add the numbers in the second column (-4 + 6 = 2):

6 | 1  -4  -20   48
  |    6
  |_________________
    1   2

Repeat the process: multiply the new number (2) by the root (6), and write the result (12) under the next coefficient (-20):

6 | 1  -4  -20   48
  |    6   12
  |_________________
    1   2

Add the numbers in the third column (-20 + 12 = -8):

6 | 1  -4  -20   48
  |    6   12
  |_________________
    1   2  -8

One last time: multiply the new number (-8) by the root (6), and write the result (-48) under the last coefficient (48):

6 | 1  -4  -20   48
  |    6   12  -48
  |_________________
    1   2  -8

Add the numbers in the last column (48 + (-48) = 0):

6 | 1  -4  -20   48
  |    6   12  -48
  |_________________
    1   2  -8   0

The last number in the bottom row, 0, is the remainder. This confirms what we already knew: since the remainder is 0, $(x-6)$ is indeed a factor. The other numbers in the bottom row (1, 2, -8) are the coefficients of our quotient, which is a quadratic polynomial. Since the original polynomial was $x^3 + ...$ , the quotient will be $1x^2 + 2x - 8$ . So, we've successfully factored $f(x)$ into:

$f(x) = (x-6)(x^2 + 2x - 8)$

We're almost there, folks! We just need to factor this quadratic expression.

Factoring the Quadratic and Finding All Factors

We've successfully used the Remainder Theorem's implication (via the Factor Theorem and synthetic division) to break down our cubic polynomial $f(x)=x^3-4 x^2-20 x+48$ into a linear factor $(x-6)$ and a quadratic factor $(x^2 + 2x - 8)$ . Now, the final frontier is to factor this quadratic expression. Remember, we're looking for two numbers that multiply to give us -8 (the constant term) and add up to give us +2 (the coefficient of the x term). Let's brainstorm some pairs of numbers that multiply to -8:

1 and -8 (sum = -7)
-1 and 8 (sum = 7)
2 and -4 (sum = -2)
-2 and 4 (sum = 2)

Bingo! The pair -2 and 4 fits the bill perfectly. They multiply to -8 and add up to +2. So, we can factor the quadratic $x^2 + 2x - 8$ as $(x - 2)(x + 4)$ .

Now, we can put it all back together. Our original function $f(x)$ can be expressed as the product of its linear factors:

$f(x) = (x-6)(x^2 + 2x - 8)$

$f(x) = (x-6)(x - 2)(x + 4)$

So, all the factors of the function $f(x)=x^3-4 x^2-20 x+48$ are $(x-6)$ , $(x-2)$ , and $(x+4)$ . This means the roots of the function are $x=6$ , $x=2$ , and $x=-4$ . This matches our understanding that a cubic polynomial has three roots.

Let's quickly check our options:

A. $(x+6)(x+8)$ - Incorrect, this only has two factors and doesn't include $(x-6)$ . B. $(x-6)(x-8)$ - Incorrect, this only has two factors and the second factor is wrong. C. $(x-2)(x+4)(x-6)$ - This is our answer! It contains all the factors we found. D. $(x+2)(x-4)(x+6)$ - Incorrect, the signs are all wrong.

Therefore, the correct option is C. $(x-2)(x+4)(x-6)$ . See how the Remainder Theorem, combined with polynomial division and basic quadratic factoring, made finding all the factors a breeze? It's all about breaking down complex problems into smaller, manageable steps. Keep practicing, and you'll be factoring like a pro in no time!