Find Factors Of Polynomial Using Remainder Theorem

Hey guys, welcome back to Plastik Magazine! Today, we're diving deep into the awesome world of algebra to tackle a super cool problem involving polynomial functions and the Remainder Theorem. If you've ever felt a bit intimidated by fancy math terms, don't sweat it! We're going to break this down in a way that's easy to grasp, so get ready to boost your math game. We're looking at a specific function, $f(x)=x^3-2 x^2-68 x-120$, and we're given a crucial piece of information: $f(-2)=0$. Our mission, should we choose to accept it, is to find all the factors of this polynomial. This isn't just about getting the right answer; it's about understanding the why behind it. We'll be using the Remainder Theorem as our trusty sidekick on this algebraic adventure. This theorem is a real game-changer when it comes to understanding the relationship between a polynomial, its factors, and its roots. So, grab your favorite beverage, get comfy, and let's unravel this mathematical mystery together. By the end of this, you'll not only know the factors but also feel way more confident tackling similar problems. Let's get started!

Understanding the Remainder Theorem and Its Power

The Remainder Theorem is a fundamental concept in algebra that provides a shortcut for evaluating polynomials and, more importantly for our current mission, for identifying factors. In simple terms, it states that when a polynomial $f(x)$ is divided by a linear factor $(x-a)$, the remainder is equal to $f(a)$. This might sound straightforward, but its implications are huge. For instance, if $f(a)=0$, it means that the remainder is zero, which directly implies that $(x-a)$ is a factor of $f(x)$. This is a direct consequence of the Factor Theorem, which is essentially a special case of the Remainder Theorem. So, our given condition, $f(-2)=0$, is a massive clue! It tells us immediately that if we substitute $x=-2$ into our function $f(x)=x^3-2 x^2-68 x-120$, the result is zero. According to the Factor Theorem, this means that $(x - (-2))$, which simplifies to $(x+2)$, must be one of the factors of $f(x)$. This is our starting point, guys. Having one factor already identified significantly simplifies the task of finding the others. Instead of trying to guess or test random factors, we now have a concrete lead. We can use this known factor to reduce the degree of the polynomial, making it easier to find the remaining factors. This process often involves polynomial division or synthetic division, both of which are powerful tools in a mathematician's arsenal. The Remainder Theorem isn't just theoretical; it's a practical tool that makes complex polynomial manipulations much more manageable. It bridges the gap between function evaluation and factorization, revealing the underlying structure of polynomials. So, when you see $f(a)=0$, always remember: boom, $(x-a)$ is a factor! It’s like finding a secret key that unlocks the rest of the puzzle.

Step-by-Step Factorization Using the Known Factor

Alright, so we've established that $(x+2)$ is a factor of $f(x)=x^3-2 x^2-68 x-120$ thanks to the Remainder Theorem and the given $f(-2)=0$. Now, how do we find the other factors? The next logical step is to divide $f(x)$ by $(x+2)$. This will give us a new polynomial of a lower degree (in this case, a quadratic), which will be much easier to factor. We can use either polynomial long division or synthetic division for this. Synthetic division is often quicker and less prone to errors once you get the hang of it. Let's go with synthetic division. We'll set up our synthetic division with the root of our factor, which is $-2$ (since the factor is $(x+2)$), and the coefficients of $f(x)$: $1$, $-2$, $-68$, and $-120$.

Here’s how synthetic division works:

1. Write down the root ($-2$) and the coefficients of the polynomial ($1 ext{ | } -2 ext{ | } -68 ext{ | } -120$).
2. Bring down the first coefficient ($1$).
3. Multiply the root ($-2$) by the number you just brought down ($1$), and write the result ($-2$) under the next coefficient ($-2$).
4. Add the numbers in the second column ($-2 + -2 = -4$).
5. Multiply the root ($-2$) by the sum ($-4$), and write the result ($8$) under the next coefficient ($-68$).
6. Add the numbers in the third column ($-68 + 8 = -60$).
7. Multiply the root ($-2$) by the sum ($-60$), and write the result ($120$) under the last coefficient ($-120$).
8. Add the numbers in the fourth column ($-120 + 120 = 0$).

The last number we got, $0$, is our remainder. This confirms, yet again, that $(x+2)$ is indeed a factor, because the remainder is zero. The other numbers we got ($1$, $-4$, $-60$) are the coefficients of the resulting quotient polynomial. Since we started with a cubic polynomial and divided by a linear factor, the quotient is a quadratic polynomial. The coefficients are $1$, $-4$, and $-60$, so the quotient is $x^2 - 4x - 60$. Our original polynomial $f(x)$ can now be expressed as the product of our known factor and the quotient: $f(x) = (x+2)(x^2 - 4x - 60)$. Our next task is to factor this quadratic expression. This is usually the easiest part, as quadratic factorization is a common skill. We need to find two numbers that multiply to $-60$ and add up to $-4$. Let's think about pairs of factors for $60$: $(1, 60), (2, 30), (3, 20), (4, 15), (5, 12), (6, 10)$. We need one positive and one negative factor to get a negative product. We're looking for a difference of $4$. The pair $(6, 10)$ looks promising. If we use $-10$ and $+6$, their product is $(-10)(+6) = -60$, and their sum is $-10 + 6 = -4$. Perfect! So, the quadratic $x^2 - 4x - 60$ factors into $(x-10)(x+6)$.

Putting It All Together: The Final Factors

We've successfully completed the factorization process, guys! By applying the Remainder Theorem and the Factor Theorem, we identified $(x+2)$ as a factor of $f(x)=x^3-2 x^2-68 x-120$. Then, we used synthetic division to divide $f(x)$ by $(x+2)$, which yielded the quadratic quotient $x^2 - 4x - 60$. Finally, we factored this quadratic expression into $(x-10)(x+6)$.

Now, we can write the complete factorization of $f(x)$ by combining all the factors we found. Remember, $f(x)$ is equal to the product of its factors. So, we have:

$f(x) = (x+2) imes (x^2 - 4x - 60)$

And since $x^2 - 4x - 60 = (x-10)(x+6)$, we can substitute this back in:

$f(x) = (x+2)(x-10)(x+6)$

These are all the linear factors of the function $f(x)$. If we were asked for the roots of the function (where $f(x)=0$), they would be $x=-2$, $x=10$, and $x=-6$.

Let's quickly check our options provided:

A. $(x+2)(x+60)$ - This is only two factors and incorrect. B. $(x-2)(x-60)$ - This doesn't include our confirmed factor $(x+2)$ and is incorrect. C. $(x-10)(x+2)(x+6)$ - This matches our derived factors exactly! Bingo! D. $(x+10)(x-2)(x-6)$ - This has incorrect signs and factors.

Therefore, the correct option is C. This problem beautifully demonstrates how the Remainder Theorem and Factor Theorem are indispensable tools for simplifying polynomial factorization. It’s like having a map to navigate through complex algebraic landscapes. Keep practicing these techniques, and you'll find yourself mastering polynomial problems in no time. Stay curious, keep learning, and we'll catch you in the next one!