Find Roots Of X^3+3x^2-x-3 Using Remainder Theorem

Dec 20, 2025 by Andrew McMorgan 51 views

Find Roots of $f(x)=x^3+3x^2-x-3$ using the Remainder Theorem

Hey guys! Today, we're diving deep into the fascinating world of polynomial roots, specifically tackling a problem that involves the Remainder Theorem. We've got this function, $f(x)=x^3+3 x^2-x-3$ , and we're told a crucial piece of information: $f(1)=0$ . This little nugget tells us that $x=1$ is already one of the roots, which is a fantastic starting point. Our mission, should we choose to accept it, is to find all the roots of this cubic function. Let's break this down step-by-step, making sure we understand every part of the process. The Remainder Theorem is our trusty sidekick here, helping us factorize the polynomial and, ultimately, uncover all its hidden roots. So, grab your thinking caps, and let's get this mathematical adventure started!

Understanding the Remainder Theorem and Its Power

Alright, let's talk about the Remainder Theorem, because it's the key player in solving this problem. In simple terms, the Remainder Theorem states that if you divide a polynomial $f(x)$ by a linear factor $(x-c)$ , the remainder will be $f(c)$ . Now, what's super cool about this is when the remainder is zero! If $f(c)=0$ , it means that $(x-c)$ is a factor of the polynomial $f(x)$ , and more importantly, $c$ is a root of the polynomial. We're given that $f(1)=0$ . What does this tell us immediately? Exactly! It tells us that $(x-1)$ is a factor of $f(x)=x^3+3 x^2-x-3$ , and $x=1$ is one of the roots. This is a massive head start. Now, since $(x-1)$ is a factor, we can use polynomial division (or synthetic division, which is usually quicker for linear factors) to divide $f(x)$ by $(x-1)$ . This will give us a resulting quadratic polynomial. Once we have that quadratic, finding its roots becomes a much simpler task, often solvable by factoring or using the quadratic formula. So, the Remainder Theorem isn't just about remainders; it's a powerful tool for uncovering factors and, consequently, the roots of polynomials. It simplifies complex problems by breaking them down into more manageable parts. We're going to leverage this theorem to its fullest potential to find all the roots of our given cubic function.

Step-by-Step Root Finding

So, we know $x=1$ is a root because $f(1)=0$ . This means $(x-1)$ is a factor of $f(x) = x^3+3 x^2-x-3$ . The next logical step is to divide $f(x)$ by $(x-1)$ to find the other factor(s). We can use synthetic division for this, which is a slick method for dividing polynomials by linear binomials. Here's how it works:

Set up the synthetic division: Write down the coefficients of $f(x)$ : $1$ (for $x^3$ ), $3$ (for $x^2$ ), $-1$ (for $x$ ), and $-3$ (for the constant term). Since we're dividing by $(x-1)$ , we use $c=1$ in the synthetic division setup.
```
1 | 1   3   -1   -3
  |________________
```
Bring down the first coefficient: Bring down the leading coefficient, which is $1$ .
```
1 | 1   3   -1   -3
  |________________
    1
```
Multiply and add: Multiply the number you just brought down ( $1$ ) by the divisor ( $1$ ), and write the result ( $1 imes 1 = 1$ ) under the next coefficient ( $3$ ). Then, add the two numbers ( $3 + 1 = 4$ ).
```
1 | 1   3   -1   -3
  |     1
  |________________
    1   4
```
Repeat the process: Multiply the new sum ( $4$ ) by the divisor ( $1$ ), and write the result ( $4 imes 1 = 4$ ) under the next coefficient ( $-1$ ). Add them together ( $-1 + 4 = 3$ ).
```
1 | 1   3   -1   -3
  |     1    4
  |________________
    1   4    3
```
Final step: Multiply the latest sum ( $3$ ) by the divisor ( $1$ ), and write the result ( $3 imes 1 = 3$ ) under the last coefficient ( $-3$ ). Add them together ( $-3 + 3 = 0$ ).
```
1 | 1   3   -1   -3
  |     1    4    3
  |________________
    1   4    3    0
```

The last number, $0$ , is the remainder. This confirms what we already knew: since the remainder is $0$ , $(x-1)$ is indeed a factor of $f(x)$ . The other numbers in the bottom row ( $1$ , $4$ , $3$ ) are the coefficients of the quotient polynomial. Since we started with a cubic polynomial and divided by a linear one, the quotient is a quadratic polynomial. The coefficients $1, 4, 3$ correspond to $1x^2 + 4x + 3$ .

So, we can rewrite $f(x)$ as: $f(x) = (x-1)(x^2 + 4x + 3)$ .

Now, we need to find the roots of the quadratic factor $x^2 + 4x + 3$ . We can do this by factoring the quadratic.

Factoring the Quadratic and Finding All Roots

We've successfully factored our cubic polynomial into $f(x) = (x-1)(x^2 + 4x + 3)$ . We already know that $x=1$ is one root from the $(x-1)$ factor. Now, we just need to find the roots of the quadratic part: $x^2 + 4x + 3 = 0$ . This is a standard quadratic equation, and we can solve it by factoring. We're looking for two numbers that multiply to $3$ (the constant term) and add up to $4$ (the coefficient of the $x$ term). Let's think...

$1 imes 3 = 3$ , and $1 + 3 = 4$ . Perfect!

So, the quadratic $x^2 + 4x + 3$ can be factored as $(x+1)(x+3)$ .

Now, we can set this factored quadratic equal to zero to find its roots:

$(x+1)(x+3) = 0$

For this product to be zero, at least one of the factors must be zero:

$x+1 = 0 ightarrow x = -1$
$x+3 = 0 ightarrow x = -3$

So, the roots from the quadratic factor are $x=-1$ and $x=-3$ . Combining these with the root we found earlier ( $x=1$ ), the complete set of roots for the function $f(x)=x^3+3 x^2-x-3$ is $x=1$ , $x=-1$ , and $x=-3$ .

Let's double-check our answer against the options provided:

A. $x=-1, x=1$ , or $x=3$ B. $x=-3, x=-1$ , or $x=1$ C. $x=-3$ or $x=1$ D. $x=-1$ or $x=3$

Our findings match option B. It's always a good idea to plug these roots back into the original equation to confirm they work.

For $x=1$ : $f(1) = (1)^3 + 3(1)^2 - (1) - 3 = 1 + 3 - 1 - 3 = 0$ . (Given)
For $x=-1$ : $f(-1) = (-1)^3 + 3(-1)^2 - (-1) - 3 = -1 + 3(1) + 1 - 3 = -1 + 3 + 1 - 3 = 0$ .
For $x=-3$ : $f(-3) = (-3)^3 + 3(-3)^2 - (-3) - 3 = -27 + 3(9) + 3 - 3 = -27 + 27 + 3 - 3 = 0$ .

All three roots satisfy the equation, confirming our solution is correct! The Remainder Theorem was instrumental in getting us started by confirming $x=1$ as a root and enabling us to factor the polynomial. From there, it was a matter of solving a simple quadratic. Pretty neat, huh?