Find The Roots Of A Polynomial Function

Hey guys, ever stumbled upon a gnarly polynomial and wondered, "What are its roots?" Well, you've come to the right place! Today, we're diving deep into a specific problem from the world of mathematics, figuring out the complete list of roots for the function $f(x)=\left(x^2+2 x-15\right)\left(x^2+8 x+17\right)$. This isn't just about memorizing formulas; it's about understanding how polynomials behave and how to break them down. We'll go step-by-step, making sure you grasp every bit of it.

Understanding Polynomial Roots

So, what exactly are roots of a polynomial function? Simply put, the roots (also called zeros or solutions) are the values of $x$ for which the function $f(x)$ equals zero. Think of it like finding where the graph of the function crosses the x-axis. For a polynomial, finding these roots is key to understanding its structure and behavior. The Fundamental Theorem of Algebra tells us that a polynomial of degree $n$ has exactly $n$ roots, counting multiplicity, in the complex number system. This is super important because it means we're not just looking for real numbers; sometimes, we have to venture into the realm of complex numbers, which involve the imaginary unit $i$ (where $i^2 = -1$). Our polynomial here is a product of two quadratic expressions, meaning it's a quartic (degree 4) polynomial. Therefore, we should expect to find exactly four roots in total.

Breaking Down the Polynomial

Our polynomial is given as $f(x)=\left(x^2+2 x-15\right)\left(x^2+8 x+17\right)$. To find the roots, we need to set $f(x) = 0$. Since $f(x)$ is a product of two factors, the entire expression will be zero if either of the factors is zero. This is the zero product property in action, guys! So, we can find the roots by solving each quadratic equation separately:

1. $x^2 + 2x - 15 = 0$
2. $x^2 + 8x + 17 = 0$

Let's tackle the first one. This quadratic expression looks factorable. We need two numbers that multiply to -15 and add up to 2. If we think about the factors of 15 (1 and 15, 3 and 5), we can see that 5 and -3 fit the bill: $5 \times -3 = -15$ and $5 + (-3) = 2$. So, we can factor the first quadratic as $(x+5)(x-3) = 0$. Setting each factor to zero gives us our first two roots:

• $x + 5 = 0 \implies x = -5$
• $x - 3 = 0 \implies x = 3$

These are two of our roots, and they are real numbers. Now, let's move on to the second quadratic equation: $x^2 + 8x + 17 = 0$. This one doesn't look as straightforward to factor with integers. When a quadratic doesn't factor easily, the quadratic formula is our best friend. The quadratic formula for an equation of the form $ax^2 + bx + c = 0$ is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

In our case, $a=1$, $b=8$, and $c=17$. Let's plug these values into the formula:

$x = \frac{-8 \pm \sqrt{8^2 - 4(1)(17)}}{2(1)}$

$x = \frac{-8 \pm \sqrt{64 - 68}}{2}$

$x = \frac{-8 \pm \sqrt{-4}}{2}$

Ah, we've hit a square root of a negative number! This is where complex roots come into play. Remember, $\sqrt{-1} = i$. So, $\sqrt{-4} = \sqrt{4 \times -1} = \sqrt{4} \times \sqrt{-1} = 2i$.

Now, substitute this back into our equation for $x$:

$x = \frac{-8 \pm 2i}{2}$

We can simplify this by dividing both terms in the numerator by 2:

$x = \frac{-8}{2} \pm \frac{2i}{2}$

$x = -4 \pm i$

This gives us our remaining two roots: $x = -4 + i$ and $x = -4 - i$. These are complex conjugate roots, which is typical when the coefficients of the polynomial are real.

The Complete List of Roots

So, putting it all together, the roots we found from the first factor were $-5$ and $3$. The roots we found from the second factor were $-4+i$ and $-4-i$. Therefore, the complete list of roots for the polynomial function $f(x)=\left(x^2+2 x-15\right)\left(x^2+8 x+17\right)$ is $-5, 3, -4+i, -4-i$. This matches option B from the choices provided. It's always satisfying when you can fully factor a polynomial and identify all its roots, isn't it?

Why Understanding Roots Matters

Guys, understanding polynomial roots isn't just an academic exercise; it has practical applications in various fields, from engineering and physics to economics and computer science. For instance, in control systems engineering, the roots of a characteristic polynomial determine the stability of a system. If the roots have positive real parts, the system is unstable. In signal processing, roots of polynomials can help in designing filters. In economics, they can be used in modeling interest rates or population growth. The ability to find and interpret these roots is a fundamental skill for anyone working with mathematical models. This problem, specifically, showcases how a polynomial can have both real and complex roots, and how the discriminant ($b^2-4ac$) of a quadratic equation tells us the nature of its roots. A positive discriminant means two distinct real roots, a zero discriminant means one repeated real root, and a negative discriminant means two complex conjugate roots. In our case, the first quadratic had a positive discriminant ($2^2 - 4(1)(-15) = 4 + 60 = 64$), giving us two distinct real roots. The second quadratic had a negative discriminant ($8^2 - 4(1)(17) = 64 - 68 = -4$), giving us two complex conjugate roots. Pretty neat, right?

Final Thoughts

So there you have it, the complete list of roots for $f(x)=\left(x^2+2 x-15\right)\left(x^2+8 x+17\right)$ is $-5, 3, -4+i, -4-i$. We successfully broke down the problem by factoring the polynomial into its constituent quadratic parts and then solving each part. We used factoring for the simpler quadratic and the quadratic formula for the one that yielded complex roots. Remember the power of the quadratic formula and the concept of complex numbers when you encounter polynomials that don't easily factor. Keep practicing, and you'll be a root-finding pro in no time! This skill is absolutely crucial for further studies in algebra and calculus. Don't shy away from those complex numbers; they're an essential part of the mathematical landscape and unlock solutions to problems that would otherwise be intractable. Keep exploring, keep learning, and happy calculating!