Real Zeros Of P(x)=x^4+21x^2-100

Hey math whizzes! Ever stared at a polynomial and thought, "How on earth do I find all the answers to this thing?" Well, you're in the right place, guys. Today, we're diving deep into a specific polynomial, P(x) = x^4 + 21x^2 - 100, and we've got a little head start: it has a zero at x = 5i. Our mission, should we choose to accept it, is to solve for the real zeros of P(x). This isn't just about crunching numbers; it's about understanding the beautiful symmetry and properties of polynomials, especially how complex zeros always come in pairs for polynomials with real coefficients. So, grab your calculators, get comfy, and let's break this down step-by-step. We'll explore how knowing one complex root unlocks the door to finding the others, and ultimately, the real ones we're after. This journey will involve some slick factoring, maybe a substitution or two, and a solid grasp of the relationship between roots and factors. Get ready to flex those mathematical muscles!

Understanding Polynomial Roots and Their Properties

Alright, let's talk about what it means for a polynomial to have a "zero." When we say P(x) has a zero at x = a, it means that if you plug 'a' into the polynomial, you get P(a) = 0. It's like finding the x-intercepts on a graph, but zeros can be real or complex numbers. Now, the Fundamental Theorem of Algebra is a big deal here, guys. It basically tells us that a polynomial of degree 'n' has exactly 'n' roots (counting multiplicity), and these roots can be real or complex. For our polynomial, P(x) = x^4 + 21x^2 - 100, the degree is 4, so we're expecting four roots in total. We're given one complex root: x = 5i. A super important property for polynomials with real coefficients (which ours is, since 1, 21, and -100 are all real numbers) is that complex roots always come in conjugate pairs. This means if a + bi is a root, then its conjugate, a - bi, must also be a root. In our case, the given root is 5i, which can be written as 0 + 5i. Its conjugate is therefore 0 - 5i, or simply -5i. So, right away, we know two of the four roots: x = 5i and x = -5i. This is huge! Knowing these two roots tells us that (x - 5i) and (x - (-5i)) = (x + 5i) are factors of our polynomial P(x). We can multiply these factors together to get a quadratic factor: $(x - 5i)(x + 5i) = x^2 - (5i)^2 = x^2 - (25i^2)$. Since $i^2 = -1$, this simplifies to $x^2 - (25)(-1) = x^2 + 25$. So, (x^2 + 25) is a factor of P(x). This means we can divide P(x) by (x^2 + 25) to find the remaining factors, which will hopefully lead us to those elusive real zeros we're hunting for. It’s like solving a puzzle, and we just found a crucial piece!

Factoring the Polynomial Using Known Roots

Okay, so we've established that (x^2 + 25) is a factor of our polynomial P(x) = x^4 + 21x^2 - 100. Now, we need to figure out what the other factor is. Since P(x) is a degree 4 polynomial and (x^2 + 25) is a degree 2 factor, the remaining factor must also be a degree 2 polynomial. We can find this missing factor using polynomial long division or synthetic division, but there's a neat shortcut here because P(x) only contains even powers of x ($x^4$ and $x^2$). This structure allows us to treat it like a quadratic equation in disguise! Let's make a substitution: let y = x^2. Then our polynomial P(x) becomes y^2 + 21y - 100. We can now try to factor this quadratic in 'y'. We're looking for two numbers that multiply to -100 and add up to 21. Think about the factors of 100: (1, 100), (2, 50), (4, 25), (5, 20), (10, 10). We need one positive and one negative factor because the product is negative. If we try 25 and -4, their product is $25 imes (-4) = -100$, and their sum is $25 + (-4) = 21$. Bingo! So, we can factor the quadratic in 'y' as (y + 25)(y - 4). Now, let's substitute back x^2 for y. This gives us (x^2 + 25)(x^2 - 4). And there you have it! We've factored P(x) into two quadratic expressions: P(x) = (x^2 + 25)(x^2 - 4). Notice how the first factor, (x^2 + 25), is exactly what we got from our complex conjugate roots. This confirms our earlier work! Now, to find all the zeros, we simply set each of these factors equal to zero. We already know that setting x^2 + 25 = 0 gives us the complex roots x = 5i and x = -5i. The real zeros will come from setting the second factor, x^2 - 4 = 0, equal to zero. This is where our real solutions lie, guys!

Finding the Real Zeros

We've successfully factored our polynomial P(x) = x^4 + 21x^2 - 100 into P(x) = (x^2 + 25)(x^2 - 4). As we discussed, the factor (x^2 + 25) yields the complex zeros x = 5i and x = -5i. Now, it's time to focus on the second factor, (x^2 - 4), because this is where our real zeros are hiding! To find these real zeros, we simply set this factor equal to zero and solve for x:

x^2 - 4 = 0

This is a straightforward equation to solve. We can add 4 to both sides to isolate the $x^2$ term:

x^2 = 4

Now, to find x, we take the square root of both sides. Remember, when you take the square root, you need to consider both the positive and negative possibilities:

x = ±√4

Calculating the square root of 4 gives us 2:

x = ±2

So, the real zeros of the polynomial P(x) = x^4 + 21x^2 - 100 are x = 2 and x = -2. We have successfully found all four zeros: two complex (5i, -5i) and two real (2, -2). This perfectly matches the degree of the polynomial, which is 4. It’s awesome how knowing one complex root can lead us to all the others, especially the real ones we’re often most interested in for graphing and real-world applications. The process involved understanding the conjugate root theorem, clever substitution to simplify the quartic equation, and basic algebraic manipulation. Keep practicing these techniques, and you'll be a polynomial-solving pro in no time!

Conclusion: A Complete Picture of P(x)'s Roots

So, there you have it, math enthusiasts! We've navigated the twists and turns of the polynomial P(x) = x^4 + 21x^2 - 100, starting with a single complex zero at x = 5i and ending with a complete understanding of all its roots. By applying the conjugate root theorem, we deduced that x = -5i must also be a zero, which in turn gave us the quadratic factor (x^2 + 25). Then, by cleverly treating the polynomial as a quadratic in x^2 (letting y = x^2), we were able to factor the remaining part into (x^2 - 4). Setting this second factor to zero, x^2 - 4 = 0, led us directly to the real zeros: x = 2 and x = -2. Our final factored form of the polynomial is P(x) = (x^2 + 25)(x - 2)(x + 2), and its four zeros are 5i, -5i, 2, and -2. This comprehensive solution showcases the power of fundamental algebraic concepts and how they interrelate. It’s a fantastic example of how seemingly complex problems can be unraveled with a systematic approach and a bit of mathematical insight. Remember, guys, every polynomial tells a story through its roots, and understanding these roots – whether real or complex – gives us a complete picture of its behavior. Keep exploring, keep questioning, and keep solving!