Math Snap

STEP 1

What is this asking?
We're given a polynomial $P(x)$ and told that $5i$ is one of its zeros.
We need to find all the other zeros and write $P(x)$ as a product of linear factors.
Watch out!
Complex roots always come in conjugate pairs!
Don't forget that if $5i$ is a root, so is $-5i$!

STEP 2

1. Find the quadratic factor.
2. Polynomial long division.
3. Factor the remaining cubic.

STEP 3

Since we know $5i$ is a zero, and complex zeros come in conjugate pairs, we know $-5i$ must also be a zero!
This means $(x - 5i)$ and $(x + 5i)$ are factors of $P(x)$.

STEP 4

Let's multiply these factors together to get a quadratic factor:
$$(x - 5i)(x + 5i) = x^2 + 5ix - 5ix - 25i^2 = x^2 - 25i^2$$ Remember that $i^2 = -1$, so we have:
$$x^2 - 25(-1) = x^2 + 25$$ So, $(x^2 + 25)$ is a factor of $P(x)$!

STEP 5

Now, we'll perform polynomial long division to divide $P(x)$ by $(x^2 + 25)$.
This will give us another factor of $P(x)$.
$$\begin{array}{cccccc} 3x^3 & -8x^2 & -5x & +10 \\ x^2+25 & 3x^5 & -8x^4 & +74x^3 & -190x^2 & -25x & +250 \\ 3x^5 & & +75x^3 \\ 0 & -8x^4 & -x^3 & -190x^2 \\ -8x^4 & & -200x^2 \\ 0 & -x^3 & +10x^2 & -25x \\ -x^3 & & -25x \\ 0 & 10x^2 & 0 & +250 \\ 10x^2 & & +250 \\ 0 & 0 & 0 \end{array}$$ Great! The division worked out perfectly, with no remainder.
This means $3x^3 - 8x^2 - 5x + 10$ is another factor.

STEP 6

We can factor the cubic $3x^3 - 8x^2 - 5x + 10$ by grouping.
$$3x^3 - 8x^2 - 5x + 10 = x^2(3x - 8) - 5(x - 2)$$ Oops, that didn't work!
Let's try factoring by grouping a different way.
$$3x^3 - 8x^2 - 5x + 10 = 3x^3 - 5x - 8x^2 + 10 = x(3x^2 - 5) - 2(4x^2 - 5)$$ Hmm, still no luck.
Let's try rational root theorem!

STEP 7

The possible rational roots are $\pm 1, \pm 2, \pm 5, \pm 10, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{5}{3}, \pm \frac{10}{3}$.
Let's try $x=2$:
$$3(2)^3 - 8(2)^2 - 5(2) + 10 = 24 - 32 - 10 + 10 = -8$$ Nope, not a root.
Let's try $x=\frac{2}{3}$:
$$3(\frac{2}{3})^3 - 8(\frac{2}{3})^2 - 5(\frac{2}{3}) + 10 = \frac{8}{9} - \frac{32}{9} - \frac{10}{3} + 10 = \frac{8-32-30+90}{9} = \frac{36}{9} = 4$$ Still no luck.
Let's try $x = \frac{5}{3}$.
$$3(\frac{5}{3})^3 - 8(\frac{5}{3})^2 - 5(\frac{5}{3}) + 10 = \frac{125}{9} - \frac{200}{9} - \frac{25}{3} + 10 = \frac{125-200-75+90}{9} = \frac{0}{9} = 0$$ Yes! $x = \frac{5}{3}$ is a root!
So $(3x - 5)$ is a factor.

STEP 8

Now we can do polynomial long division again, dividing $3x^3 - 8x^2 - 5x + 10$ by $(3x - 5)$.
$$\begin{array}{cccc} x^2 & -x & -2 \\ 3x-5 & 3x^3 & -8x^2 & -5x & +10 \\ 3x^3 & -5x^2 \\ 0 & -3x^2 & -5x \\ -3x^2 & +5x \\ 0 & -10x & +10 \\ -10x & +10 \\ 0 & 0 \end{array}$$ So we have $x^2 - x - 2 = (x-2)(x+1)$.

$P(x) = (x-5i)(x+5i)(3x-5)(x-2)(x+1)$