Finding zeros?

Question

2 problems, i have the answers but i would like to know how to do them for the final exam...

Given that one zero is 2-3i, find all zeros of P(x) = x^3 - 9x^2 +33x -65 (answer is 2+/- 3i or 5)

Find all zeros: f(x)= 10x^4 + x^3 - 12x^2 - x + 2

Thank you, I will reward best answer.

Answer 1

If one zero is 2-3i, another must be 2+3i, because complex zeroes always come in conjugate pairs. This means that a factor of P(x) is (x-2-3i)(x-2+3i)=x^2 - 4x +13.

Divide P(x) by this and you get the third factor is (x-5), so the third zero is 5.

To find all the zeroes of f(x), using the rational roots theorem, the possible rational zeroes are +/-1, +/-2, +/-5, +/-10, +/-1/2 or +/-5/2.

Using synthetic division, fortunately it turns out that +/- 1 are rational zeroes. After factoring out (x-1) and (x+1), the reduced polynomial is 10x^2+x-2. That factors to (5x-2)(2x+1), so the remaining zeroes are 2/5 and -1/2.

Hope that helps.

Answer 2

You know one of the zeroes of P(x) is 2-3i, and complex zerores always come in pairs - if one zero is 2-3i, another must be its complex conjugate, 2+3i. Thus you know two factors of the equation are (x-2+3i) and (x-2-3i), respectively. Multiplying these factors, you get (x²-2x-3ix-2x+4+6i+3ix-6i-9i²), which simplified is (x²-4x+13). Dividing your original function by this to get the third factor, you have

1 -4 13|1 -9 33 -65
______1 -4 13
_______-5 20 -65
_______-5 20 -65
_____________0

Thus, the result of your division is (x-5) with a remainder of zero. (x-5) is therefore your third factor, and your third zero is at x=5. The final answer is x=5, 2+3i, or 2-3i.

For the bottom question:

f(x)=10x^4+x³-12x²-x+2

Here, it's a good idea to start by looking for any rational roots this polynomial might have. You know from the rational root theorem that any rational zeroes will be of the form ±p/q, where p is a factor of the last number in the polynomial and q is a factor of the first. The roots you need to try are ±1/10, ±1/5, ±2/5, ±1/2, ±1, and ±2. As it turns out, all of your roots are in this form, and are 1, -1, -1/2, and 2/5 (and if you have access to a graphing calculator, by all means graph the function so you don't have to try every last one of them). If you had been unable to find all of your roots this way, a good strategy would be to divide the polynomial by the factors corresponding to the roots you did find, and evaluate the polynomial that remained.

In the general case, there's no easy method of solving quartic equations exactly like the quadratic formula is for quadratic equations. Well, okay, there is - you can find the formulas at http://planetmath.org/encyclopedia/QuarticFormula.html - but I would under no circumstances recommend trying to memorize them. All you can really do is guess and check - that's one of the frustrating things about higher-degree polynomials.