find the zeros of 5x(x2-4x-1)
the only example the book gives is -2x4+2x2 and the teacher hasn't gone over this in detail in class. I'm lost talk me thru it please.
The "zeroes" of f(x) are those values of x for which f(x) = 0. Another word for the same thing is "roots." Are you clear on the definition?
A polynomial of degree n is defined as \(\displaystyle \displaystyle P_n(x) = \sum_{i = 1}^{n+1}a_ix^{n + 1 - i}, a_1 \ne 0.\)
That is, a polynomial in x of degree n is a sum, where each summand is a number times a power of x, and the highest power is n. Very simple.
Do you understand what a polynomial is?
A polynomial of degree n has AT MOST n distinct real zeroes.
A polynomial with real zeroes can be factored into linear terms, where each linear term is x MINUS the real root.
Finding the zeroes of a polynomial is frequently a matter of factoring it. (There are general ways to solve quadratics, cubics, and quartics that do not require factoring.)
So let's look at the book's example. \(\displaystyle Where\ does\ -2x^4 + 2x^2 = 0?\)
Let's factor
\(\displaystyle -2x^4 + 2x^2 = 2(x^2)(x^2 - 1) = 0.\)
But we want differences with x.
\(\displaystyle 2(x^2)(x^2 - 1) = 2(x - 0)^2(x - 1)(x + 1) = 0.\)
\(\displaystyle \text{That means the zeroes are: }0,\ 1,\ and -1.\)
Lets check
\(\displaystyle -2(0)^4 + 2(0)^2 = -2 * 0 + 2 * 0 = 0 + 0 = 0.\ OK.\)
\(\displaystyle -2(1)^4 + 2(1)^2 = - 2 * 1 + 2 * 1 = -2 + 2 = 0.\ OK.\)
\(\displaystyle -2(-1)^4 + 2(-1)^2 = -2 * 1 + 2 * 1 = -2 + 2 = 0.\ OK.\)
OK The problem you gave is ALREADY partially factored. Complete factoring. Can you figure out the zeroes now?
