Math problem: How do you find the zeros of.....?

Question

x^4-2x^3-5x^2+8x+4

Answer 1

To find the zeros, what you have to do is first define the function as p(x).

p(x) = x^4 - 2x^3 - 5x^2 + 8x + 4.

The numbers you want to test are p(1), p(-1), p(2), p(-2), p(4), p(-4).

If ANY one of them gives you a result of 0 and you get a value of c, then (x - c) is a factor, and you do long division.

p(1) = 1 - 2 - 5 + 8 + 4 = NOT zero.
p(-1) = 1 + 2 - 5 - 8 + 4 = NOT zero
p(2) = 16 - 16 - 20 + 16 + 4 = 0

GREAT! 2 is a root, so (x - 2) is a factor.

Do long division, and then repeat the process over again. The key point to know is to look at the constant value (in this case it's 4), and test ALL integer factors of 4.

Answer 2

Let f(x) = x^4-2x^3-5x^2+8x+4
f(2) = 2^4 - 2*2^3 -5^2^2 + 8*2 + 4
= 0 so x - 2 is a factor

So f(x) = (x - 2)(x^3 - 5x - 2)

f(-2) = -4*(-8 + 10 - 2) = 0 so (x + 2) is a factor

Thus f(x) = (x - 2)(x + 2)(x^2 - 2x -1)
Now zeros of x^2 - 2x -1 are (2 ± √(4 + 4)/2 = 1 ± √2

So all zeros are, 2, -2, 1 ± √2 and 1 - √2

Answer 3

Substitute in some values of x which are factors of the constant term (4) to see if you can make it 0.
You'll find that x = 2 and x = -2 work. So we know that (x+2) and (x-2) are factors.
Now we can do a long division: divide (x+2) then (x-2) into that expression. Actually we can multiply those together first to make (x^2 - 4) and then divide that.
You'll get
(x^2 - 4)(x^2 - 2x - 1).
Thus, by the quadratic formula, the other roots are (2 ± sqrt(8))/2 = 1 ± sqrt(2).

So, 1±sqrt(2), ±2.