Finding the zeros of the polynomial...?

Question

I needed to find the zeros of this polynomial:

1/4(t2-2t+15)

So, I graphed it, and there are none.

How do I show this algebraically?

Thank you.

Answer 1

Use the quadratic formula. You'll find that the value under the radical is negative.

Answer 2

when you use the quadratic formula, the solutions com out to be "i". This means that there are imaginary zeros.

So the graph will not show any zeros. To demonstrate this algebraically, you need to demonstrate knowledge of the quadratic formula.

[.5 +/- sqrt (.25- 15) ]/.5=

1 +/- sqrt (-59)= 1 +/- i sqrt (59)

These are the zeros: 1 +/- i sqrt (59)

Answer 3

If you meant
1/(4(t² - 2t + 15)) then you're correct, it has no zeros (except at infinity). But if it's supposed to be
(1/4)(t² - 2t +15) then it -does- have complex zeros at
t = 1 ± i√14

HTH

Doug

Answer 4

just prove that t^2-2t+15 is always greater or smaller than cero

t^2-2t+15>0, t^2-2t>-15, t(t-2)>-15

read http://en.wikipedia.org/wiki/Quadratic_equation

because if the denominator is always positive or negative then it never crosses the x axis so the equation has no (real) roots

Answer 5

P(x) has one 0 at -a million potential it could factorized into P(x) = Q(x)*(x+a million). you are able to then use long branch to locate Q(x). on the different hand, via watching P(x), it is likewise possible to work out P(x)=(x^2-a million)(x^2-2)=(x-a million)(x+a million) *(x-sqrt(2))(x+sqrt(2)). So the different 3 zeros are a million, sqrt(2), and -sqrt(2). (b) somewhat confusing merely via looking. yet long branch ought to artwork.

Answer 6

By showing the graph. Presumably it shows that the line representing the polynomial does not intersect the x axis.

Answer 7

yes