Intercept of quartic function

[needshelp22]

I am having a little trouble figuring out all of the x-intercepts for this quartic function:
h(x)=x^4+2x^3-5x^2+2x+6

I understand that there needs to be four solutions to the problem. I graphed it and got two of the solutions (or so, I think) I got, -.85 and -3.464 by using the trace button and getting as close to zero. The only thing is that I learned to take the real numbers and use synthetic division to find the non-real zeros. That didn't work because of the decimals.Overall, I am just confused.

 

[mmm4444bot]

Hi needhelp22:

It is true that a fourth-degree polynomial has four roots, but this does not necessarily mean that the roots are four distinct numbers.

There can be repeated roots (see "multiplicity").

Also, if any two roots represent a pair of Complex conjugates, then there are no x-intercepts associated with those numbers because they are not Real.

The given polynomial has such a pair of Complex roots: 1.16 + 0.84i and 1.16 - 0.84i.

The approximations that you found represent two Real roots. They look good to me (rounded to two decimal places), although I find that the smaller root rounds to -3.47.

Cheers,

~ Mark

 

[BigGlenntheHeavy]

\(\displaystyle There \ are \ two \ real \ solutions \ (which \ you \ already \ found), \ the \ other \ two \ are \ imaginary.\)

\(\displaystyle Real: \ x \ \dot= \ -3.465 \ and \ x \ \dot= \ -.848.\)

\(\displaystyle Imaginary: \ x \ \dot= \ 1.157+.838i \ and \ x \ \dot= \ 1.157-.838i\)