finding zeros of a polynomial, where it crosses x-axis?

Question

Question Details: 0 = x^3 + 4x + 3

help me please!!!! I don't know how to find the zero, or how after you do synthetic division I'm supposed to find the zero, and my Algebra II book does not explain it very well so help me please!!!!!!!

Someone said its miswritten but I wrote it exactly like in the book, maybe the books miswritten?

Answer 1

Beware of the 3rd post that goes through a long factorization.. it is wrong in the 3rd step you don't get x^2 (x+3), you would get x^2(x+1) and then you can't factor the x=3 from both terms.

As you gave the problem, the first real root is -0.67429 approx. The following synthetic div will be a nightmare because I think theproblem is mistated.

But really I think you made a mistake in the problem statement. If it is -4x instead of +4x then x=1 would be one easy root by guessing and you can use (x-1) to do synthetic division and get x^2 + x -3 for which you can find the roots using the quadratic formula Ans -1/2 +or - 1/2* sqrt(13)

It would be almost impossible to write steps here for synthetic division.. need to paste a picture of the steps and explanations.

The link will get you to a graphic calc online.

Answer 2

I never got the hang of synthetic division either. You have several choices, but the first thing you should try is to GRAPH THE POLYNOMIAL. This behaves fairly similar to the y = x^3 curve, which has one real root (the root being the x-value when the curve crosses the Y-AXIS , just as it was in Algebra I). A quick examination indicates that the root is between 0 and -1. So you can try synthetic division, which probably won't do too much good, or a fancy trial and error method.

Answer 3

one 0 is (0,0) with the aid of fact 0 = 0 - 0. to discover different zeros, set f(x) = 0. 0 = x^4 - 25x^2 x^4 = 25x^2 x^2 = 25 x = 5, -5. So your 3 zeros are: (-5,0), (0,0), and (5,0). to be certain what they do with admire to the x-axis, i might draw out the graph and notice what happens (probable swifter in case you have paper). i did no longer draw it out, yet from only plugging in a million and -a million into f(x), it sounds like f(a million) and f(-a million) are the two -24, as a effect (0,0) touches the x-axis and turns around. For the different zeros, i might plug in -6, -4, 4, and six, and notice what you get; if -4 and -6 / 4 and six are the two the two valuable or adverse, than the 0 only touches the x-axis; if the two -4 or -6 / 4 or 6 is valuable at the same time as the different is adverse, than it crosses the x axis. wish this permits! :)

Answer 4

You are looking for all values of x where y=0.
since the expression for x is equal to zero, this means that y is zero.

All you have to do is solve the given equation for all values of x and these will be the zeros that you seek

Answer 5

TI -83 says X=0 at Y= 3. Not sure on this one. Sorry

Answer 6

x^3 + 4x + 3 = 0

x^3 + 3x + x + 3 = 0

x^2(x + 3) + 1(x +3) = 0

(x +3)(x^2+1) = 0

x + 3 = 0 or

x^2 + 1 = 0

x = -3 or

x^2 = -1

x = ±√-1

x = ± i

so three zeros of given expression are -3, i, -i

Answer 7

put the problem in to a graphing calculator (you can use one online if you dont have one) and where ever it crosses the x-axis are the zeros

If you are using a TI calculator e-mail me and i can tell you how to find the zeros exactly with only a couple of keys