okay the question is: find all the real roots of 5x^3-17^2-x+30=0

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- Sep 2nd 2008, 02:44 AMrequalAnother simple polynomial question
okay the question is: find all the real roots of 5x^3-17^2-x+30=0

- Sep 2nd 2008, 02:57 AMMoo
Hello,

Try this : Cubic function - Wikipedia, the free encyclopedia - Sep 2nd 2008, 03:14 AMrequal
Is there any other way of doing this ala factor theorem, or long divison? Because I have no clue of what a "cubic function is".(Doh) If there isnt, can sum1 explain it to me?(Talking)

PS- I just realised that this was in the wrong forum- feel free to move this to the high school section if u must - Sep 2nd 2008, 03:15 AMmathceleb
Do you have a typo in that? Should that be $\displaystyle 5x^3-17x^2-x+30$

If so, check out my cubic calculator that shows you the math:

http://www.mathcelebrity.com/cubic.php

Since the discriminant is less than 0, there are 3 real roots. - Sep 2nd 2008, 03:23 AMrequal
okayyyyyyyyyyy... I dont understand one bit of the theorem. I guess my teacher made a typo or somethin(Headbang). Thanks anyway(Clapping)

- Sep 2nd 2008, 04:24 AMticbol
In solving quadratic equations that cannot be factored or hard to factor, the Quadratic Formula is very handy and useful and simple.

In soving cubic equations that cannot be factored, or are hard to factor, the Cubic Formula is hardly useful and "simple" is a term that is alien to its use.

For cubic equations then I usually resort to iteration or repeatition or trial and error. (The synthetic division is faster for integer values of x).

By iteration, I mean I try values for x that will make the given cubic equation true.

I always start with x=1. Then x=2, x=3, etc. Or x = -1, x = -2, x = -3, etc.

So, if x = 1,

5x^3 -17x^2 -x +30 = 0

5(1^3) -17(1^2) -1 +30 = 17 .......not zero.

If x = 2,

5(2^3) -17(2^20 -2 +30 = 0 ........hey, it is zero already!

Meaning, 2 is a root, and (x -2) is a factor.

So, use long division, or synthetic division, to find the qudratic remainder.

You shold find it as

5x^2 -7x -15 = 0

You cannot factor that anymore, so use the Quadratic Formula to find the other two roots.

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It does not always happen that an integer value of x will make the original cubic equation true. That is why watch the results as you test for integer values of x.

If x = 2 came out not zero, say it came out (- 3), then that means 2 is more than the correct value of x to make the cubic equation true or zero. That means x = 1.8 might be the correct root. So test x = 1.8 in the cubic equation. See if it will make the cubic equation to become zero.

Etc...

Iteration, repeatition, trial and error.

Later on , you'd come to established methods of iteration, like the Newton's method, to minimize trial and error.

By the way, if x = 2 resulted in, say, 4 for the above cubic, then try x = 3. You must go beyond zero, or to the negative, before you try non-integer values of x. - Sep 2nd 2008, 06:20 AMmathceleb
- Sep 4th 2008, 09:10 AMShyam
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