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i have to draw a graph of this, stationary points dont matter but all else does.

i know what it would look like because it is a cubic it will have 3 roots.

but i don't know how to find them, i forgot :P

can someone help please :)

- Thread starter DeanBH
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- #1

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i have to draw a graph of this, stationary points dont matter but all else does.

i know what it would look like because it is a cubic it will have 3 roots.

but i don't know how to find them, i forgot :P

can someone help please :)

- #2

symbolipoint

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For this equation it's not that hard.

The first thing I usually do is try to find roots manually. To find roots you need to solve y = 0.

[tex]x^3-4x^2-7x+10=0[/tex]

Just try a few integers ranging from -3 to 3 or something and you will probably find one or two values for x that yield 0.

Let's say you found 1 of these values,*x = a*.

You now know that (*x - a*) is a factor in the factorized form of the equation:

[tex](x-a)(...) = x^3-4x^2-7x+10[/tex]

To find the remaining (...) you can use Long Division:

[tex](...) = \frac{x^3-4x^2-7x+10}{x-a}[/tex]

An easier way, which afaik only works if you found two roots manually, let's say*x = b* and *x = c*, you can do this:

[tex](x-b)(x-c)(x-y) = x^3-4x^2-7x+10[/tex] for some unknown*y*.

Factor out the brackets and you can find the value of*y* easily.

The first thing I usually do is try to find roots manually. To find roots you need to solve y = 0.

[tex]x^3-4x^2-7x+10=0[/tex]

Just try a few integers ranging from -3 to 3 or something and you will probably find one or two values for x that yield 0.

Let's say you found 1 of these values,

You now know that (

[tex](x-a)(...) = x^3-4x^2-7x+10[/tex]

To find the remaining (...) you can use Long Division:

[tex](...) = \frac{x^3-4x^2-7x+10}{x-a}[/tex]

An easier way, which afaik only works if you found two roots manually, let's say

[tex](x-b)(x-c)(x-y) = x^3-4x^2-7x+10[/tex] for some unknown

Factor out the brackets and you can find the value of

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- #4

HallsofIvy

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That's why the first thing I would try is to "look for" simple zeros.

Very, very large hint: what is y(1)?

- #5

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i thought there might have been a method. thx

- #6

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Another hint, this equation allows you to find two roots easily so you don't need to use Long Division if you don't know how that works with polynomials.

EDIT

Too late...

What do you mean by a method?

Perhaps you mean some formula like the quadratic formula [tex]x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]?

If so, these formulas do exist for third (cubic) and fourth (quartic) degree polynomials, but they are really hard and you won't even try to remember them.

Check out Wikipedia for derivations: http://en.wikipedia.org/wiki/Cubic_function

If you encounter the need to find the roots of a cubic in school or on an exam, the roots are usually integers or very easy numbers.

In real world problems however this is usually not the case and the method I described above does not work. Instead you would probably use a computer who does it in seconds...

EDIT

Too late...

What do you mean by a method?

Perhaps you mean some formula like the quadratic formula [tex]x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]?

If so, these formulas do exist for third (cubic) and fourth (quartic) degree polynomials, but they are really hard and you won't even try to remember them.

Check out Wikipedia for derivations: http://en.wikipedia.org/wiki/Cubic_function

If you encounter the need to find the roots of a cubic in school or on an exam, the roots are usually integers or very easy numbers.

In real world problems however this is usually not the case and the method I described above does not work. Instead you would probably use a computer who does it in seconds...

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- #7

symbolipoint

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Try division using binomials of (x-1), (x-2), (x-5), (x+1), (x+2), (x+5), (x-10), (x+10). Use synthetic division if you know it because it is faster. When you divide and obtain a remainder of zero, you have found a linear binomial factor.