# Homework Help: How do i find the roots of this?

1. May 6, 2008

### DeanBH

y = X^3 - 4x^2 -7x + 10

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. May 6, 2008

### symbolipoint

You need to factor the expression into linear binomials or a quadratic and a linear binomial. Study synthetic division and rational roots theorem.

3. May 6, 2008

### Nick89

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.
$$x^3-4x^2-7x+10=0$$

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:
$$(x-a)(...) = x^3-4x^2-7x+10$$

To find the remaining (...) you can use Long Division:
$$(...) = \frac{x^3-4x^2-7x+10}{x-a}$$

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:
$$(x-b)(x-c)(x-y) = x^3-4x^2-7x+10$$ for some unknown y.

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

Last edited: May 6, 2008
4. May 6, 2008

### HallsofIvy

There is no easy way to factor any polynomial- just look at the factors of constant term and then "trial and error". And cubics are much harder than quadratics.

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

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

5. May 6, 2008

### DeanBH

I ended up trial and erroring it to find that its -2 1 and 5, but yeh..

i thought there might have been a method. thx

6. May 6, 2008

### Nick89

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 $$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$?
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...

Last edited: May 6, 2008
7. May 6, 2008

### symbolipoint

Some students begin learning about such polynomials and their zeros in Intermediate Algebra; most learn about this in PreCalculus/College Algebra.

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.

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