Polynomials (1 Viewer)

[WildestDreams]

In polynomials, how do you figure out if a question has a double or triple root? Using the long division method?

 

[InteGrand]

In polynomials, how do you figure out if a question has a double or triple root? Using the long division method?

What was the question? There may be shortcuts available.

 

[fluffchuck]

In extension 2, if a polynomial has a double root, this root is also true for its first derivative, if it has a triple root, this root is true for its first and second derivative and so on.

But keep in mind, you can only use this in extension 2.

 

[WildestDreams]

The question was to factorise P(x)=−x^3 +x^2 +5x+3

 

[WildestDreams]

In extension 2, if a polynomial has a double root, this root is also true for its first derivative, if it has a triple root, this root is true for its first and second derivative and so on.

But keep in mind, you can only use this in extension 2.

Ohh so does this only occur in Ext 2?!?! So I don't need it for Ext 1?

 

[WildestDreams]

 

[fluffchuck]

P(x)=−x^3 + x^2 + 5x + 3

P(x)=−(x^3 - x^2 - 5x - 3)

Factors of the constant term include: 1, -1, 3, -3

Test for x= 1, -1, 3, -3

P(1)=−1^3 + 1^2 + 5(1) + 3
P(1)=8
Since P(1) is not 0, it is not a zero.

P(-1)= -(-1)^3 + (-1)^2 +5(-1) + 3
P(-1)= 0
Since P(1)=0, x=1 is a zero.

Now you just long divide P(x)=−(x^3 - x^2 - 5x - 3) by x-1 hope I helped

 

[fluffchuck]

Ohh so does this only occur in Ext 2?!?! So I don't need it for Ext 1?

This is a technique in extension 2, keep in mind there are techniques to solve this in extension 1, I have demonstrated how to do it above

 

[InteGrand]

The question was to factorise P(x)=−x^3 +x^2 +5x+3

 

[WildestDreams]

Ohhh that's actually a smart way to look at it! Thank you so much, I shall attempt to use that method

Would you do the same for a triple root?

 

[InteGrand]

Ohhh that's actually a smart way to look at it! Thank you so much, I shall attempt to use that method

Would you do the same for a triple root?

Probably. If it was 4U, we could differentiate P(x) and sub. in a root we've found already to see if it's also a root of the derivative, which would show us whether or not it's a double root (and if it was a double root, we could differentiate again and sub. in to the second derivative to check if it's a triple root, and so on).

With the above method, it doesn't matter whether the roots are repeated or not. Once we've found one root of a cubic, we can factorise it by inspection to get a quadratic factor, and usually quadratics are easy to factorise by inspection.

You can also go through checking all the factors of -3 (or whatever the relevant number is) to see if you can first find any other nice roots that way.

 

[Paradoxica]

Another method is using residues.

Suppose P(x) has a root α.

Now, you don't know if it has a, say, triple root of α.

Then what you can do, is take the expression

P(x)/(x-α)³, and take the limit using L'Hôpital's Rule, and, if it does happen to be a triple root, the limit will approach some non-zero constant.

If it approaches 0/0, then that means the root has multiplicity higher than 3, and if it vanishes, then it has multiplicity lower than 3.

The same technique works for roots of any multiplicity, not just 3. Just make sure you divide by the necessary (x-α)^n, where n is the multiplicity.

However, differentiating can be quite tedious after a while, and it would be much better to just perform the factorisation by inspection.