polynomial question (1 Viewer)

[sasquatch]

Show that 1 and 2 are zeros of the polynomail P(x) = x⁴ - 2x³ + 5x² - 16x +12 and hence that (x-1)(x-2) is a factor of P(x).

I showed that 1 and 2 are the zeros of the polynomail and hence the factors are (x - 1) and (x - 2). But how do you show that (x-1)(x-2) is a factor?

Does this make sense

When x = 1 or x = 2

P(x) = 0

(x - 1)(x - 2)A(x) = 0
(x - 1)(x - 2) = 0

.:. (x - 1)(x - 2) is a factor of P(x).

 

[Templar]

P(x)=(x-1)Q(x)
P(1)=0
P(2)=Q(x)
=0
Therefore Q(x)=(x-2)R(x)
P(x)=(x-1)(x-2)R(x)

 

[sasquatch]

P(x)=(x-1)Q(x)
P(1)=0
P(2)=-Q(x)
=0
Therefore Q(x)=(x-2)R(x)
P(x)=(x-1)(x-2)R(x)

I dont get that. P(2) = - Q(x), Is the - sign a typo?

If P(x) = (x - 1)Q(x) then

P(2) = (2 - 1)Q(x)
= Q(x)

Therefore Q(x)=(x-2)R(x)
P(x)=(x-1)(x-2)R(x)

I dont get tat bit either. Could you please explain? Thanks.

 

[Riviet]

You could also try using long division to divide p(x) by (x-1)(x-2). When there's no remainder, you can deduce that it is divisible.

 

[sasquatch]

hmm i tried that but i couldnt get anything usable..maybe i screwed it up.

When you divide by say (x -1)(x-2), you expand it right?

EDIT: Yeah i screwed it up i did a +3 instead of a -3 somewhere. Thanks all.

 

[sasquatch]

Also id like to ask if this is true about roots, as the book doesnt actually tell you.

Are:
Double roots are min/max turning points and,
Triple roots are points of inflexion?

Thanks in advance.

 

[Riviet]

When you divide by say (x -1)(x-2), you expand it right?

Yup.

Are:
Double roots are turning points and,
Triple roots are points of inflexion?

Double roots in any polynomial means there is a turning point right on an x-intercept, so the curve touches the x-axis at exactly that one point and bounches back off it, \./ or /^.\, where the . is the single point that it touches.

I'm not sure about a triple root, but it's not a point inflexion. However, a single root is a point of inflexion.

 

[sasquatch]

I'm not sure about a triple root, but it's not a point inflexion. However, a single root is a point of inflexion.

Just to test it out, i derived and ect. and for the question i was doing it turned out to be a horizontal point of inflexion.

Also your statement of a single root being a point of inflexion is false, such as condsidering e.g. y = (x + 1)(x - 1), we know for a fact that it as a parabola. Parabolas do NOT have any points of inflexion, but in this case it has two 'single' roots as you put it x = 1, x = -1.

Refering to triple roots being points of inflexion, consider the following:

P(x) = x³ + 3x² + 3x + 1

it ends up simplifying to
P(x) = (x+1)³

So we see there is a triple root when x = -1

Using calculus:

dP(x) / dx = 3(x +1)²
d²P(x) / dx² = 6(x + 1)

When d²P(x) / dx² = 0

x = -1
I cant be bothered to show its sign on both sides, but i have and it is a point of inflexion.

When x = -1
dP(x) / dx = 0

.:. (-1, 0) is a horizontal point of inflexion.

It also works as such for y = x³ so i was just wondering if it works in all cases that it is either a point of inflexion or horizontal point of inflexion, considering the result of two examples.

 

[Riviet]

Just to test it out, i derived and ect. and for the question i was doing it turned out to be a horizontal point of inflexion.

Also your statement of a single root being a point of inflexion is false, such as condsidering e.g. y = (x + 1)(x - 1), we know for a fact that it as a parabola. Parabolas do NOT have any points of inflexion, but in this case it has two 'single' roots as you put it x = 1, x = -1.

Refering to triple roots being points of inflexion, consider the following:

P(x) = x³ + 3x² + 3x + 1

it ends up simplifying to
P(x) = (x+1)³

So we see there is a triple root when x = -1

Using calculus:

dP(x) / dx = 3(x +1)²
d²P(x) / dx² = 6(x + 1)

When d²P(x) / dx² = 0

x = -1
I cant be bothered to show its sign on both sides, but i have and it is a point of inflexion.

When x = -1
dP(x) / dx = 0

.:. (-1, 0) is a horizontal point of inflexion.

It also works as such for y = x³ so i was just wondering if it works in all cases that it is either a point of inflexion or horizontal point of inflexion, considering the result of two examples.

If you knew, then why did you ask?

I meant to say this. A single root means the curve cuts through the x-axis at that one point, it may or may not be a point of inflexion, you have to test that yourself. A triple root usually means a point of inflexion, a quick test to make sure it is always helps.

 

[sasquatch]

No no no, i wasnt asking cuz i "knew". Before i did the question i suspected such because of the relation between min/max t.p.s and double roots, and then it turned out to work out in that question. I was asking generally, do triple roots mean a point of inflexion (or horizontal point of inflexion).

 

[Riviet]

I see... they're actually very useful for fully factorised polynomials of any degree, so a quartic root would be a very steep double root, a quintuple root would be a very steep point of inflexion, etc. Nice observations too.

 

[sasquatch]

I have another question!!!! What do the terms "arithmetic progression" and "geometric progression" mean, relating to polynomials?

 

[insert-username]

I have another question!!!! What do the terms "arithmetic progression" and "geometric progression" mean, relating to polynomials?

I would assume that they're the same to the series definition, i.e. an arithmetic progression increases/decreases by a constant, and a geometric progression increases/decreases exponentially.


I_F

 

[sasquatch]

Hmm.. that sucks for me then. I havent done series and sequence. I saw the book im using has an example with arithmetic progression, (sorry for asking before checking) but nothing on geometric progression.

When you say increasing exponentially do you mean a^x, where a is a constant? Or..well could you give me an example.

 

[insert-username]

An arithmetic series is given by:

T_n = a + (n-1)d, where T_n is the nth term, a is the first term, n is a positive integer, and d is the 'common difference' - how much the series is increasing by. An example is:

2, 4, 6, 8, 10, etc.

This is written as T_n = 2 + (n-1)2 - it is increasing by 2 at a time.


A geometric series is given by:

T_n = ar^n-1, again with T_n being the nth term, a being the first term, n being a positive integer, and r being the 'common ratio' - what the term is multiplied by. An example is:

2, 4, 8, 16, 32, 64, etc.

This can be written as T_n = 2 x 2^n-1 - each term of this series is double the previous term (r = 2, so each term doubles).

Hope that helps.


I_F

 

[sasquatch]

Hmmmm why the hell do they have to ask stuff like that in polynomials?!?! I dont really understand it.. ah well maybe i can just skip the question.. Its only one question.

 

[insert-username]

What was the question that referred to the progressions?


I_F

 

[sasquatch]

The questions were:

Solve the equation x^3 - 12x^2 + 39x - 28 = 0 if the roots are in arithmetic progression. (Following the example i was able to do this).

Solve the equation x^3 - 14x^2 + 56x - 64 = 0 if the roots are in geometric progression.

 

[Riviet]

The questions were:

Solve the equation x^3 - 12x^2 + 39x - 28 = 0 if the roots are in arithmetic progression. (Following the example i was able to do this).

Solve the equation x^3 - 14x^2 + 56x - 64 = 0 if the roots are in geometric progression.

For the first one, let the roots be a-d, a, and a+d.

For the second, let the roots be a/r, a, and ar.

Then use sum of roots, product of roots, and sum of the product of the roots two at a time and solve the equations simultaneously for a and d.

 

[sasquatch]

Oh yeah the book had that a - d, a, a + d for the arithmetic progression. I jsut didnt get the geometric progression question. Thanks.