factorising cubic equation (1 Viewer)

[horndog huang 2]

i) Show that the curves y=x^2+4 and y=x^3 intersect at the point ( 2, 8)

HOw do i factorise x^3-x^2-4=0

THanks

 

[InteGrand]

i) Show that the curves y=x^2+4 and y=x^3 intersect at the point ( 2, 8)

HOw do i factorise x^3-x^2-4=0

THanks

By inspection, 2 is a root. So (x – 2) is a factor. Now you can use polynomial division or inspection to get a remaining quadratic factor; you should end up with x³ – x² – 4 = (x – 2)(x² + x + 2). It turns out there will be no other real roots as the quadratic factor has negative discriminant.

But in order to show the curves intersect at that point, you can just sub. in x = 2 to both curves and show they both have y = 8.

 

[davidgoes4wce]

You could also follow the 'rational roots theorem' by considering the positive and negative of the trailing and leading coefficient. (in this case the leading coefficient being 1 and the trailing coefficient is 4)

Consider all the factors of the trailing term ('p'), 4 (including the negatives) are (+/- 1, +/-2, +/-4) and factors of the leading term (q), 1 (including the negatives) are (+/-1 only)

Diving the trailing term by the leading term p/q your only possibilities are: +1, -1,+2,-2,+4 and -4. In this case, x=2 , (x-2) is your only solution.

(x-2)(x^2+x+2) is your factorised form. But keep in mind that (x^2+x+2) there are no solutions when y=0.

 

[Mr_Kap]

You could also follow the 'rational roots theorem' by considering the positive and negative of the trailing and leading coefficient. (in this case the leading coefficient being 1 and the trailing coefficient is 4)

Consider all the factors of the trailing term ('p'), 4 (including the negatives) are (+/- 1, +/-2, +/-4) and factors of the leading term (q), 1 (including the negatives) are (+/-1 only)

Diving the trailing term by the leading term p/q your only possibilities are: +1, -1,+2,-2,+4 and -4. In this case, x=2 , (x-2) is your only solution.

(x-2)(x^2+x+2) is your factorised form. But keep in mind that (x^2+x+2) there are no solutions when y=0.

Yeh, because a 2 unit student is gonna wanna learn that. haha

 

[davidgoes4wce]

Yeh, because a 2 unit student is gonna wanna learn that. haha

I just read a bit about that from one of the online sources. I was studying 'Polynomials' last year and I'll admit it , this time last year had 'no idea' about it and wanted to figure an alternative way to solve for this.

 

[InteGrand]

Yeh, because a 2 unit student is gonna wanna learn that. haha

They should at least know it for the case of a monic polynomial shouldn't they? This way they know what to test for possible roots, namely the (positive and negative) factors of the constant term. That was how I saw 2 was a root, because I tested some factors of the constant term 4.

 

[Drongoski]

i) Show that the curves y=x^2+4 and y=x^3 intersect at the point ( 2, 8)

HOw do i factorise x^3-x^2-4=0

THanks

The best approach is not to solve for the 2 equations simultaneously; that is what you were trying to do.

Instead, all you need to do is to show that the point (2,8) lies on both curves, viz:


- 8 = 2² + 4

- 8 = 2³



Edit

InteGrand has in fact pointed this out; I did not see it earlier.

 

[psyc1011]

I required knowledge of x - 2 being a factor of the polynomial by factor theorem demonstarted by Integrand, then it sparked this idea:

x³ - x² - 4 = x³ - 2x² + x² - 4

Then factorising each pair

x²(x - 2) + (x - 2)(x + 2) = 0

(x - 2)(x² + x + 2) = 0

This is a bit hard to see..

The reason on why I splitted (in my first line) that up is I wanted to factorise (x - 2) since it's a factor

 

[horndog huang 2]

Oh i see, this is still a bit confusing but i just divide by the factors of the constant term to find a root right?

 

[davidgoes4wce]

The reason I say the 'rational roots' theorem is in some instances, you can see solutions which are all rational terms for a cubic. I have seen instances of this but in this question it is a whole number.

 

[Mr_Kap]

Oh i see, this is still a bit confusing but i just divide by the factors of the constant term to find a root right?

If you are doing two unit you do not need to know how to factorise cubics like this. that involves factor theorem and all that. For the first question you would do as people above have done and just sub the point (2,8) in the equation and show it equals 0.

 

[horndog huang 2]

ye doing 2u, but i needed to roots to find the area between two curves lol

 

[braintic]

It is a SHOW question for god sake. Why is everyone trying to solve? Just show by substitution. Solving this equation is not a 2 unit question.

 

[InteGrand]

It is a SHOW question for god sake. Why is everyone trying to solve? Just show by substitution. Solving this equation is not a 2 unit question.

Some people did say to just sub. in x = 2. And the reason everyone tried to solve is probably that the original poster was asking how to factorise the expression (also in thread title).

 

[DatAtarLyfe]

Just as an fyi, if you do have to factorise a cubic or even quartic function, inspection (which is what integrand said) is the best way but i think they only teach that in mx2 polynomials

 

[InteGrand]

Just as an fyi, if you do have to factorise a cubic or even quartic function, inspection (which is what integrand said) is the best way but i think they only teach that in mx2 polynomials

I don't think they usually teach inspection specifically, it just comes from practice.

 

[DatAtarLyfe]

I mean i was taught by my teachers so idk

 

[psyc1011]

Just as an fyi, if you do have to factorise a cubic or even quartic function, inspection (which is what integrand said) is the best way but i think they only teach that in mx2 polynomials

you learn factor theorem in MX1 though

 

[DatAtarLyfe]

you learn factor theorem in MX1 though

You can only implement factor theorem if you know one of the factors. Factoring by inspection is a way to find the factors but yeh you still need to use factor theorem in order to do it

 

[psyc1011]

You can only implement factor theorem if you know one of the factors. Factoring by inspection is a way to find the factors but yeh you still need to use factor theorem in order to do it

Thats not the only time to use factor theorem. Factor theorem can be a quick way to check that a guessed factor is a factor. Guesses are usually -2, -1, 1, 2 (kinda leads to david's post but 2u dont need it.)