Watching these videos, I've started to realize that the guess-and0check can be simplified when the linear term shares a factor with the constant term, as here where our polynomial becomes: 0 = 8*(x/2)^3 - 4*(x/2)^2 + 20*(x/2) - 24 = 2*(x/2)^2 - (x/2)^2 + 5*(x/2) - 6 Here, it's now easy to see that ((x/2)-1), equivalently (x-2), is a factor; but the analysis above is a big help when the original reduction is by a factor of the linear root, rather than the root itself, by simplifying the mental math required in quick guess-and-check.
Got x=2 by guess✓ then noted that for the right limb of the parabola, the slope of the cubic will always be greater than the slope of the quadratic (compare the derivatives) so they will never meet again. Only one solution
After 1.52 I saw that X=2 was a solution thus I did long division. (X^3 - X^2 +10X -24) / (X-2) = X^2 +X +12 And this kwadratic had no zeros. Thus X=2 was the only zero. The deriv also had no zero s thus also no extremes Then the function is monotoom rising with only one zero in X=2.
At the end of each of these kinds of problems in which you show the graph, is there any way that you could sometimes show the complex solution on the complex plane?
Matt Parker has some videos where he talks about this. Basically, we’re taking a 2D number (which is what complex numbers are, 2D) and mapping it onto another 2D-number. This means that we need a 4D graph to illustrate the function, which isn’t very easy to visualize.
@Umbra451 I've seen this accomplished in Complex Analysis books. They did it by showing the domain as one graph and the range as another graph, side by side.
This is a very common question in my country school exam. If you can't pull out the common factor then you just guess a root and use Horner's algorithm. Cubic or quaric equation tend to grow really fast so the root is usually close to 0
Sorry for the question but at 8:29 when he says, “let’s put everything on the same side.” What happened to the (x-2) from the right side? To put it all on the same side, wouldn’t you divide by (x-2)(x-8)? It seems that the (x-2) from the RHS disappeared, and the x-8 was added to the LHS as (-x+8). Can someone please explain? Thank you.
Watching these videos, I've started to realize that the guess-and0check can be simplified when the linear term shares a factor with the constant term, as here where our polynomial becomes:
0 = 8*(x/2)^3 - 4*(x/2)^2 + 20*(x/2) - 24
= 2*(x/2)^2 - (x/2)^2 + 5*(x/2) - 6
Here, it's now easy to see that ((x/2)-1), equivalently (x-2), is a factor; but the analysis above is a big help when the original reduction is by a factor of the linear root, rather than the root itself, by simplifying the mental math required in quick guess-and-check.
Got x=2 by guess✓ then noted that for the right limb of the parabola, the slope of the cubic will always be greater than the slope of the quadratic (compare the derivatives) so they will never meet again. Only one solution
Nice question...👌👌👌.
After 1.52 I saw that X=2 was a solution thus I did long division. (X^3 - X^2 +10X -24) / (X-2) = X^2 +X +12 And this kwadratic had no zeros. Thus X=2 was the only zero. The deriv also had no zero s thus also no extremes Then the function is monotoom rising with only one zero in X=2.
There's an element of guess and check in method 2. But yes, if it works it's much quicker
At the end of each of these kinds of problems in which you show the graph, is there any way that you could sometimes show the complex solution on the complex plane?
Matt Parker has some videos where he talks about this. Basically, we’re taking a 2D number (which is what complex numbers are, 2D) and mapping it onto another 2D-number. This means that we need a 4D graph to illustrate the function, which isn’t very easy to visualize.
@Umbra451 I've seen this accomplished in Complex Analysis books. They did it by showing the domain as one graph and the range as another graph, side by side.
@@erikroberts8307but how do you know which point in the domain corresponds to which point in the range then?
@@Umbra451😊l8😊íooooppoopppppppiooopopoooooopppopop9ooppl😊😊😊😊😊😊😊09😅00o9😅ooooo00pppppppppppppppppp😅ppppp0ppp00po99l
Excelente
I used the first method.
This is a very common question in my country school exam. If you can't pull out the common factor then you just guess a root and use Horner's algorithm. Cubic or quaric equation tend to grow really fast so the root is usually close to 0
Which grade you from?
@@bgpcusercheater5174 this is a typical 9th grade question for gifted students and 10th grade question for all students
@@NovaH00alright okay thnx
Sorry for the question but at 8:29 when he says, “let’s put everything on the same side.” What happened to the (x-2) from the right side? To put it all on the same side, wouldn’t you divide by (x-2)(x-8)? It seems that the (x-2) from the RHS disappeared, and the x-8 was added to the LHS as (-x+8). Can someone please explain? Thank you.
You dont divide, you substract and take common factor
Just plug in 2 after rearranging then factor
2^3+1=9
(2-5)^2=9
So 2 is a root and then it will be easy to find another 2 roots. 🤴🤴🤴🤴🤴🤴
Is this a gcse question?
Yea but WHY 99--if youwantasquareWHY NOT 4 its a perfect square-so dontyou agree ppl would think of that also
(x-2)(x^2+x-12)=0