Solving A Nice Cubic Equation | #Factoring #Algebra

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If you need to post a picture of your solution or idea:
x^3+1=(x-5)^2
#PolynomialEquations #Polynomials
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PLAYLISTS 🎵 :

[pietergeerkens]

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.

[misterdubity]

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

[erikroberts]

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?

[RubenHogenhout]

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.

[mcwulf]

There's an element of guess and check in method 2. But yes, if it works it's much quicker

[NovaH]

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

[tryingtomakeanamebelike]

Just plug in 2 after rearranging then factor

[ospninja]

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.

[alextang]

2^3+1=9
(2-5)^2=9
So 2 is a root and then it will be easy to find another 2 roots. 🤴🤴🤴🤴🤴🤴

[leif]

Yea but WHY 99--if youwantasquareWHY NOT 4 its a perfect square-so dontyou agree ppl would think of that also