Can You Factor and Solve This Cubic Equation in Under 60 Seconds? - Fast Math Trick

[@timeonly1401]

In the first cubic ( 10x^3 + 11x^2 - 2x - 3 ), we see that 10 & 11 and 2 & 3 pairs each differ by 1. So, immediately checked x=1 and x=-1 for roots: clearly 1 doesn't work since 10 & 11 are both + (no way to add to zero), but if we negate the coefficients of the odd-powered terms, we get -10+11+2-3 =-13+13=0 (yay!). Synthetic division verifies -1 as a root, and gives the quadratic 10x^2 + x - 3, from which we get the other 2 roots using the quadratic formula. In the 2nd cubic ( 15x^3 - 31x^2 + 4 ), we see that neither x=1 nor x=-1 are roots. So we use other clues... p(x) has 2 sign-changes => either 2 or 0 R+ roots, and p(-x) has 1 sign-change => 1 R- root (guaranteed!!). Using the rational root theorem to get candidate rational roots +/- (1,2,4)/(1,3,5,15), we 1st try +/- the "nicer" ones {2, 4, 1/3, 2/3}. Got lucky with +2 (not pure luck, if you notice that 15 is about 1/2 of 31, and using x=2 gives us a good chance of getting a root; using synthetic division verifies 2 as a root). Then quadratic formula to get remaining roots. Before discovering all the wonderful interesting math problems videos on KZbin (I now subscribe to 3 or 4 different ones; I can't get enough of them!!), i never needed any techniques beyond what learned from secondary school. These interesting problems and the techniques shown by the various presenters are excellent for expanding on my math toolset! I am very grateful!! Thx.

[@glasssmirror2314]

Please sir if you can give us a lecture to find Possible Divisors on fractions and also finding the three possible numbers which would work for the cubic equation.

[@justme9818]

Hello. -There are so many permutations of factors of the constant that could sum up to the coefficient of the squared term. -How can you know you're choosing the correct ones?

[@PreMath]

Dear just me!, it comes with lots of practice and drill. Math is just any other language! You must repeat it over and over again if you want to be fluent. Thank you so much for taking the time to leave this comment. I'm sure you are an awesome student. Keep it up. Best regards :)

[@marceloboda4218]

I want to know, how can i find 3/5 when you put 1,(1/2.5)

[@lefleurtobias3034]

Hi Sir. Please explain how you substituted ( 1/5.3 ) with ( 2/5 - 1/3 ). I know how to get the possible divisors, but not the penultimate step. Valuable time is lost if I do not know the manipulation to ( 2/5 - 1/3 ). I like this method a lot.

[@planqbutter]

heyyy how do i get possible divisors of a fraction please help me!!

[@humester]

In this case the following approach is simpler than trying to figure out so many possible divisors and then that 1/(2x5) = (3/5) - (1/2): 10 - 2 = 11 - 3 Since the sums of the alternate coefficients are equal, (x + 1) is a factor and x = -1 is a solution. We next use synthetic division to get a quadratic equation: -1 | 10 11 -2 -3 | -10 -1 3 | 10 1 -3 0  Remainder is zero 10x^2 + 1x - 3 = 0 10 x (-3) = -30, so we want numbers that multiply to give -30, and add to give (+1) 30 = 5 x 6 so (-5) and (+6) fit, giving 10x^2 - 5x + 6x - 3 = 0 5x(2x - 1) + 3(2x - 1) = 0 (2x - 1)(5x + 3) = 0 Therefore, the factors are: (x + 1)(2x - 1)(5x + 3), and the solution set is {-1, - (3/5), +(1/2)}

[@kfjfkeofitorhf9520]

X-3 10×(-8)+

[@yosiganish1218]

On the last one he didn’t explain the last stage which is very important for understanding Thank you

[@PreMath]

You are very welcome Yosi Ganish! Take care dear and all the best😃

[@alster724]

Factor/Remainder and the Rational Zero Theorems did the trick here

[@SuperYoonHo]

Wow that was amazing1 thanks a lot mr love your videos and love your channel its the best best best love and prayers from south korea

[@SuperYoonHo]

YoonHo

[@dipalishroff8856]

How can we choose the possible divisors.....

[@vaibhavsawroop6238]

How to know possible no. And how to use friction 3/5

[@neelayadagiri74]

Sir plz tell what to do when constant is +1/-1 and when coefficient of x cube is 1...

[@banazamalsadiwala7516]

Amazing

[@theophonchana5025]

#CubicEquation #factoring #factor

[@edmondscott7444]

Seems very complicated. I would use factor theorem to find 1 root, then divide cubic expression by factor to get a quadratic.

[@youlookweirddude747]

How to solve in less than 1 minutes but makes 10 minute video. Oh he's gooood

[@archiedavid1126]

how to get possible divisor/s?

[@planqbutter]

hi can someone help me how do u get possible divisors of a fraction like the first one??

[@himeshsingh9179]

What if the last term is missing i.e D ??

[@antoinelebon5357]

Your factoring seems strange to me, because x=2 is a counter example to x+0.6 becoming 5x+3 since you have 2+(3/5)= (5*2)+3, which is false!

[@samundrakhadka8330]

How _ came 11/10= 3/5+1/2=2×3 6+5×1=5=5+6=11=11/10 may be there + not _

[@theophonchana5025]

x - 0.5 = 0 x = 0.5

[@theophonchana5025]

x - 2 = 0 x = 2

[@theophonchana5025]

x + 1 = 0 x = -1

[@msafasharhan]

Difficult

[@paddle_shift]

Than you for not explaining the most important part of the solution!

[@raziasiddique3253]

You don't know anything go to play group to learn English and math

[@theophonchana5025]

x - 0.4 = 0 x = 0.4