Dear Professor, I just want to say your videos are real helpful to college students and also for refreshing forgotten topics tips and tricks. I hope you get more recognition, views, and subscribers with this channel
A little trick I like to use to avoid getting confused is to write the products to the side, instead of underneath the polynomial, then multiply by -1 and writing that under the polynomial. It might not work for everyone, but I find it simpler to keep from getting confused with my operations.
Thank you so much sir. I have been struggling a lot with Calculus and this helped me a lot as I struggle with long division the most. It helped me understand better. Thank you
Nice. You solved it so easily. In our book we have to do partial fraction. To convert improper fraction to proper, we divide the denominator with nominator. Then "Let" the denominator into "A1", "A2", etc. Then put some number in "x" to find thr values of A1, A2, etc and then out the value of A1, A2, etc in equation 1. But i think in america they use A,B,C,D instead of A1, A2,... Was a fun chapter ngl. The one about "theory of quadratic equation" was giving me a headache tho, ngl lol.
Another trick for reducing improper polynomial fractions: add zero in a fancy way. Given x^2/(x + 1) to reduce. Create a quadratic of x*(x + 1), to make the factor (x + 1) appear upstairs: x*(x + 1) = x^2 + x Add zero in a fancy way, i.e. add "+ x - x", to form a term we can cancel: (x^2 + x - x)/(x + 1) = (x + 1)*x/(x + 1) + -x/(x + 1) = x - x/(x + 1) Repeat the step of adding zero (i.e. add "+1 - 1"): x - (x + 1 - 1)/(x + 1) = x - [(x + 1)/(x + 1) - 1/(x + 1)] = x - (x + 1)/(x + 1) + 1/(x + 1) Result: x - 1 + 1/(x + 1)
I have already uploaded a video on synthetic division. Long duvision is usually helpful to learn before learning the synthetic division. Thanks for pointing it out.
Why short cut ? It is better to make a little practice or mental floss. Can you try to solve this problem ( the circumference of the hind wheel of a wago⁴n is 5 ft more than that of the front wheel. If the hind wheel makes 150 fewer revolutions than the front wheel in going one mile, find the circumference of each wheel . Thanks for ur effort ! ).
@@khalifmohamed2486 In 1 rotation, a wheel's axle moves a distance equal to its circumference, assuming a simple flat road. Assign wheel 1 to the front, and wheel 2 to the rear. Let C1 and C2 be their circumferences, and N1 and N2 be their two numbers of rotations. Assign constants, T= 150 for the difference in turns, and F = 5 ft, and D = distance of 1 mile or 5280 ft, to avoid writing big numbers at intermediate steps. This allows us to create the following equations: N1*C1 = D N2*C2 = D The other two constraints given are: C2 = C1 + F N1 = N2 + T 4 equations, 4 unknowns, a fully constrained system we can solve. Consolidate the equations, so we only have one equation with one unknown and only constants with it: C2 = C1 + F D/N2 = C1 + F D/(N1 - T) = C1 + F D/(D/C1 - T) = C1 + F Multiply to clear denominators: D = (C1 + F)*(D/C1 - T) C1*D = (C1 + F)*(D - T*C1) Shuffle to the same side and expand: (C1 + F)*(D - T*C1) - C1*D = 0 T*C1^2 + F*T*C1 - D*F = 0 Quadratic formula for C1. Plug in data: C1 = (-F +/- sqrt((F*T)^2 + 4*T*D*F))/(2*T) C1 = (-5 +/- sqrt((5*150)^2 + 4*150*5280*5))/(2*150) Solutions: C1 = 11 ft & C1 = -16 ft. Only C1 = 11 ft is a realistic solution. From the given relationship to C2, we can determine that C2 = 16 ft.
@@mrhtutoring Thank you for going through the long division. For some of us the most important thing isn't WHAT is the answer, it is the HOW you get the answer.
