High School Math Solutions – Polynomial Long Division Calculator

Polynomial long division is very similar to numerical long division where you first divide the large part of the number, multiply the answer by the divisor, subtract and continue dividing the reminder until the reminder is smaller than the divisor.

With polynomials you do the same, only now you have to divide the higher order terms (first term of the numerator divide by the first term of the denominator).


Let’s start with a simple example  (click here):

Here’s another example this time with multiple iterations (click here):

Cheers,
Michal

High School Math Solutions – Partial Fractions Calculator

Partial fractions decomposition is the opposite of adding fractions, we are trying to break a rational expression into simpler fractions.  It takes a lot of work, but is extremely useful with integrals for instance (simplification can be a good strategy).  We start by factoring the denominator (if the numerator order is higher than the denominator we start with long division), then we write the partial fraction for each of the factors (watch out for high order factors), multiply and solve for the coefficients using the factors zeros.  Step by step examples can be really helpful here.

Let’s start with an example  (click here):

Here’s a more advanced example with high order factors (click here):

From here simply solve the equation and plug in the solutions to get the partial fractions.


Here’s an example where the order of the numerator is higher than the denominator (click here):

Cheers,
Michal

Advanced Math Solutions – Integral Calculator, trigonometric substitution

In the previous posts we covered substitution, but standard substitution is not always enough. Integrals involving radicals for instance, we want to get rid of the square root.
Here’s how:

• For \sqrt{a^2-bx^2}, let x=\frac{\sqrt{a}}{\sqrt{b}}\sin \left(u\right) and use the identity 1-\sin^2(u)=\cos^2(u)
• For \sqrt{a^2+bx^2}, let x=\frac{\sqrt{a}}{\sqrt{b}}\tan \left(u\right) and use the identity 1+\tan^2(u)=\sec^2(u)
• For \sqrt{bx^2-a^2}, let x=\frac{\sqrt{a}}{\sqrt{b}}\sec \left(u\right) and use the identity \sec^2(u)-1=\tan^2(u)

Here’s an example of tangent substitution (click here):

From here simply cancel, integrate and substitute back


Here’s an example of sine substitution (click here):

 
Here’s an example of secant substitution (click here):

We covered almost all there is to know about substitution.  In the next post we will cover inverse trigonometric functions.


Cheers,
Michal