STEP Tricks and Hacks Thread

I highly recommend reading through the majority of this thread. I recall it containing many helpful methods for STEP.

Properties of Roots of Polynomials
Im pretty sure this hasn't been explicitly (me thinks) covered already and pops up everywhere so it's definitely worth looking at!
Consider the equation $x^n+a_{n-1}x^{n-1}+...+a_{1}x+a_{0}=0$.
If the roots of this equation are $\alpha_{1}, \alpha_{2}, ..., \alpha_{n}$,
our polynomial can be expressed as $(x-\alpha_{1})(x-\alpha_{2})...(x-\alpha_{n})=0$.

By considering how each of the coefficients of the polynomial is formed, we see $a_{0}=\alpha_{1}\alpha_{2}... \alpha_{n}, a_{n-1}=-\sum_{r=0}^n \alpha_{r}$
This can be extended to other coefficients too using basic combinatorics.
Some common examples/questions using complex numbers:
Factorise as a product of real linear and quadratic polynomials.
Write as a product of real quadratic polynomials.

Strong induction v/s weak induction

I swear if you don't get at least S,1 in the steps I'll kill you!

I think Zacken will be the next DFranklin

I don't get the RHS of the a_0 part. Is that supposed to be a product sign? Also you can't pull out -1 out of the product it is (-1)^n depending on n even or odd.


Posted from TSR Mobile

Summing arctans is useful:

$\arctan a + \arctan b = \arctan \left( \frac{a + b}{1-ab} \right) + n\pi$

Which can be derived from the tangent compound angle formula.

Could you quote this without proof in a step question?
(Unless they asked you to prove it of course)

Lol.
Zackens good but not that good

Yes - think about what the proof is; this follows immediately from the tangent addition formula

Thanks, yeah It's come up in a step question before I think somewhere, just wondering if you could save time by quoting it straight off

When x is small;

$\displaystyle \sin(x) \approx x$

$\displaystyle \tan(x) \approx x$

$\displaystyle \cos(x) \approx 1-\frac{x^2}{2}$

again probably not useful, but better to post it and not need it and not post it and need it

this post is more useful i think, and im surprised it hasnt been posted about yet; (maybe everyone knows it already)

$\displaystyle \left(a^n + b^n \right) = \left(a-b\right)\left(a^{n-1} - a^{n-2}b + a^{n-3}b^2 - a^{n-4}b^3 + ... - ab^{n-2} + b^{n-1}\right)$

^^ this is only valid when n is even. Notice how you alternate signs the whole way down the right hand side bracket, starting and ending with plus signs

$\displaystyle \left(a^n - b^n\right) = \left(a-b\right)\left(a^{n-1} + a^{n-2}b + a^{n-3}b^2 + ... + ab^{n-2} + b^{n-1}\right)$

valid for all integer n, and notice how the right hand bracket is all plus signs

This came up on Q2 of the 2015 Oxford MAT.