Recent content by MiddleEast

Exactly, thats my point it is at least non-standard way to simplify polynomial without standard long/synthetic vision method. This could help with higher level math questions when coe's are arbitrary constants.

It is given in the question that x=2 is a root, and solution 1 was straightforward by dividing f by (x-2) and factor resulting quadratic equation to get f completely factored out as product of linear terms. Author said division could be long or synthetic. Up to here, it is pretty standard...

Not deriving, main goal is rewriting polynomial in terms of (x-a) in each term in one single step so that (x-a) will be factored out.

You wont start by writing A(x-2)+B(x-2)+C(x-2), you start by writing (x-2), divide first term of f by x (first term in x-2) and you would get "2x^2(x-2)", then ON SAME LINE continue writing 2x^2(x-2)+ [space] (x-2) .. now try to figure out what to put in the space by considering x^2 term coming...

Thanks for reply, actually I found the way. Basically start with A(x-2) + B (x-2) + C (x-2) where A,B & C could be variables/constants. That A must be 2x^2 to get "2x^3" in original function, that A when multiplied by -2 will give -4x^4, now you have to choose B as x so that -4x^4 & B*x will...

Thanks for reply. I know that, basically divide f by (x-2). But am interested in the way they did it algebraically. Cause they gave two solutions for that problem in the book, first solution is the standard one which is dividing f by (x-2) which is a factor of f. The second solution they did as...

How can you rewrite polynomial in terms of (x-a) instead of x? One thing came to mind is rewrite each x as x-a+2 (So it is x-2+2 in our example) but this will take long time and a lot of algebra steps, how did they do it very fast in the attached picture? thanks

Thank you for your reply. Appreciated but it is algebraic. I need the one which uses euler formula.

Hello, This is actually not homework. I was google searching for "proving trig identities from geometric point of view), found one of the result which proves trig identities using Euler formula. I really liked it. Easier, quicker & simple. But when the author speak about sum to product formulas...

Thanks for quick reply. It is simple to start with R.H.S to L.H.S. As per solution, they move from L.H.S to R.H.S, thats my question how to play with it? Tried to get different common factors over and over, no success. It is supposed to move from L.H.S to R.H.S.

Hello, While following problem solution found this $$ 4a^4 + 8 a^3 + 8 a^2 + 4a + 1 = ( 2a (a+1) + 1 )^2 $$ Trying to figure out how did author do it but failed. Anyone?

Explanation is somehow terse. But problems are good. He divides them to exercises & problems. exercises are almost straightforward questions but problems make you think. In general, I would consider axler books between dry & rigorous levels. I took cohen as main book & axler as supplementary for...

Phase 1 (Gentle introduction) : Trigonometry by Lial. (Know concept in general without any in-depth explanation, easy problems to get used with subject + has good geometry refresher in first chapter). Phase 2 (Better problem set) : Trigonometry chapters in precalculus by david cohen. Much better...

Oops. It seems like I found this thread while searching. Anyway, david cohen deserves mentioning.

Precalculus by david cohen is a gem, big gem. It even has chapters/sections not found in other books. Highly recommended.