Math Is Fun Forum

16 is a power of 2 so it's worth re-writing this with powers of 2

You can extract 2x out of that first expression leaving (2x)^1/3 which becomes a common factor in both terms.

I started the second and got to wondering if that last term should be (8x^3y^3)^(1/3).

It simplfies nicely if that power is 1/3 rather than 1/2.

Bob

From the (newtonian) equations of motion for a object  falling from rest with  an acceleration g (ok to take this as 10m/s/s) the final velocity is given by v = 0.5gt^2

So when t = 0.1   v = 0.05
              t = 0.2   v = 0.2
              t = 0.3   v = 0.45

................................................

              t = 1     v = 5

The v values aren't in arithmetic progression so that formula won't work.

Bob

A long time ago, when I was preparing for my A level exams (age 18) I did two things for practice.

Starting with an A4 piece of paper I expanded that as a determinant into 27 terms each with three letters.

Then I reversed that using a different order to make, say, this:

which demonstrates a property of determinants.

It's hard work and you have to be carefully accurate not to miss a term.

Starting with a quadratic with integers coefficients but which won't factorise easily I would do the formula 'in my head'. ie. remember each calculation answer and carry it forward to the next stage in my head. It's hard work but these two things had two big advantages. (1) It trains the brain to do hard stuff and (2) it teaches you to be very careful with the algebra.

I think of it as similar to an athlete doing weight training.  They may not be using weights in their sport but it develops muscle, flexibilty and stamina. Now I'm reminded I'm going to give it a go again to try and keep away from dementia.

Bob

Consider x^3 + 1.  The factor theorem can be used here. Set x = -1 and x^3 + 1 evaluates to zero. This means (x + 1) is a factor.

x^3 + 1 = (x+1)(x^2 -x +1)

Bob

When you do numeric long division, you're more likely to succeed with errors if you keep the thousands, hundreds, tens and units neatly in columns.  The zero place holders (yes, that's the term) help with the algebraic version. I did this by hand first on a sheet of plain A4 paper. It took about a third of the page to complete.

Bob

Division is defined as the inverse process to multiplication.

So, for example, we know that  6 x 8 = 48. This leads to 48 ÷ 8 = 6.

To attach a meaning to 48 divided by 8 we can ask "What number times 8 gives 48?"

So, a/0 would mean asking "What number times 0 gives a?" You cannot find a number to answer this question.

Instead of 0, you could consider a very small number instead of 0.  Let's choose a = 6 and divide by 0.1. We get an answer of 60.

Divide by 0.01 and we get 600. Divide by 0.001 and  we get 6000.  So as the divisor gets smaller the answer gets bigger so we might say as the divisor tends to zero the answer tends to infinity. This is useful when trying to sketch a graph.

What about 0 ÷ 0 ?

For this the question would be "What number times 0 gives 0?"  This time there's no shortage of answers as every number times 0 gives 0. So we have to say the answer is indeterminate.

In differential calculus  for a general point (x,y) we construct a chord by joining {x,f(x)} to {x + h, f(x+h)} and calculate it's gradient: {f(x+h) - f(x)} / {h}

As h tends to zero this gradient may tend to a limit and that enables us to assume the gradient at the point is that limit.  It works for lots of functions, so in those special circumstances 0/0 can be given a value.

Bob

You might spot an immediate factorisation but if not here's how to force it.

Let y = x^3 and the expression becomes y^2 + 2y +1 which can be factored into two identical factors,.

Putting x back into one of those and you're down to a cubic.  Substitute x = 1 and the expression is zero so x-1 nust be a factor. Extracting that leaves a quadratic which cannot be factored further.

Bob