Generally, you would estimate by rounding all the numbers that affect the solution.

This would usually be used to make a quick but not accurate answer. Sometimes I use an estimate to double check that my accurate answer is in the correct ball park. If my estimate and my accurate answer are very close then I can be a bit more confident that I am correct, if they are considerably different then I know to recheck both answers to see where I went wrong. 

What is the answer to (3x+5)(3x-5)?

Firstly, are you trying to expand these brackets or solve this for x?

is this question factorable 2y^2+9y+4

2y²+9y+4

I must admit, I nearly replied saying I didn't think this could be factorised, but luckily I got it in the end.

First we would look to see if there are common factors between the terms, there are none in this case. I then made the mistake of just looking for two numbers that could be added together to make 9 and multiplied together to make 4. Luckily I then laid out my brackets like this;

(2y??)(y??)

and remembered that the 2 before the "y" was also going to affect the two numbers that I was looking for.

The answer is yes, it is factorable. If you follow the layout of the brackets and their contents that I have given, hopefully you can fill in the gaps where I have question marks. If not, please reply and I or someone else will give you further help.

Ah yes, I was in a rush as was just on my way out and the first time I tried to publish it just went back to the question so had to retype it all. Must have retyped it incorrectly as it obviously doesn't work in that format with + and - in wrong order. Should have read;

x²+7x=18

x²+7x-18=0

(x-2)(x+9)=0  So x=+2 or x=-9

I always expand the brackets after factorising to double check and usually substitute the algebraic terms for my answers to double check as well. I rarely trust myself enough to just put the answers without double checking them. 

x²+7x=18

x²+7x-18=0

(x-2)(x+9)=0  So x=+2 or x=-9   Edited because I had the + and - the wrong way round originally, now correct. Thanks to Melody for pointing that out.

x²+9x-2x-18

x²+7x-18

By the time I saw this there were other answers so too late but I purposefully avoided looking at them, no point in cheating. I only saw the first two posts, the question and Melody's hints. I doubt I would have done this so easily without those hints Melody, so thanks. I think this is the first time I have encountered a quadratic trinomial in a practical ish situation. Up till now I have only encountered them as isolated exercises. It is useful to see how they can actually be used and be of benefit.

Don't worry, the + and - are pretty close together, easy to do.

Ah yes, so the trinomial was right but the quadratic comes from the term being squared not the four terms in two brackets as I thought. So I would say the question you have set below is a quadratic trinomial.

Let's try this one;

x²-x-6     Not too difficult

= (x-3)(x+2)

= x²+2x-3x-6

= x²-x-6

That's definitely a lot simpler Melody, but I was going for a fraction as requested, not a mixed fraction.

Depends what the poster wanted, but now they have both so should be happy, hopefully.

Lets say your fraction is $${\frac{{\mathtt{1}}}{{\mathtt{2}}}}$$ for example. If you type 1/2= you will get the answer 0.5 which is the decimal equivelant of $${\frac{{\mathtt{1}}}{{\mathtt{2}}}}$$.

So lets say for example you wanted to answer 10*1/2=

If you typed in 10*1/2=  you would effectively be typing 10*0.5= and the answer would be 5

I will try to demonstrate the method for this with a simpler number, then go onto your question.

If we take the number 1 and divide it by 2 we get half of 1 which is 0.5      $$\mathrm{\ }$$$${\frac{{\mathtt{1}}}{{\mathtt{2}}}} = {\mathtt{0.5}}$$

So we can see that as a fraction and a decimal, and we could have probably guessed the answer.

In this example we are saying that 0.5 is a fraction of 1, so could be written like this $${\frac{{\mathtt{0.5}}}{{\mathtt{1}}}}$$

Sadly $${\frac{{\mathtt{0.5}}}{{\mathtt{1}}}}$$ is what we call an improper fraction, in that the top number (Numerator) or bottom number (denominator) are not whole numbers.

To convert $${\frac{{\mathtt{0.5}}}{{\mathtt{1}}}}$$ to a proper fraction, where the numerator and denominator are both whole numbers we multiply both terms by 10 until we loose the decimal point.

$${\frac{{\mathtt{0.5}}{\mathtt{\,\times\,}}{\mathtt{10}}}{\left({\mathtt{1}}{\mathtt{\,\times\,}}{\mathtt{10}}\right)}}$$ = $${\frac{{\mathtt{5}}}{{\mathtt{10}}}}$$  This is now a proper fraction but needs to be reduced or simplified down to it's simplest form by finding a number (their greatest common divisor) that could divide both the numerator and denominator to give us whole numbers for the numerator and denominator.

$${\frac{{\mathtt{5}}}{{\mathtt{10}}}}$$ divided by 5 is simplified to $${\frac{{\mathtt{1}}}{{\mathtt{2}}}}$$ which we know as half of 1.

Now to answer your question; what is 276.2 as a fraction?

We are saying that 276.2 is 276 *1 and a fraction of 1 so would be written

$${\frac{{\mathtt{276.2}}}{{\mathtt{1}}}}$$   As in my example, we need to convert this to a proper fraction, so we multiply both the numerator and denominator by 10 until we lose the decimal point.

$${\frac{\left({\mathtt{276.2}}{\mathtt{\,\times\,}}{\mathtt{10}}\right)}{\left({\mathtt{1}}{\mathtt{\,\times\,}}{\mathtt{10}}\right)}}$$ = $${\frac{{\mathtt{2\,762}}}{{\mathtt{10}}}}$$ We now have a proper fraction with no decimal points. 

This could be a lengthy way to do it if we had more numbers after the decimal point, so we have a simpler method. In this process we can see that we had to multiply both the numerator and denominator by 10, once. This coincides with the amount of numbers after the decimal point in the original 276.2

So, if we said that for every number after the decimal point, we have to multiply both the numerator and the denominator by 10. This is effectively the same as in other math problems where we whish to move the decimal point.

Now all we have to do is simplify $${\frac{{\mathtt{2\,762}}}{{\mathtt{10}}}}$$ to it's simplest form by dividing it by it's greatest common divisor which is 2, $${\frac{{\mathtt{2\,762}}}{{\mathtt{10}}}}$$ divided by 2 equals $${\frac{{\mathtt{1\,381}}}{{\mathtt{5}}}}$$, which is your answer.