Factorising

hi Tenacious

Okay lets take a look    

7x^2+10x-360

I am looking for 2 numbers that multiply to give  7*-360 = -7*360    I already know that one will be pos and the other neg because when they are multipied together the answer will be negative.

and the two numbers have to add to give +10     That is positive so the the bigger one will be positive.

They are only 10 apart so the numbers are quite close.

I did play around with this but it was just too horrible so I am going to take a different tact

-----------------------------------------------------------------

I am going to look at

7x^2+10x-360=0

I am just going to start by looking at the discriminant of the quadratic formula - that is the bit under the square root.

$$\triangle = 100+4*7*360 = 10180$$    This is not a perfect square so there are no rational roots.

So none of those answers are correct!

The problem that you gave us cannot be factored using whole numbers; it will need square roots.

There could be a mistake when the test was typed.

There is a formula -- you will probably be given it later -- that shows why square roots are necessary:  The formula is:  b² - 4·a·c

When you get the problem reduced to this form:  7x² + 10x - 360,

'a' is the coefficient of the squared term = 7

'b' is the coefficient of the linear term = 10

'c' is the constant = -360

Each of these values includes any negative sign that precedes the number.

So:  b² - 4·a·c  =  10² - 4·7·-360 =  100 + 10080  =  10180

Now, if this number is a perfect square, the problem can be factored into whole numbers; if this number is not a perfect square, the problem cannot be factored. Since this number is not a perfect square, the problem cannot be factored into whole numbers.

So, there must be a typo somewhere.

Thanks for the answers guys.

I'm glad I was right in firstly doubting any of their answers were correct and secondly doubting that it could be done. At least this confirms that I am on the right lines. It's surprising how many errors you can find in these online tests. When it comes to those that have supplied them for free, I do let them know if possible but I am grateful that they have made them available. This one was from a few sample questions to tempt you to pay for online service of test papers etc. They shall not be receiving any money from me.  

geno3141 - I am vaguely aware of the basic formula where a,b and c are related to the separate terms but I have not encountered the formula you have given before. I shall try to make sense of it but it is a bit past where I'm at at the moment. I shall try to work my way up to it. Thanks.

Melody - I tried a few different ways of doing this but yours is yet another for me to try. Thanks again.

I'm glad I got as close as I did but was still aware that it seemed as though it couldn't be done.

Thanks.

Hi Tenacious,

If you know what you want to learn then we can help point you in the right direction.  There are heaps of good resources out there and we love to help out when people get stuck. 

Thanks Melody,

At the moment I'm just doing any online tests at the GCSE (senior/high school)/University entry level and anything I have trouble with, I dig into until I have a better understanding of it or until my brain starts to fry, then I go back to the tests to find my next challenge. I'm at the brain frying point with factorising at the moment so I'm letting what I have learned so far sink in but this one turned up in a sample test and I felt it was at a level that I should have been capable of so couldn't resist challenging it.

I'm trying to cover everything in maths up to a similar level, going up in increments across the board so I don't get to confident in some areas at the expense of others. I don't see this being a quick task but it should be worth it.

Thanks.

It sounds like a good approach.

Are you doing this for general interest or are you hoping to work your way up to enter some course or something like that?

Factorising is a big, important topic.  I would like to see what you can do.

Could you factorise these for me please  (only if you want too - obviously)

$$\\1)\;\;5xy^4-10xty+5x^7t^2\\\\
2)\;\; x^2+x+6$$

It is a mixture of self improvement so that I am more capable for day to day maths but with the added incentive that being more capable at maths could help at future employment and possibly lead to gaining qualifications which could help with future employment opportunities. I haven't even looked into what qualification I could go for as it's not my immediate plan.

I'm not sure I should be admitting this on this forum but I don't actually enjoy maths much, possibly because I'm not very good at it, possibly why I'm not very good at it? I do get a sense of satisfaction from finally working something out, it's like solving a puzzle, or doing well in a mock test I have found, but I still wouldn't call it fun.

I didn't get a chance to do any yesterday but I have been puzzling of your two questions for a while today.

1) I spent time trying to get this into two brackets as I've become used to that format. Eventually I cottoned onto the common factors they all share and then it started to sink in.

5xy⁴-10xty+5x⁷t²

= 5xy⁴-10txy+5t²x⁷

= 5x(y⁴-2ty+t²x⁶) That was quite challenging, I think it's correct.

2) This looks simple due to only having a few characters but I'm stumped by it. There are no common factors across all three terms so it can't be a term multiplying the contents of a set of brackets as in my answer to question one, so I presume it has to be a pair of brackets?

x²+x+6   As it has positive terms only I have to use positive terms only in the brackets? I could use two negatives to achieve the +6 but then I would have a -x.

= (x+?)(x+?) = x²+x?+?

= (x+2)(x+3) = x²+5x+6 or (x+1)(x+6) = x²+7x+6  Easy enough to get the first and last term but not the second one correct.

= (x-2)(x+3) = x²+x-6   Got the first and second term but not the last.

Tried lots more alternatives on this, including negative exponents, but still not got it. I intend to keep trying, but at least you have an answer for question one.