Factorising Quadratics
- Thread starter mufc802
- Start date
Well I was hoping to see your method, but mine is:
20x^2-17x-63 > x^2-17x-1260 (Taking the 20 from A to C)
-45 & 28 multiply to -1260 and add to -17
Divide by the highest common number:
(20x-45) /5 = (4x-9)
(20x+28) /4 = (5x+7)
In this case you only have the following factors to work with:
20: 1*20, 2*10, 4*5
63: 1*63, 3*21, 7*9
/edit:
Strange way of thinking but ok.
Where does this come from? 1260 has loads of factors. Seems like an overly hard way to do things, unless you are able to look up all possible factors of a number.
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I just look and think what makes 63?
You can have either 21 x 3 or 9 x 7 or 63 x 1.
Then look at 20 and you can have 5 x 4 or 20 x 1 or 10 x 2.
Then you think of combinations of the above two lines that will give a difference of 17.
Quickly you realise 9 x 7 and 5 x 4 work.
Although I'm writing this as if it's step by step it doesn't really work like that when I'm thinking about it. I just look at it and if it's relatively simple the obvious factors just come into my head. For this one I probably went a little more step by step in my head, discarding the factors I know are unlikely to work and keeping the more likely one (i.e. 9 x 7) in my head. Then just play about with the factors of 20.
You can have either 21 x 3 or 9 x 7 or 63 x 1.
Then look at 20 and you can have 5 x 4 or 20 x 1 or 10 x 2.
Then you think of combinations of the above two lines that will give a difference of 17.
Quickly you realise 9 x 7 and 5 x 4 work.
Although I'm writing this as if it's step by step it doesn't really work like that when I'm thinking about it. I just look at it and if it's relatively simple the obvious factors just come into my head. For this one I probably went a little more step by step in my head, discarding the factors I know are unlikely to work and keeping the more likely one (i.e. 9 x 7) in my head. Then just play about with the factors of 20.
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How do you quickly realise their the ones that work though? If I do it this way I have no idea and do one by one..
I don't really know. Practice and just seeing it I guess. If you find a method that works for you stick with it - as long as you get the right answer it doesn't matter how you get there.
Ok, consider the general equation:
ax^2 + bx + c = (mx+p)*(nx+r)
You only have the following factors to work with:
a=20: 1*20, 2*10, 4*5
c=63: 1*63, 3*21, 7*9
We know that 'm' and 'n' are the factors of 'a' that we need to chose from the ones shown above.
We know that 'p' and 'r' are the factors of 'c' that we need to chose from the ones shown above.
Remember that any of 'm', 'n', 'r' and 'p' can be negative, or positive.
If 'a' is negative, we know one of 'm' and 'n' is negative.
If 'c' is negative, we know one of 'p' and 'r' is negative.
We know that (p*n)+(m*r)=b.
It looks complicated, but it's actually very simple, and as nydryl says, once you know all the factors of 'a' and 'c', you get a feel for which ones work to give 'b'.
/edit - so in this example we have:
20x^2-17x-63 and:
ax^2 + bx + c = (mx+p)*(nx+r)
Factors (of 'a' and 'c'):
a=20: 1*20, 2*10, 4*5
c=63: 1*63, 3*21, 7*9
c=-63 so one of 'p' and 'r' is negative.
We know that 'b' is -17, so if b=(m*r)+(n*p)
Just work through what combinations of these give -17
.
.
.
Again it looks complicated, but once you get used to it, it's really easy.
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Thats the reason I like my way, there's no 20x and 63 to get factors for both, just one set of factors just for 1260. I just can't do it with two lots of factors in there.
@div0, I understand that method, the one we were actually taught in the lesson, but you have to somehow guess either 1x20, 2x10 or 4x5 then work out the other factors for 63 as well, and have to make sure the a factors multiply by the c factors to add up to the value of b. With my method you only need the factors of 1260, to also add/subtract to 17.
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