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I don't really understand how to factor polynomials. It's mostly because we have a teacher who doesn't teach. Can anyone help?
Problems like:
7x²-14x
19x³-38x²
8j³k-4jk³-7
It's like a another language to me.
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well the general method, is to find the largest common factor of all the expressions:
so for example:
with 7x[sup]2[/sup] - 14x, you can immediately see that you can divide both sides by 7 without going into any decimal expansions:
and then by x keeping it even more simple:
and that would be the expressino factorised
similarly for the second one: 19 is a factor, and x^2 is a factor
Last edited by luca-deltodesco (2007-02-03 10:31:21)
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Thanks a bunch, I tihnk I kind of get it. Is it the same thing when it comes to problems with the polynomials being divided?
You have to look for common variables and coeficients:
7x² - 14x
7x.x - 7*2.x //7x appears in both terms, we can take it out
7x(x-2)19x³ - 38x² //notice that 38 = 19*2
19x².x - 19*2x²
19x² (x-2)8j³k - 4jk³ - 7 //never let multiple variables scare you ; ) Notice that 8 = 4*2
4*2j.k.j² - 4j.k.k² -7 //we find one 'j' and one 'k' in each term. Leave that '7' alone
4j.k(2j²-k²) - 7
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i guess you might make a comparison to division, only that when factoring, youre trying to find the largest expression that will divide into the original, rather than knowing what the divisor is already
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there's one other problem that I am have trouble with. I can't find a common factor in it.
a²+7a-18
Is there a way to factor it or is it prime?
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technically, factorising means reducing to linear equations.
so like
so technically, you cant factorise that equation as simple as that, sometimes you dont have a common factor in it, in which case it will be in the case of a quadratic, in the form
since its a[sup]2[/sup] a,c = 1, then you have to work out values of b and d to give the 7a and -18, such that bd = -18 and b+d = 7
so in this case you would have
which you can see if you expand the brackets you get
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Those are a bit trickier. Before, you could just look at every term to see if they all had a common factor, but here that method doesn't work.
It will indeed factorise, into (a+9)(a-2), but that fact that it does is less obvious.
For quadratics (expressions in the form ax² + bx +c, where x is the variable and a, b, c are constants), there is at least a fairly easy way to see whether or not it will factorise.
Work out the value of b² - 4ac, where a, b and c are as described above, and if the result is a square number then the expression will factorise.
So, using yours, we have 7² - 4*1*(-18) = 49 - (-72) = 121 = 11², which shows that it factorises.
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often you can just work out what the values in the brackets are, but as a help:
if the quadratic is in the form
ax^2 + bx - c, then factorised, it will be in the form (ax + b)(cx - d), bc > da
ax^2 + bx + c, then factorised, it will be in the form (ax + b)(cx + d)
ax^2 - bx - c, then factorised, it will be in the form (ax + b)(cx - d), bc < da
ax^2 - bx + c, then factorised, it will be in the form (ax - b)(cx - d)
where a,b,c,d are all positive.
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So I have another problem. The next set of problems I have it says to simplify. It's the polynomial things agin except they are set up for divison. I'll try to put one on here.
x²+7x-18
_________
x²+4x-45
They look like that, but you don't divide anything.
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just factorise both expressions
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Okay, thanks so much. This forum has helped me a lot.
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The hardest part, however, is to find the factor.
X'(y-Xβ)=0
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