algebra
posted by Fidelia .
Okay, how do you factor a polynomial when there is a number in front of the highest x term.
For example:
how would you factor 2x^211x+15
i know the answer is (2x5)(x3), right? So you put the 2x on one side and then you put an x on the other. And then you know that 15*2=30, so you find numbers that multiply to equal 30 and add to equal 11 and you get 6 and 5 and then you take have of the 6 to get a 3, and you put it on the x side because x is half of 2x. Is this why (2x5)(x3) is the answer? Is that how you do it for every problem when there is a number in front of the x^2?
(2x5)(x3) is the answer because
2*1 = 2, 6 + 5 = 11, and 5*x = 15, the three coefficients of the binomial answer. I don't agree with or understand your reasoning. There is a bit of trial and error in factoring. If nothing seems to work, use the quadratic equation
[b +/ sqrt (b^2  4ac)]/2a for the roots.
can you plz explain how to do it then?
Fidelia, it looks like you are learning a method called decomposition of the middle term.
I will try to explain it in the simplest way I know how
First you multiply the first number by the last, 2*15=30, you did that.
now make a column of pairs of numbers which when multiplied give you +30
They must be either both positive or both negative to get a positive 30, but the middle number is 11, so obviously the pair of numbers are both negative
2 15
3 10
5 6
15 1
Which pair adds up to 11?
it is 5 and 6
so now we know that 11x must be split up into 5x and 6x, so....
2x^2  11x + 15
=2x^2  5x  6x + 15
Now use common factoring in "pairs"
=x(2x  5)  3(2x  5) , now we see another common factor of (2x5)
=(2x5)(x3)
Now isn't that easy?
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