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Why is it difficult to factor a trinomial with a leading coefficient that is not equal...
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For a trinomial wit coefficnets not one colud be product of not one coefficents,and is hard to compare with a known form.
I.e, if the trinomial :
`x^2-12x-28 ` (1)
To find factors, have to find its roots.
Note that the trinomial looks almost like a square, but isn't.
That doesn't mean we couldn't manage in order to take advantage of "looks like a a square". Indeed a square of a sum `x` and a number, is a trinomial with first term is a `x` square, a double product by `x` and numbers, and number square. So, is easy to catch that in the binomial square `(x+n)^2` `rArr 2xn=-12x`
so that `n=-6`
Now we can write: `x^2-12x-28=x^2-12x +36-28 -36`
We added and subtract `36` as `n^2` ,according what we did find out by `12x`
Since we wanna find the roots have to find the solution of equation:
Writing as above:
So that we have two solution: `x_1=14;x_2=-2`
It points out : `x^2-12-28=(x-14)(x+2)`
Note that we proceed negletting the first coefficient, relative at `x^2`
Suppose now `12x^2-10x+7`
is given: We could not say `2xn=-10x`
For isn't true , we have to divide by 12.
`x^2-5/6 x +7/12=0`
It's clear, that it's very hard to writeasa binomial square plus a number.
Posted by oldnick on May 10, 2013 at 2:47 AM (Answer #1)
It is not difficult but little complicated. You can change leadin coefficient to 1 ,by factoring out it.For example
`` Because operations in integer are easy and in fraction complicated ,so you have to be care full when working with coefficients like (b/a) and (c/a) because may not be these are integers.
On contrary to this
If coefficients m and n are not integer then again it will same as above.
Posted by pramodpandey on May 10, 2013 at 4:19 AM (Answer #2)
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