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Why is it difficult to factor a trinomial with a leading coefficient that is not equal...

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monique06 | Valedictorian

Posted May 10, 2013 at 12:55 AM via web

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Why is it difficult to factor a trinomial with a leading coefficient that is not equal to one?

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oldnick | Valedictorian

Posted May 10, 2013 at 2:47 AM (Answer #1)

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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)

is given.  

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:

                                     `x^2-12x-28=0`

Writing as above:

                                    `x^2-12x+36-28-36=0`

                                    `x^2-12x+36-64=0`

                                     `(x-6)^2=64`

                                     `x-6=+-8`

                                     `x=6+-8`

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.

  

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pramodpandey | College Teacher | Valedictorian

Posted May 10, 2013 at 4:19 AM (Answer #2)

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It is not difficult but little complicated. You can change leadin coefficient to 1 ,by factoring out it.For example

`ax^2+bx+c=a(x^2+(b/a)x+(c/a))`

`` 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

`x^2+mx+n`   ,

If coefficients m and n are not integer then again it will same as above.

 

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