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



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oldnick's profile pic

Posted on (Answer #1)

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:


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.


pramodpandey's profile pic

Posted on (Answer #2)

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

`x^2+mx+n`   ,

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


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