Why not just take out as a factor?
This would leave
...
I keep reasoning like this:
-8(-h^2+3h+40), find some numbers that mult. to -40 and add to 3, 8*-5=-40 and 8+-5=3, so -8(-h+8)(h-5) but this makes 8h^2-24h+320=0 which is a different statement.
there seems to be something wrong with the -8(-h+8)(h-5) step. Somehow the 8 and -5 are the zeros and the real answer is 8(h-8)(h+5)=0 but that contradicts how every other quadratic is factored -normally c=mn and b=m+n and the "m"s and "n"s are always here (x+m)(x+n) away from the zeros.
And why not tell the OP how to factorise rather than just do it for them.
If a polynomial with integer coefficients has rational roots they are a ratio of a factor of the constant term to a factor of the leading coefficient.
So for we know all the rational roots are factors of , these are . Now check through this list to find the rational roots, then you can turn these into the factors.
CB
Why should the sign of the coefficient matter? The textbook agrees with you but still isn't a positive coefficient (like +a) just a negative coefficient (-(-a))? So if the text is right, we could never factor since positive numbers are implicitly negative. Obviously I either don't know what coefficients are or there's no reason why one should worry about the signs -and I don't know which to choose.
(sorry this was a response to kylehk)
Hi bournouli,
sorry for any misunderstanding about the 'positive' conclusion; it just could simplify the factorization for us but not necessarily to enforce a '+' sign.
Just reread ur post, now I guess I find where u went wrong:
you transformed the original equation to -8(-h^2+3H+40)=0, it's ok for the '-h^2';
now suppose the factorization result is (-h+a)(h+b)=0
spread it we'll get -h^2++(a-b)h+ab=0; compare it with -h^2+3H+40=0
then we get two equations: a-b=3, ab=40 (exactly where u went wrong, as u considered ab=-40 in ur post)
a=8, b=5.
back to (-h+a)(h+b)=0, two solutions should be h=a or h=-b, i.e., h=8, or h=-5.
hope it's clear for u now
suppose the factorization result is (-h+a)(h+b)=0; spread it we'll get -h^2+(a-b)h+ab=0;
as -h^2+(a-b)h+ab=0 is just another expression of the equation -h^2+3h+40=0, the coefficient of corresponding items (i.e., h^2, h, and the constant item) should be equivalent, right?
-h^2 + (a-b)h + ab=0
-h^2 + 3 h + 40=0
Factoring is just the following process...
We do not know "h", so we do not know "h+8" etc,
hence we multiply out "term-by-term".
Your equation is
which is also saying
Both are equal, so if you subtract them, the result is 0.
You can subtract either side, so you can have
or
Hence, if you a "minus" on the squared term, you can always change all the signs
when you have an equality.
so you can factor using
Hence, you need the factors of -40 that combine to give -3.
If you have "negative h squared", your factors look like this...
and you must arrive at
so you need the factors of 40,
such that multiplying out the brackets leaves you with +3h
so
Ah yes.
The "a", "b" and "c" in the quadratic refer to those specific multipliers of x^2 and x and the constant at the end.
This allows direct placement of those constants into certain formulae,
such as the quadratic formula and the perpendicular distance from a point to a line formula.
If any of those constants are negative, then a negative value goes into those formulae.
For example 2x^2-3x+4 has a=2, b=-3, c=4
while -3x^2+2x+5 has a=-3, b=2, c=5.
You'll get used to this.