# Thread: How do you factor 8h^2-24h-320=0

1. ## How do you factor 8h^2-24h-320=0

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.

2. Why not just take out $\displaystyle \displaystyle 8$ as a factor?

This would leave $\displaystyle \displaystyle 8(h^2 - 3h - 40) = 0$

$\displaystyle \displaystyle 8(h-8)(h+5) = 0$...

3. Originally Posted by bournouli
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.
this method applies only when the coefficient of the quadratic item is positive, i.e., u need to simplify the original equation like this:
8(h^2-3h+40)=0;
now I think u can handle it well and get the right answers

4. Originally Posted by Prove It
Why not just take out $\displaystyle \displaystyle 8$ as a factor?

This would leave $\displaystyle \displaystyle 8(h^2 - 3h - 40) = 0$

$\displaystyle \displaystyle 8(h-8)(h+5) = 0$...
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 $\displaystyle x^2-3h-40$ we know all the rational roots are factors of $\displaystyle 40$, these are $\displaystyle \pm1, \pm 2, \pm 4, \pm 8, \pm 5, \pm 10, \pm 20$. Now check through this list to find the rational roots, then you can turn these into the factors.

CB

5. Originally Posted by CaptainBlack
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 $\displaystyle x^2-3h-40$ we know all the rational roots are factors of $\displaystyle 40$, these are $\displaystyle \pm1, \pm 2, \pm 4, \pm 8, \pm 5, \pm 10, \pm 20$. Now check through this list to find the rational roots, then you can turn these into the factors.

CB
The OP had given a full solution, which was nearly correct except for the sign error I pointed out ....

6. 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)

7. Originally Posted by bournouli
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

8. Originally Posted by 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
for this, a-b=3, ab=40 how did you get the positive 40? I want to take the -1 coefficient on h^2 and mult. that by b to get -40 but that always leads to antinomies so that last part isn't clear. Thanks for the help though.

9. Originally Posted by bournouli
for this, a-b=3, ab=40 how did you get the positive 40? I want to take the -1 coefficient on h^2 and mult. that by b to get -40 but that always leads to antinomies so that last part isn't clear. Thanks for the help though.
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

10. thanks kylehk,

but why do you think my original method failed -it seemed so straightforward?

Is it only applicable when a=1?

11. Originally Posted by bournouli
thanks kylehk,

but why do you think my original method failed -it seemed so straightforward?

Is it only applicable when a=1?
sorry, that was kinda stupid, I mean is it only applicable when a=positive number? It probably isn't because you explicitly stated that the sign doesn't matter but there is truly no other way I can find that explains why I was wrong.

12. Originally Posted by bournouli
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.

Where you have gone wrong here is....

you should be finding the factors of 40 that differ by 3, since "after" multiplying out

$\displaystyle (h+?)(-h+?)$

one of the numbers will multiply by "-h"

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.
Factoring is just the following process...

$\displaystyle (5)3=15$

$\displaystyle (7-2)(1+2)=15$

$\displaystyle 7(1+2)-2(1+2)=7(1)+7(2)-2(1)-2(2)=7+14-2-4=21-6=15$

$\displaystyle 5(6)=30$

$\displaystyle (3+2)(10-4)=3(10-4)+2(10-4)=30-12+20-8=50-20=30$

$\displaystyle (h-8)(h+5)=?$

We do not know "h", so we do not know "h+8" etc,
hence we multiply out "term-by-term".

$\displaystyle (h-8)(h+5)=h(h+5)-8(h+5)=h^2+5h-8h-40=h^2-3h-40$

$\displaystyle 8h^2-24h-320=0\Rightarrow\ 8\left(h^2-3h-40\right)=0$

$\displaystyle 8\ne\ 0\Rightarrow\ h^2-3h-40=0$

which is also saying

$\displaystyle h^2-(3h+40)=0$

$\displaystyle h^2=3h+40$

Both are equal, so if you subtract them, the result is 0.
You can subtract either side, so you can have

$\displaystyle h^2-3h-40=0$

or

$\displaystyle 3h+40-h^2=0$

Hence, if you a "minus" on the squared term, you can always change all the signs
when you have an equality.

$\displaystyle h^2-3h-40=0$

$\displaystyle h(h)=h^2$

so you can factor using

$\displaystyle (h+?)(h+?)=h^2-3h-40$

Hence, you need the factors of -40 that combine to give -3.

If you have "negative h squared", your factors look like this...

$\displaystyle (h+?)(-h+?)$

and you must arrive at

$\displaystyle -h^2+3h+40=0$

so you need the factors of 40,
such that multiplying out the brackets leaves you with +3h

$\displaystyle (h+a)(-h+b)=h(-h+b)+a(-h+b)$

$\displaystyle =-h^2+hb-ah+ab=-h^2+h(b-a)+ab$

so

$\displaystyle b-a=3$

$\displaystyle ab=40$

13. Originally Posted by Archie Meade
Factoring is just the following process...

$\displaystyle (5)3=15$

$\displaystyle (7-2)(1+2)=15$

$\displaystyle 7(1+2)-2(1+2)=7(1)+7(2)-2(1)-2(2)=7+14-2-4=21-6=15$

$\displaystyle 5(6)=30$

$\displaystyle (3+2)(10-4)=3(10-4)+2(10-4)=30-12+20-8=50-20=30$

$\displaystyle (h-8)(h+5)=?$

We do not know "h", so we do not know "h+8" etc,
hence we multiply out "term-by-term".

$\displaystyle (h-8)(h+5)=h(h+5)-8(h+5)=h^2+5h-8h-40=h^2-3h-40$

$\displaystyle 8h^2-24h-320=0\Rightarrow\ 8\left(h^2-3h-40\right)=0$

$\displaystyle 8\ne\ 0\Rightarrow\ h^2-3h-40=0$

which is also saying

$\displaystyle h^2-(3h+40)=0$

$\displaystyle h^2=3h+40$

Both are equal, so if you subtract them, the result is 0.
You can subtract either side, so you can have

$\displaystyle h^2-3h-40=0$

or

$\displaystyle 3h+40-h^2=0$

Hence, if you a "minus" on the squared term, you can always change all the signs
when you have an equality.

$\displaystyle h^2-3h-40=0$

$\displaystyle h(h)=h^2$

so you can factor using

$\displaystyle (h+?)(h+?)=h^2-3h-40$

Hence, you need the factors of -40 that combine to give -3.

If you have "negative h squared", your factors look like this...

$\displaystyle (h+?)(-h+?)$

and you must arrive at

$\displaystyle -h^2+3h+40=0$

so you need the factors of 40,
such that multiplying out the brackets leaves you with +3h

$\displaystyle (h+a)(-h+b)=h(-h+b)+a(-h+b)$

$\displaystyle =-h^2+hb-ah+ab=-h^2+h(b-a)+ab$

so

$\displaystyle b-a=3$

$\displaystyle ab=40$
Wow, that helped.

I suppose my fallacy was in assuming that the "a" in ab=40 was the same as the "a" in ax^2+bx+c so that if it was -ax^2+bx+c one had to use -1 or something?

14. 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.