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

**bournouli** · March 25th 2011, 10:22 PM

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.

**Prove It** · March 25th 2011, 10:31 PM

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

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

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

**kylehk** · March 25th 2011, 10:32 PM

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

**CaptainBlack** · March 25th 2011, 10:37 PM

Originally Posted by Prove It

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

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

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

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

CB

**Prove It** · March 25th 2011, 10:42 PM

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 $x^2-3h-40$ we know all the rational roots are factors of $40$ , these are $\pm1, \pm 2, \pm 4, \pm 8, \pm 5, \pm 10, \pm 20$ . 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 ....

**bournouli** · March 25th 2011, 10:43 PM

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)

**kylehk** · March 25th 2011, 11:01 PM

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

**bournouli** · March 25th 2011, 11:23 PM

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.

**kylehk** · March 25th 2011, 11:34 PM

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

**bournouli** · March 26th 2011, 08:35 AM

thanks kylehk,

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

Is it only applicable when a=1?

**bournouli** · March 26th 2011, 10:28 AM

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.

**Archie Meade** · March 26th 2011, 01:14 PM

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

$(h+?)(-h+?)$

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...

$(5)3=15$

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

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

$5(6)=30$

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

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

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

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

Your equation is

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

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

which is also saying

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

$h^2=3h+40$

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

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

or

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

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

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

$h(h)=h^2$

so you can factor using

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

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

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

$(h+?)(-h+?)$

and you must arrive at

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

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

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

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

so

$b-a=3$

$ab=40$

**bournouli** · March 26th 2011, 02:49 PM

Originally Posted by Archie Meade

Factoring is just the following process...

$(5)3=15$

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

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

$5(6)=30$

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

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

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

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

Your equation is

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

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

which is also saying

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

$h^2=3h+40$

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

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

or

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

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

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

$h(h)=h^2$

so you can factor using

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

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

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

$(h+?)(-h+?)$

and you must arrive at

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

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

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

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

so

$b-a=3$

$ab=40$

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?

**Archie Meade** · March 26th 2011, 03:05 PM

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.

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

LinkBack

Thread Tools

Display

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

Similar Math Help Forum Discussions

Histograms scale factor.. bins statistics and scale factor multiplication...?

factor analysis , extract second factor loading

factor analysis , extract second factor loading

Factor 3x^2 +2x -6

How do i factor this?

Search Tags