Factoring

[Redskins]

I have been working on this for over an hour. how do you factor h2  9hs + 9s2? I don't think you it can be factored becauseyou cannot get a factor of 9 and a sum of -9.

 

[Denis]

I have been working on this for over an hour. how do you factor h2  9hs + 9s2? I don't think you it can be factored becauseyou cannot get a factor of 9 and a sum of -9.

Post this way: h^2 - 9hs + 9s^2
You're correct: that's not EASILY factorable.

But h^2 - 10hs + 9s^2 would be; typo perhaps?

 

[Redskins]

Unfortunately, it is not a typo and I have searched everywhere and can't find an explanation to a problem when the products don't equal a certain sum. I have never ran into this before. h^2-9hs+9s^2 The only solution I can come up with is that it is not a prime. Anyone else have a suggestion?

 

[Denis]

At what "stage" are you in this factoring/solving section?
Have you been exposed to the quadratic equation?

 

[Redskins]

Yes I am. I am in factoring complete squares, difference of squares, solving quadratic equations with the zero factor, Pathagorean theory and all sortd of goodies. This is why this question doesn't make sense. Like i said the only solution I know of is it is not prime and cannot be factored.

 

[Denis]

Well, all I can think of is set it equal to 0: h^2 - 9hs + 9s^2 = 0 ; then use quadratic formula:

h = [9s +- 3sSQRT(5)] / 2

Factor from above results; let x = SQRT(5):

[h - (9s + 3sx)/2] [h - (9s - 3sx)/2]

 

[Redskins]

I'll give it a try and see where I do from here. Thanks for the help. I'm glad I'm not the only one who thought this problem was a little out there.

 

[galactus]

A nice trick to tell whether or not a quadratic is factorable or not is to check the discriminant. If it IS a perfect square, then it IS factorable.

\(\displaystyle h^{2}-9sh+9s^{2}\)

The discriminant is \(\displaystyle b^{2}-4ac\)

Plug them in and see what we get:

\(\displaystyle (-9s)^{2}-4(1)(9s^{2})=45s^{2}\)

Is this a perfect square?. \(\displaystyle \sqrt{45s^{2}}=3\sqrt{5}s\)

Nope. So, it is NOT factorable.

This little trick may be handy before you spend time wrestlng around with it and find it is not factorable.

Try it with one we know is factorable. Say, \(\displaystyle x^{2}-2x-15\)

\(\displaystyle (-2)^{2}-4(1)(-15)=64\)

64 is certainly a perfect square, so it is factorable.