Factoring a polynomial with two variables

[sofrustrated52]

Hi, I am a college student who desperately needs all the help I can get. I am trying to factor a polynomial with two variables: h^2 - 9hs + 9s^2
Here is what I have so far h(-3 + hs)s(3+3). The instructions say to factor the polynomial, so I am assuming this is how it is done, but something tells me it is wrong. Can you help me :?:

 

[BigGlenntheHeavy]

\(\displaystyle a^2-2ab+b^2 \ = (a-b)^2\)

\(\displaystyle h^2-9hs+9s^2 \ is \ prime, \ however, \ h^2-6hs+9s^2 \ = \ (h-3s)^2.\)

 

[galactus]

This does not factor.

If the middle term were 6hs or 10hs instead of 9hs, then it would factor.

 

[sofrustrated52]

What I really need is someone to show me step by step because that is the only way I can grasp the concept. It is bad enough I am already flunking this course and it is only week 2 with three more weeks to go.

 

[galactus]

Here is a trick to see if one of these is factorable or not before you waste time trying and it isn't.

For a quadratic, \(\displaystyle ah^{2}+bh+c\),

Check the discriminant, \(\displaystyle b^{2}-4ac\). If it is a perfect square, then it is factorable.

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

In this case, we have \(\displaystyle (\underbrace{-9s}_{\text{b}})^{2}-4(\underbrace{1}_{\text{a}})(\overbrace{9s^{2}}^{\text{c}})=45s^{2}\)

Not a perfect square, therefore, not factorable.

Let's say you had \(\displaystyle h^{2}-10sh+9s^{2}\)

This is factorable because the discriminant is \(\displaystyle (-10s)^{2}-4(1)(9s^{2})=64s^{2}\).

64 is a perfect square because \(\displaystyle (8s)^{2}=64s^{2}\)

Now, factor it:

Take out the s because h is the variable and s is the constant. Then, factor and put it back when finished.

\(\displaystyle h^{2}-10h+9\)

What two numbers when multiplied equal 9 and when added equal -10?.

How about -9 and -1

So, we have \(\displaystyle (h+9)(h+1)\)

h+9=0 when h=-9 and h+1=0 when h=-1.

Now, put the s back:

\(\displaystyle (h-9s)(h-s)\)

There it is:

 

[sofrustrated52]

Thanks galactus. Having the problem worked out step by step makes it easier for me to understand. Hope your good at word problems because that one is just around the corner.

 

[lookagain]

Hi, I am a college student who desperately needs all the help I can get. I am trying to factor a polynomial with two variables: h^2 - 9hs + 9s^2

sofrustrated52,

you can also write it as \(\displaystyle \ \ 9s^2 - (9h)s + h^2\) as a quadratic in terms of \(\displaystyle s.\)

Then, for \(\displaystyle b^2 - 4ac,\)

\(\displaystyle a = 9\)

\(\displaystyle b = -9h\)

\(\displaystyle c = h^2\)


And \(\displaystyle \ \ b^2 - 4ac \ \ = \ \ (-9h)^2 - 4(9)(h^2) \ \ = \ \ 81h^2 - 36h^2\ \ =\ \ 45h^2,\) which is not a perfect square.

This also shows that the expression does not factor.