Factoring Polynomials - Difference of Cubes

[coughsyrup78]

I am having trouble factoring this polynomial.

h⁶ - 64

Here is what I've done so far:

h⁶ - 64 <-- difference of two cubes, if a = h² and b = 4, then

(h² - 4)(h⁴ + 4h² + 16) <-- use difference of squares on first polynomial, if a = h and b = 2, then

(h + 2)(h - 2)(h⁴ + 4h² + 16) <--how do I factor the last polynomial?

Thanks.

 

[srmichael]

I am having trouble factoring this polynomial.

h⁶ - 64

Here is what I've done so far:

h⁶ - 64 <-- difference of two cubes, if a = h² and b = 4, then

(h² - 4)(h⁴ + 4h² + 16) <-- use difference of squares on first polynomial, if a = h and b = 2, then

(h + 2)(h - 2)(h⁴ + 4h² + 16) <--how do I factor the last polynomial?

Thanks.

I would have approached it as a difference of two squares:

\(\displaystyle h^6-64=(h^3)^2-8^2\)

\(\displaystyle (h^3+8)(h^3-8)\)

then factor each of the two cubics:

\(\displaystyle (h+2)(h^2-2h+4)(h-2)(h^2+2h+4)\)

rearrange factors for ease of reading:

\(\displaystyle (h+2)(h-2)(h^2-2h+4)(h^2+2h+4)\)

 

[coughsyrup78]

Interesting, the reason why I did it the way I did was because all of the other problems in the section were done the same way, and nowhere does the book mention approaching it the way you did.

So my question now is, is it possible to do it the way I was trying?

 

[Deleted member 4993]

I am having trouble factoring this polynomial.

h⁶ - 64

Here is what I've done so far:

h⁶ - 64 <-- difference of two cubes, if a = h² and b = 4, then

(h² - 4)(h⁴ + 4h² + 16) <-- use difference of squares on first polynomial, if a = h and b = 2, then

(h + 2)(h - 2)(h⁴ + 4h² + 16) <--how do I factor the last polynomial?

Thanks.

(h + 2)(h - 2)(h⁴ + 4h² + 16)

(h + 2)(h - 2)(h⁴ + 2*4*h² + 4² - 4h²)

(h + 2)(h - 2)[(h² + 4)² - (2h)²]

and continue.....

 

[coughsyrup78]

(h + 2)(h - 2)(h⁴ + 4h² + 16)

(h + 2)(h - 2)(h⁴ + 2*4*h² + 4² - 4h²)

(h + 2)(h - 2)[(h² + 4)² - (2h)²]

and continue.....

I don't understand what you did there.

What form is the polynomial h⁴ + 4h² + 16 in?

Difference of squares, binomial squares (perfect square trinomials), sum of cubes, difference of cubes...

Or is it just a regular ax² + bx + c type of polynomial?

I can't see how it fits any of the forms that I've learned, and there's no GCF to factor out either.

 

[JeffM]

I don't understand what you did there.

What form is the polynomial h⁴ + 4h² + 16 in?

Difference of squares, binomial squares (perfect square trinomials), sum of cubes, difference of cubes...

Or is it just a regular ax² + bx + c type of polynomial?

I can't see how it fits any of the forms that I've learned, and there's no GCF to factor out either.

Two common mathematical "tricks" are to add 0 to an expression or to multiply an expression by 1.

\(\displaystyle h^4 + 4h^2 + 16 =\)

\(\displaystyle h^4 + 4h^2 + 16 + 0 = \)

\(\displaystyle h^4 + 4h^2 + 16 + (4h^2 - 4h^2) = \)

\(\displaystyle h^4 + 4h^2 + 4h^2 + 16 - 4h^2 = \)

\(\displaystyle (h^4 + 2 * 4h^2 + 4^2) - 4h^2 =\)

\(\displaystyle (h^2 + 4)^2 - (2h)^2 = \)

\(\displaystyle \{(h^2 + 4) - 2h\} * \{(h^2 + 4) + 2h\}.\)

Edit: Thanks Bob for catching the typo. I do proof my posts, but somehow ...

 

[DrPhil]

I don't understand what you did there.

What form is the polynomial h⁴ + 4h² + 16 in?

Difference of squares, binomial squares (perfect square trinomials), sum of cubes, difference of cubes...

Or is it just a regular ax² + bx + c type of polynomial?

I can't see how it fits any of the forms that I've learned, and there's no GCF to factor out either.

Subhotosh Kahn and JeffM both showed how to make the polynomial equal to the difference of two squares:

\(\displaystyle h^4 + 4h^2 + 16 = (h^4 + 8 h^2 + 16) - 4h^2 = (h^2 + 4)^2 - (2h)^2 \)

Since the original expression was a difference of sixth powers, you can start with either the squares or the cubes. Doing the difference of squares first happened to be easier, because the cofactors were then easy to do as sum of cubes and difference of cubes. It is not against the rules to start over if you hit a hard place.

 

[Bob Brown MSEE]

Two common mathematical "tricks" are to add 0 to an expression or to multiply an expression by 1.


\(\displaystyle \{(h^2 + 4) - 2h\} * \{(h^2 + 4) - 2h\}.\)

Very nice jeff,
correction (typo): \(\displaystyle \{(h^2 + 4) - 2h\} * \{(h^2 + 4) + 2h\}.\)

 

[coughsyrup78]

Thanks everyone for your help.

I scratch my head at why the book would put a problem like that in the exercises. All the book shows us how to do is find A and B and then plug it into one of the forms like a² + 2ab + b² = (a + b)²

There's nothing even remotely mentioned about what you guys are doing, and I was certainly never going to figure any of that out intuitively. And since there's only one problem like this in the exercises for me to practice, it's kind of pointless figuring out a new technique just to do this one problem.

Sigh...

 

[JeffM]

Thanks everyone for your help.

I scratch my head at why the book would put a problem like that in the exercises. All the book shows us how to do is find A and B and then plug it into one of the forms like a² + 2ab + b² = (a + b)²

There's nothing even remotely mentioned about what you guys are doing, and I was certainly never going to figure any of that out intuitively. And since there's only one problem like this in the exercises for me to practice, it's kind of pointless figuring out a new technique just to do this one problem.

Sigh...

I suspect all the book expected as an answer was:

\(\displaystyle h^6 - 64 = (h^3)^2 - 8^2 = (h^3 - 8)(h^3 + 8).\)

If your book teaches difference and sum of cubes, then it may expect you to factor further after first factoring as a difference of squares.

Because you started with difference of cubes (which is perfectly correct), you ended up on a more challenging road.

 

[Bob Brown MSEE]

Since the original expression was a difference of sixth powers, you can start with either the squares or the cubes. Doing the difference of squares first happened to be easier, because the cofactors were then easy to do as sum of cubes and difference of cubes. It is not against the rules to start over if you hit a hard place.

This was your question:
(h⁴ + 4h² + 16) <--how do I factor?

It is likely that you would not have been asked that in isolation,
However you did know
1) difference of cubes
2) difference of squares

The best advice in this case is the quote from Dr Phil.
Your question was a good one though.

 

[Deleted member 4993]

Thanks everyone for your help.

I scratch my head at why the book would put a problem like that in the exercises..

In my opinion, every book should have some problems (minority) where you have to scratch your head. Otherwise it will be spoon-feeding...