Factoring Question

  • B
  • Thread starter gede
  • Start date
  • #1
18
0
Since this is not available in my algebra textbook, how do you factorize the ##a^4 + b^4## and ##a^4 - b^4##?

I also would like to know how do you obtain ##a^3 - b^3 = (a - b)(a^2 + ab + b^2)## and ##a^3 + b^3 = (a + b)(a^2 - ab + b^2)##?
 

Answers and Replies

  • #2
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,798
1,670
Since this is not available in my algebra textbook, how do you factorize the ##a^4 + b^4## and ##a^4 - b^4##?

I also would like to know how do you obtain ##a^3 - b^3 = (a - b)(a^2 + ab + b^2)## and ##a^3 + b^3 = (a + b)(a^2 - ab + b^2)##?
Factoring a4 + b4 is tricky.

Factoring a4 - b4 should be obvious if you know how to factor a2 - b2

For the factoring of the cubics (or polynomials in general), you can always use polynomial long division.
 
  • #3
35,235
7,053
SteamKing said:
Factoring a4 + b4 is tricky.
And if you're limited to factors with real coefficients, it's not factorable at all.
 
  • #4
299
20
And if you're limited to factors with real coefficients, it's not factorable at all.

Actually it is. It factors as [itex](a^2 + \sqrt{2}ab + b^2)(a^2 - \sqrt{2}ab + b^2)[/itex]
 
  • #5
member 587159
1) a^4 - b^4 = ...

Substituting a^2 = x and b^2 = y

=> x^2 - y^2 = ...

2) a^4 + b^4 = a^4 + 2a^2*b^2 + b^4 - 2a^2*b^2 = (a^2 + b^2)^2 - 2a^2*b^2

Now use a^2 - b^2 = (a-b)(a+b) to find the answer

3) a^3 - b^3 = (a-b)(a^2 + ab + b^2)
=> (a^3 - b^3 )/(a-b) = (a^2 + ab + b^2)

Use euclidean division or Horner (let a^3 be the variable, let b^3 be the constant)
 
  • #6
There's a trick in factoring (a3 - b3) and any expression alike of this. (Only works for this kind of binomial).

1) The first factor, which is a binomial, is always the cube root of the two terms.
(The blank space represents the missing terms and its operation/ sign).

a3 - b3 = (a - b)( _ _ _ )

2) The second factor, which is a trinomial, always constitutes three terms as follows.
a. The first term is square of the first term in the binomial: (a - b)(a2 _ _ )
b. The middle term is reverse sign of the product of the two term: (a - b)(a2 + ab _)
c. The last term is the square of the second term in the binomial: (a - b)(a2 + ab + b2)

And that completes the factoring of (a3 - b3)
 
  • #7
member 587159
There's a trick in factoring (a3 - b3) and any expression alike of this. (Only works for this kind of binomial).

1) The first factor, which is a binomial, is always the cube root of the two terms.
(The blank space represents the missing terms and its operation/ sign).

a3 - b3 = (a - b)( _ _ _ )

2) The second factor, which is a trinomial, always constitutes three terms as follows.
a. The first term is square of the first term in the binomial: (a - b)(a2 _ _ )
b. The middle term is reverse sign of the product of the two term: (a - b)(a2 + ab _)
c. The last term is the square of the second term in the binomial: (a - b)(a2 + ab + b2)

And that completes the factoring of (a3 - b3)

This is true, although the point of mathematics is not to remember some tricks. With some experience, it takes 30 seconds to proof this formula using general calculus.
 
  • #8
This is true, although the point of mathematics is not to remember some tricks. With some experience, it takes 30 seconds to proof this formula using general calculus.

Nonetheless, it is a basic format on how to factor such kind of expressions. It's not a point of remembering a trick, it is one way of knowing a proof how such expression can be branched down into its respective factors.
 
  • #9
35,235
7,053
This is true, although the point of mathematics is not to remember some tricks.
It's a huge time saver to remember a few "tricks", such as the Quadratic Formula and how to factor the difference of squares and the sum or difference of cubes.
Math_QED said:
With some experience, it takes 30 seconds to proof this formula using general calculus.
You don't need calculus to derive ##a^3 - b^3 = (a - b)(a^2 + ab + b^2)##. Also, since the OP is studying algebra, it's reasonable to assume that he hasn't been exposed to the techniques of calculus.
 
  • #10
22,129
3,297
This is true, although the point of mathematics is not to remember some tricks. With some experience, it takes 30 seconds to proof this formula using general calculus.

How would you prove this using calculus?
 
  • #11
18
0
Please tell me the method of factoring of ##a^n + b^n## and ##a^n - b^n##?
 
  • #12
member 587159
Please tell me the method of factoring of ##a^n + b^n## and ##a^n - b^n##?

1) Define a function F: R -> R: a -> a^n + b^n => F(a) = a^n + b^n
b and n are natural numbers, n is odd

We see: F(-b) = -b^n + b^n = 0 => a + b is a divisor of F(a) (fundamental theorem of algebra)
That's one factor, you can find what's left after the division using horner's rule or using euclidean division. Then you can try to find those factors.

Same thing when you have Z(a) = a^n -b^n
b and n natural numbers, n odd

2) Define a function G: R -> R: a -> a^n - b^n => G(a) = a^n - b^n
b and n are natural numbers, n is not odd

Use a^2 - b^2 = (a-b)(a+b) to find factors.

3) Define a function H: R -> R: a -> a^n + b^n => H(a) = a^n + b^n
b and n are natural numbers, n is not odd

This is the hardest one. You have to manipulate the expressions. For example,
a^4 +1 = a^4 + 1 + 2a^2 - 2a^2 = (a^2 +1)^2 - 2a^2 = (a^2 + 1 + SQRT(2)a)(a^2 + 1 - SQRT(2)a)
 
  • #13
35,235
7,053
1) Define a function F: R -> R: a -> a^n + b^n => F(a) = a^n + b^n
b and n are natural numbers, n is odd

We see: F(-b) = -b^n + b^n = 0 => a + b is a divisor of F(a) (fundamental theorem of algebra)
That's one factor, you can find what's left after the division using horner's rule or using euclidean division. Then you can try to find those factors.

Same thing when you have Z(a) = a^n -b^n
b and n natural numbers, n odd

2) Define a function G: R -> R: a -> a^n - b^n => G(a) = a^n - b^n
b and n are natural numbers, n is not odd

Use a^2 - b^2 = (a-b)(a+b) to find factors.

3) Define a function H: R -> R: a -> a^n + b^n => H(a) = a^n + b^n
b and n are natural numbers, n is not odd

This is the hardest one. You have to manipulate the expressions. For example,
a^4 +1 = a^4 + 1 + 2a^2 - 2a^2 = (a^2 +1)^2 - 2a^2 = (a^2 + 1 + SQRT(2)a)(a^2 + 1 - SQRT(2)a)
None of these techniques use calculus, which is what @micromass asked about. For ##a^4 + 1##, if by factoring, one means splitting the polynomial into linear factors (i.e., first degree factors) with real coefficients, it's not factorable.

That was what I was thinking when I said that ##a^4 + b^4## wasn't factorable. To clarify my statement, ##a^4 + b^4## isn't factorable into linear factors with real coefficients.
 

Related Threads on Factoring Question

  • Last Post
Replies
15
Views
2K
  • Last Post
Replies
8
Views
1K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
2
Views
2K
Replies
1
Views
1K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
1
Views
665
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
14
Views
2K
  • Last Post
Replies
2
Views
2K
Top