how do you expand and simplify in the easiest way:

(x-1)(x+3)(x-3)(x-7)

and

factorise:

(1-a^2)(1-b^2) - 4ab

and X^4 + 4

Printable View

- April 13th 2014, 07:59 AMrichardkim4668two more questions
how do you expand and simplify in the easiest way:

(x-1)(x+3)(x-3)(x-7)

and

factorise:

(1-a^2)(1-b^2) - 4ab

and X^4 + 4 - April 13th 2014, 08:39 AMSlipEternalRe: two more questions
You multiply it out. What do you mean "the easiest way"? I guess you can do it term-by-term using the Distributive Property:

$\begin{align*}(x-1)(x+3)(x-3)(x-7) & = x(x+3)(x-3)(x-7) - 1(x+3)(x-3)(x-7) \\ & = x\cdot x(x-3)(x-7) + x\cdot 3(x-3)(x-7) - 1(x+3)(x-3)(x-7) \\ & = \vdots\end{align*}$

For $(1-a^2)(1-b^2) - 4ab$, multiply out, then factor.

For $x^4+4$, it is irreducible over the reals. Are you allowed to factor over some larger field, like the complex numbers? - April 13th 2014, 08:39 AMJeffMRe: two more questions
I am afraid that products generally require mechanical working out and that few tricks are available to reduce the work. Factoring is always a matter of "seeing."

HOWEVER, all these problems involve a difference of powers and so permit reduction of work for your first problem.

$\displaystyle p^n - q^n \equiv (p - q) * \sum_{i=0}^{n}p^{(n - i)}q^i \implies p^2 - q^2 \equiv (p - q)(p + q).$

In terms of the first problem, you can use the difference of squares in reverse to reduce the problem from one involving four factors to one involving three factors.

A clue for the third problem. Do you see that:

$x^4 + 4 = x^4 - (-4).$ Now you have a difference of fourth powers.

EDIT: As slipeternal pointed out, $x^4 - (-4)$ is not reducible over the reals, but it is in terms of complex numbers. - April 13th 2014, 11:27 AMSorobanRe: two more questions
Hello, richardkim4668!

The factoring problems are quite tricky.

Quote:

Factor: .

Expand: .

We have: .

. . . .

Factor: . . . Difference of squares!

Factor: .

. . . .

Quote:

Factor: .

Add and subtract

We have: . . . Difference of squares!

Therefore: . - April 13th 2014, 12:04 PMJeffMRe: two more questions
Wow soroban. Very impressed.

- April 13th 2014, 01:36 PMsakonpure6Re: two more questions
Soroban, how did you know to add and subtract 4x??

- April 13th 2014, 02:38 PMSorobanRe: two more questions
Hello, sakonpure6!

Quote:

Soroban, how did you know to add and subtract 4x^2?

Many years ago, I saw someone else factor .

I said it was impossible; we have theof two squares.*sum*

Then they added and subtracted

. . . Difference of squares!

And factored: .

. . . . . . . . .

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Since then I can factor the sum of two even powers.

(We sort of "complete the square" for the middle term.)

. . . . . .

. . . . . .