# two more questions

• April 13th 2014, 06:59 AM
richardkim4668
two 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, 07:39 AM
SlipEternal
Re: 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, 07:39 AM
JeffM
Re: two more questions
Quote:

Originally Posted by richardkim4668
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

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, 10:27 AM
Soroban
Re: two more questions
Hello, richardkim4668!

The factoring problems are quite tricky.

Quote:

Factor: . $(1-a^2)(1-b^2) - 4ab$

Expand: . $1 - a^2 - b^2 + a^2b^2 - 4ab$

We have: . $a^2b^2 - 2ab + 1 - a^2 - 2ab - b^2$

. . . . $=\;(a^2b^2 - 2ab + 1) - (a^2 + 2ab + b^2)$

Factor: . $(ab - 1)^2 - (a + b)^2$ . .
Difference of squares!

Factor: . $\big([ab-1] - [a+b]\big)\big([ab-1] +[a+b]\big)$

. . . . $=\; (ab-1 - a - b)(ab-1 +a+b)$

Quote:

Factor: . $x^4 + 4$

Add and subtract $4x^2\!:\;\;x^4 \,{\color{red}+\,4x^2} + 4 \,{\color{red}-\, 4x^2}$

We have: . $(x^2 + 2)^2 - (2x)^2$ . .
Difference of squares!

Therefore: . $(x^2 + 2 - 2x)(x^2+2 + 2x)$
• April 13th 2014, 11:04 AM
JeffM
Re: two more questions
Wow soroban. Very impressed.
• April 13th 2014, 12:36 PM
sakonpure6
Re: two more questions
Soroban, how did you know to add and subtract 4x??
• April 13th 2014, 01:38 PM
Soroban
Re: 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 $x^4 + 1$.

I said it was impossible; we have the sum of two squares.

Then they added and subtracted $2x^2.$

. . $x^4 + 2x^2 + 1 - 2x^2 \;=\;(x^2+1)^2 - (\sqrt{2}x)^2$ .
Difference of squares!

And factored: . $(x^2 + 1 - \sqrt{2}x)(x^2+1 + \sqrt{2}x)$

. . . . . . . . . $=\;(x^2 - \sqrt{2}x + 1)(x^2 + \sqrt{2}x + 1)$

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

Since then I can factor the sum of two even powers.
(We sort of "complete the square" for the middle term.)

$x^8 + 9 \;=\;x^8 \,{\color{red}+\,6x^4} + 9\,{\color{red}-\,6x^4}$

. . . . . . $=\;(x^4 + 3)^2 - (\sqrt{6}x^2)^2$

. . . . . . $=\;(x^4 - \sqrt{6}x^2 + 3)(x^4 + \sqrt{6}x^2 + 3)$