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
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?
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.
Hello, richardkim4668!
The factoring problems are quite tricky.
Factor: .$\displaystyle (1-a^2)(1-b^2) - 4ab$
Expand: .$\displaystyle 1 - a^2 - b^2 + a^2b^2 - 4ab$
We have: .$\displaystyle a^2b^2 - 2ab + 1 - a^2 - 2ab - b^2$
. . . . $\displaystyle =\;(a^2b^2 - 2ab + 1) - (a^2 + 2ab + b^2)$
Factor: .$\displaystyle (ab - 1)^2 - (a + b)^2$ . . Difference of squares!
Factor: .$\displaystyle \big([ab-1] - [a+b]\big)\big([ab-1] +[a+b]\big) $
. . . . $\displaystyle =\; (ab-1 - a - b)(ab-1 +a+b)$
Factor: .$\displaystyle x^4 + 4$
Add and subtract $\displaystyle 4x^2\!:\;\;x^4 \,{\color{red}+\,4x^2} + 4 \,{\color{red}-\, 4x^2}$
We have: .$\displaystyle (x^2 + 2)^2 - (2x)^2$ . . Difference of squares!
Therefore: .$\displaystyle (x^2 + 2 - 2x)(x^2+2 + 2x)$
Hello, sakonpure6!
Soroban, how did you know to add and subtract 4x^2?
Many years ago, I saw someone else factor $\displaystyle x^4 + 1$.
I said it was impossible; we have the sum of two squares.
Then they added and subtracted $\displaystyle 2x^2.$
. . $\displaystyle x^4 + 2x^2 + 1 - 2x^2 \;=\;(x^2+1)^2 - (\sqrt{2}x)^2$ . Difference of squares!
And factored: .$\displaystyle (x^2 + 1 - \sqrt{2}x)(x^2+1 + \sqrt{2}x)$
. . . . . . . . . $\displaystyle =\;(x^2 - \sqrt{2}x + 1)(x^2 + \sqrt{2}x + 1)$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Since then I can factor the sum of two even powers.
(We sort of "complete the square" for the middle term.)
$\displaystyle x^8 + 9 \;=\;x^8 \,{\color{red}+\,6x^4} + 9\,{\color{red}-\,6x^4}$
. . . . . . $\displaystyle =\;(x^4 + 3)^2 - (\sqrt{6}x^2)^2$
. . . . . . $\displaystyle =\;(x^4 - \sqrt{6}x^2 + 3)(x^4 + \sqrt{6}x^2 + 3)$