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: .
Expand: .
We have: .
. . . .
Factor: . . . Difference of squares!
Factor: .
. . . .
Factor: .
Add and subtract
We have: . . . Difference of squares!
Therefore: .
Hello, sakonpure6!
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 the sum of two squares.
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.)
. . . . . .
. . . . . .