Hi, nellycrane!
I think you will be better off if I give you an easier problem like \(x^2-4 \) . You are probably well aware that this is a difference of squares. \((x+2)(x-2)\) would be the corresponding factorization.
Now, let's say that you want to factor \(x^2-30\) . This is not possible if you restrict yourself to the rational number set. However, it can be factored as \((x+\sqrt{30})(x-\sqrt{30})\).
In general, \(a^2-b^2=\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)\) . You can apply this knowledge to the problems at hand.
\(x^2+50\\ a=x^2,b=-50;\\ (\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})\\ (\sqrt{x^2}+\sqrt{-50})(\sqrt{x^2}-\sqrt{-50}) \)
You can simplify this to get the factorization amongst the complex numbers. Good luck!
This problem is easier than it looks at first glance. If you utilize clever algebraic manipulation, this problem becomes simpler.
\(x-y=16\) | \(xy=23\) |
\((x-y)^2=16^2\) | |
\(\boxed{1}\hspace{1mm}x^2-2xy+y^2=256\) | \(\boxed{2}\hspace{1mm}2xy=46\) |
Notice what I have done. I have manipulated the information I know about these real numbers, x and y , and I am manipulating it in a way that is much more convenient for this particular problem. The only thing left to do is add the equations together.
\(\boxed{1}\hspace{1mm}x^2-2xy+y^2=256\\ \boxed{2}\hspace{5mm}+2xy\hspace{10mm}=46\\ \overline{\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad}\\ \hspace{18mm}x^2+y^2=302\)
Expand the binomials on the left-hand side of the equation and simplify:
\((a-1)^3=a^3-3a^2+3a-1\\ (a-1)^2=a^2-2a+1\) | Let's put this together into one cohesive expression. |
\(a^3-3a^2+3a-1+a^2-2a+1\) | Combine like terms. |
\(a^3-2a^2+a\) | Factor out an a . |
\(a(a^2-2a+1)\) | The trinomial is indeed factorable and happens to factor into a perfect square. |
\(a(a-1)(a-1)\) | |
\(a(a-1)^2\) | |