+2446 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!
+2446 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.
\(\boxed{1}\hspace{1mm}x^2-2xy+y^2=256\\ \boxed{2}\hspace{5mm}+2xy\hspace{10mm}=46\\ \overline{\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad}\\ \hspace{18mm}x^2+y^2=302\)
+2446 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. |
| \(a(a-1)(a-1)\) | |
| \(a(a-1)^2\) | |
