From Förberedande kurs i matematik 1

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-	{{NAVCONTENT_START}}	+	We can split up the first term on the left-hand side, <math>x(x^{2}-2x)</math>, into factors by taking <math>x</math> outside the bracket, <math>x(x^{2}-2x) = x\cdot x\cdot (x-2)</math> and writing the other term as <math>x\cdot (2-x) = -x(x-2)</math>. From this we see that both terms contain <math>x(x-2)</math> as common factors and, if we take out those, the left-hand side becomes
-	<center> [[Bild:2_3_3f.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>\begin{align}
		+	x(x^{2}-2x) + x(2-x)
		+	&= x^{2}(x-2) - x(x-2)\\[5pt]
		+	&= x\bigl(x(x-2)-(x-2)\bigr)\\[5pt]
		+	&= x(x-2)(x-1)\,\textrm{.}
		+	\end{align}</math>}}
		+
		+	The whole equation can be written as
		+
		+	{{Displayed math||<math>x(x-2)(x-1) = 0</math>}}
		+
		+	and this equation is satisfied only when one of the three factors <math>x</math>, <math>x-2</math> or <math>x-1</math> is zero, i.e. the solutions are <math>x=0</math>, <math>x=2</math> and <math>x=1</math>.
		+
		+	Because it is not completely obvious that <math>x=1</math> is a solution of the equation, we check that <math>x=1</math> satisfies the equation, i.e. that we haven't calculated incorrectly:
		+
		+	:*''x''&nbsp;=&nbsp;1: <math>\ \text{LHS} = 1\cdot (1^{2}-2\cdot 1) + 1\cdot (2-1) = 1\cdot (-1) + 1\cdot 1 = 0 = \text{RHS.}</math>

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Current revision

We can split up the first term on the left-hand side, \displaystyle x(x^{2}-2x), into factors by taking \displaystyle x outside the bracket, \displaystyle x(x^{2}-2x) = x\cdot x\cdot (x-2) and writing the other term as \displaystyle x\cdot (2-x) = -x(x-2). From this we see that both terms contain \displaystyle x(x-2) as common factors and, if we take out those, the left-hand side becomes

\displaystyle \begin{align} x(x^{2}-2x) + x(2-x) &= x^{2}(x-2) - x(x-2)\\[5pt] &= x\bigl(x(x-2)-(x-2)\bigr)\\[5pt] &= x(x-2)(x-1)\,\textrm{.} \end{align}

The whole equation can be written as

\displaystyle x(x-2)(x-1) = 0

and this equation is satisfied only when one of the three factors \displaystyle x, \displaystyle x-2 or \displaystyle x-1 is zero, i.e. the solutions are \displaystyle x=0, \displaystyle x=2 and \displaystyle x=1.

Because it is not completely obvious that \displaystyle x=1 is a solution of the equation, we check that \displaystyle x=1 satisfies the equation, i.e. that we haven't calculated incorrectly:

• x = 1: \displaystyle \ \text{LHS} = 1\cdot (1^{2}-2\cdot 1) + 1\cdot (2-1) = 1\cdot (-1) + 1\cdot 1 = 0 = \text{RHS.}