Solution 2.3:3f

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