Solution 2.1:4a

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Current revision (08:54, 23 September 2008) (edit) (undo)
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First, we multiply the second bracket by
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First, we multiply the second bracket by ''x'' from the first bracket,
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<math>x</math>
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from the first bracket,
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{{Displayed math||<math>(\bbox[#FFEEAA;,1.5pt]{\strut x}+\bbox[#FFFFFF;,1.5pt]{\strut 2})(3x^{2}-x+5) = \bbox[#FFEEAA;,1.5pt]{\strut x\cdot 3x^{2}-x\cdot x+x\cdot 5}+{}\rlap{\cdots}\phantom{\bbox[#FFEEAA;,1.5pt]{\strut 2\cdot 3x^{2}-2\cdot x+2\cdot 5}\,\textrm{.}}</math>}}
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<math>\left( x+2 \right)\left( 3x^{2}-x+5 \right)=x\centerdot 3x^{2}-x\centerdot x+x\centerdot 5+...</math>
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Then, do the same for 2 from the first bracket
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{{Displayed math||
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<math>(\bbox[#FFFFFF;,1.5pt]x+\bbox[#FFEEAA;,1.5pt]{\strut 2})(3x^{2}-x+5) = \secondcbox{#FFFFFF;}{\strut x\cdot 3x^{2}-x\cdot x+x\cdot 5}{3x^{3}-x^{2}+5x}+\bbox[#FFEEAA;,1.5pt]{\strut 2\cdot 3x^{2}-2\cdot x+2\cdot 5}\,\textrm{.}</math>}}
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Then, do the same for
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Now, collect together ''x''³-, ''x''²-, ''x''- and the constant terms
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<math>2</math>
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from the first bracket:
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{{Displayed math||
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<math>3x^{3}+(-1+6)x^{2}+(5-2)x+10=3x^{3}+5x^{2}+3x+10\,\textrm{.}</math>}}
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<math>\left( x+2 \right)\left( 3x^{2}-x+5 \right)=3x^{3}-x^{2}+5x+2\centerdot 3x^{2}-2\centerdot x+2\centerdot 5</math>
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The coefficient in front of ''x''² is 5 and the coefficient in front of ''x'' is 3.
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Now, collect together
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<math>x^{3}</math>-,
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<math>x^{2}</math>-,
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<math>x</math>- and the constant terms:
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<math>3x^{3}+\left( -1+6 \right)x^{2}+\left( 5-2 \right)x+10=3x^{3}+5x^{2}+3x+10</math>
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The coefficient in front of
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<math>x^{2}</math>
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is
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<math>5</math>
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and the coefficient in front of x is
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<math>3</math>.
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Current revision

First, we multiply the second bracket by x from the first bracket,

\displaystyle (\bbox[#FFEEAA;,1.5pt]{\strut x}+\bbox[#FFFFFF;,1.5pt]{\strut 2})(3x^{2}-x+5) = \bbox[#FFEEAA;,1.5pt]{\strut x\cdot 3x^{2}-x\cdot x+x\cdot 5}+{}\rlap{\cdots}\phantom{\bbox[#FFEEAA;,1.5pt]{\strut 2\cdot 3x^{2}-2\cdot x+2\cdot 5}\,\textrm{.}}

Then, do the same for 2 from the first bracket

\displaystyle (\bbox[#FFFFFF;,1.5pt]x+\bbox[#FFEEAA;,1.5pt]{\strut 2})(3x^{2}-x+5) = \secondcbox{#FFFFFF;}{\strut x\cdot 3x^{2}-x\cdot x+x\cdot 5}{3x^{3}-x^{2}+5x}+\bbox[#FFEEAA;,1.5pt]{\strut 2\cdot 3x^{2}-2\cdot x+2\cdot 5}\,\textrm{.}

Now, collect together x³-, x²-, x- and the constant terms

\displaystyle 3x^{3}+(-1+6)x^{2}+(5-2)x+10=3x^{3}+5x^{2}+3x+10\,\textrm{.}

The coefficient in front of x² is 5 and the coefficient in front of x is 3.

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