Solution 2.1:4a
From Förberedande kurs i matematik 1
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| - | {{ | + | First, we multiply the second bracket by ''x'' 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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| + | Then, do the same for 2 from the first bracket | ||
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| + | {{Displayed math|| | ||
| + | <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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| + | Now, collect together ''x''³-, ''x''²-, ''x''- and the constant terms | ||
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| + | {{Displayed math|| | ||
| + | <math>3x^{3}+(-1+6)x^{2}+(5-2)x+10=3x^{3}+5x^{2}+3x+10\,\textrm{.}</math>}} | ||
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| + | The coefficient in front of ''x''² is 5 and the coefficient in front of ''x'' is 3. | ||
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.
