Solution 3.4:1d
From Förberedande kurs i matematik 2
m (Lösning 3.4:1d moved to Solution 3.4:1d: Robot: moved page) |
m |
||
| (One intermediate revision not shown.) | |||
| Line 1: | Line 1: | ||
| - | {{ | + | We start by adding and taking away <math>x^2</math> in the numerator, so that, in combination with <math>x^3</math>, we obtain the expression <math>x^3+x^2 = x^2(x+1)</math> which can be simplified with the denominator <math>x+1</math>, |
| - | < | + | |
| - | {{ | + | {{Displayed math||<math>\begin{align} |
| + | \frac{x^3+x+2}{x+1} | ||
| + | &= \frac{x^3+x^2-x^2+x+2}{x+1}\\[5pt] | ||
| + | &= \frac{x^3+x^2}{x+1} + \frac{-x^2+x+2}{x+1}\\[5pt] | ||
| + | &= \frac{x^2(x+1)}{x+1} + \frac{-x^2+x+2}{x+1}\\[5pt] | ||
| + | &= x^2 + \frac{-x^2+x+2}{x+1}\,\textrm{.} | ||
| + | \end{align}</math>}} | ||
| + | |||
| + | The term <math>-x^2</math> in the remaining quotient needs to complemented with | ||
| + | <math>-x</math> so that we get <math>-x^2-x = -x(x+1)</math>, which is divisible by | ||
| + | <math>x+1</math>, | ||
| + | |||
| + | {{Displayed math||<math>\begin{align} | ||
| + | x^2 + \frac{-x^2+x+2}{x+1} | ||
| + | &= x^2 + \frac{-x^2-x+x+x+2}{x+1}\\[5pt] | ||
| + | &= x^2 + \frac{-x^2-x}{x+1} + \frac{2x+2}{x+1}\\[5pt] | ||
| + | &= x^2 + \frac{-x(x+1)}{x+1} + \frac{2x+2}{x+1}\\[5pt] | ||
| + | &= x^2 - x + \frac{2x+2}{x+1}\,\textrm{.} | ||
| + | \end{align}</math>}} | ||
| + | |||
| + | The last quotient divides perfectly and we obtain | ||
| + | |||
| + | {{Displayed math||<math>x^2-x+\frac{2x+2}{x+1}=x^2-x+2\,\textrm{.}</math>}} | ||
| + | |||
| + | A quick check of whether | ||
| + | |||
| + | {{Displayed math||<math>\frac{x^3+x+2}{x+1} = x^2-x+2\,\textrm{.}</math>}} | ||
| + | |||
| + | is the correct answer is to investigate whether | ||
| + | |||
| + | {{Displayed math||<math>x^3+x+2 = (x^2-x+2)(x+1)</math>}} | ||
| + | |||
| + | holds. If we expand the right-hand side, we see that the relation really does hold | ||
| + | |||
| + | {{Displayed math||<math>\begin{align} | ||
| + | (x^2-x+2)(x+1) = x^3+x^2-x^2-x+2x+2 = x^3+x+2\,\textrm{.} | ||
| + | \end{align}</math>}} | ||
Current revision
We start by adding and taking away \displaystyle x^2 in the numerator, so that, in combination with \displaystyle x^3, we obtain the expression \displaystyle x^3+x^2 = x^2(x+1) which can be simplified with the denominator \displaystyle x+1,
| \displaystyle \begin{align}
\frac{x^3+x+2}{x+1} &= \frac{x^3+x^2-x^2+x+2}{x+1}\\[5pt] &= \frac{x^3+x^2}{x+1} + \frac{-x^2+x+2}{x+1}\\[5pt] &= \frac{x^2(x+1)}{x+1} + \frac{-x^2+x+2}{x+1}\\[5pt] &= x^2 + \frac{-x^2+x+2}{x+1}\,\textrm{.} \end{align} |
The term \displaystyle -x^2 in the remaining quotient needs to complemented with \displaystyle -x so that we get \displaystyle -x^2-x = -x(x+1), which is divisible by \displaystyle x+1,
| \displaystyle \begin{align}
x^2 + \frac{-x^2+x+2}{x+1} &= x^2 + \frac{-x^2-x+x+x+2}{x+1}\\[5pt] &= x^2 + \frac{-x^2-x}{x+1} + \frac{2x+2}{x+1}\\[5pt] &= x^2 + \frac{-x(x+1)}{x+1} + \frac{2x+2}{x+1}\\[5pt] &= x^2 - x + \frac{2x+2}{x+1}\,\textrm{.} \end{align} |
The last quotient divides perfectly and we obtain
| \displaystyle x^2-x+\frac{2x+2}{x+1}=x^2-x+2\,\textrm{.} |
A quick check of whether
| \displaystyle \frac{x^3+x+2}{x+1} = x^2-x+2\,\textrm{.} |
is the correct answer is to investigate whether
| \displaystyle x^3+x+2 = (x^2-x+2)(x+1) |
holds. If we expand the right-hand side, we see that the relation really does hold
| \displaystyle \begin{align}
(x^2-x+2)(x+1) = x^3+x^2-x^2-x+2x+2 = x^3+x+2\,\textrm{.} \end{align} |
