Solution 3.4:1c
From Förberedande kurs i matematik 2
m (Lösning 3.4:1c moved to Solution 3.4:1c: Robot: moved page) |
m |
||
| (One intermediate revision not shown.) | |||
| Line 1: | Line 1: | ||
| - | {{ | + | If we focus on the leading term <math>x^3</math>, we need to complement it with <math>ax^2</math> in order to get a sub expression that is divisible by the denominator, |
| - | < | + | |
| - | {{ | + | {{Displayed math||<math>\frac{x^3+a^3}{x+a} = \frac{x^3+ax^2-ax^2+a^3}{x+a}\,\textrm{.}</math>}} |
| + | |||
| + | With this form on the right-hand side, we can separate away the first two terms in the numerator and have left a polynomial quotient with <math>-ax^2+a^3</math> in the numerator, | ||
| + | |||
| + | {{Displayed math||<math>\begin{align} | ||
| + | \frac{x^3+ax^2-ax^2+a^3}{x+a} | ||
| + | &= \frac{x^3+ax^2}{x+a} + \frac{-ax^2+a^3}{x+a}\\[5pt] | ||
| + | &= \frac{x^2(x+a)}{x+a} + \frac{-ax^2+a^3}{x+a}\\[5pt] | ||
| + | &= x^2 + \frac{-ax^2+a^3}{x+a}\,\textrm{.} | ||
| + | \end{align}</math>}} | ||
| + | |||
| + | When we treat the new quotient, we add and take away <math>-a^2x</math> to/from <math>-ax^2</math> in order to get something divisible by <math>x+a</math>, | ||
| + | |||
| + | {{Displayed math||<math>\begin{align} | ||
| + | x^2+\frac{-ax^2+a^3}{x+a} | ||
| + | &= x^2 + \frac{-ax^2-a^2x+a^2x+a^3}{x+a}\\[5pt] | ||
| + | &= x^2 + \frac{-ax^2-a^2x}{x+a} + \frac{a^2x+a^3}{x+a}\\[5pt] | ||
| + | &= x^2 + \frac{-ax(x+a)}{x+a} + \frac{a^2x+a^3}{x+a}\\[5pt] | ||
| + | &= x^2 - ax + \frac{a^2x+a^3}{x+a}\,\textrm{.} | ||
| + | \end{align}</math>}} | ||
| + | |||
| + | In the last quotient, the numerator has <math>x+a</math> as a factor, and we obtain a perfect division, | ||
| + | |||
| + | {{Displayed math||<math>x^2 - ax + \frac{a^2x+a^3}{x+a} = x^2 - ax + \frac{a^2(x+a)}{x+a} = x^2-ax+a^2\,\textrm{.}</math>}} | ||
| + | |||
| + | If we have calculated correctly, we should have | ||
| + | |||
| + | {{Displayed math||<math>\frac{x^3+a^3}{x+a} = x^2-ax+a^2</math>}} | ||
| + | |||
| + | and one way to check the answer is to multiply both sides by <math>x+a</math>, | ||
| + | |||
| + | {{Displayed math||<math>x^3+a^3 = (x^2-ax+a^2)(x+a)\,\textrm{.}</math>}} | ||
| + | |||
| + | Then, expand the right-hand side and we should get what is on the left-hand side, | ||
| + | |||
| + | {{Displayed math||<math>\begin{align} | ||
| + | \text{RHS} | ||
| + | &= (x^2-ax+a^2)(x+a)\\[5pt] | ||
| + | &= x^3+ax^2-ax^2-a^2x+a^2x+a^3\\[5pt] | ||
| + | &= x^3+a^3\\[5pt] | ||
| + | &= \text{LHS.} | ||
| + | \end{align}</math>}} | ||
Current revision
If we focus on the leading term \displaystyle x^3, we need to complement it with \displaystyle ax^2 in order to get a sub expression that is divisible by the denominator,
| \displaystyle \frac{x^3+a^3}{x+a} = \frac{x^3+ax^2-ax^2+a^3}{x+a}\,\textrm{.} |
With this form on the right-hand side, we can separate away the first two terms in the numerator and have left a polynomial quotient with \displaystyle -ax^2+a^3 in the numerator,
| \displaystyle \begin{align}
\frac{x^3+ax^2-ax^2+a^3}{x+a} &= \frac{x^3+ax^2}{x+a} + \frac{-ax^2+a^3}{x+a}\\[5pt] &= \frac{x^2(x+a)}{x+a} + \frac{-ax^2+a^3}{x+a}\\[5pt] &= x^2 + \frac{-ax^2+a^3}{x+a}\,\textrm{.} \end{align} |
When we treat the new quotient, we add and take away \displaystyle -a^2x to/from \displaystyle -ax^2 in order to get something divisible by \displaystyle x+a,
| \displaystyle \begin{align}
x^2+\frac{-ax^2+a^3}{x+a} &= x^2 + \frac{-ax^2-a^2x+a^2x+a^3}{x+a}\\[5pt] &= x^2 + \frac{-ax^2-a^2x}{x+a} + \frac{a^2x+a^3}{x+a}\\[5pt] &= x^2 + \frac{-ax(x+a)}{x+a} + \frac{a^2x+a^3}{x+a}\\[5pt] &= x^2 - ax + \frac{a^2x+a^3}{x+a}\,\textrm{.} \end{align} |
In the last quotient, the numerator has \displaystyle x+a as a factor, and we obtain a perfect division,
| \displaystyle x^2 - ax + \frac{a^2x+a^3}{x+a} = x^2 - ax + \frac{a^2(x+a)}{x+a} = x^2-ax+a^2\,\textrm{.} |
If we have calculated correctly, we should have
| \displaystyle \frac{x^3+a^3}{x+a} = x^2-ax+a^2 |
and one way to check the answer is to multiply both sides by \displaystyle x+a,
| \displaystyle x^3+a^3 = (x^2-ax+a^2)(x+a)\,\textrm{.} |
Then, expand the right-hand side and we should get what is on the left-hand side,
| \displaystyle \begin{align}
\text{RHS} &= (x^2-ax+a^2)(x+a)\\[5pt] &= x^3+ax^2-ax^2-a^2x+a^2x+a^3\\[5pt] &= x^3+a^3\\[5pt] &= \text{LHS.} \end{align} |
