Further Maths - Partial Fractions
What is the simple case of partial fractions?
Where the denominator is $(ax + b)(cx + d)$.
What is the harder case of partial fractions?
Where the denominator is $(ax + b)(cx + d)^2$
\[\frac{5x}{(x + 2)(x - 3)}\]
What’s the first step to finding the partial fractions?
Rewriting as
\[\frac{A}{x + 2} + \frac{B}{x - 3}\]\[\frac{A}{x + 2} + \frac{B}{x - 3}\]
How could you add these two fractions together?
\[\frac{5x}{(x + 2)(x - 3)} = \frac{A(x - 3) + B(x + 2)}{(x+2)(x-3)}\]
How could you simplify this?
\[5x = A(x - 3) + B(x + 2)\]
If you’re solving this, what could you set $x$ equal to in order to make one of the unknowns dissapear?
- $3$
- $-2$
\[5x = A(x - 3) + B(x + 2)\]
If $x = 3$, what is $B$ equal to?
If the denominator is
\[(ax + b)(cx + d)^2\]
, how many partial fractions would there be?
If the denominator is
\[(ax + b)(cx + d)^2\]
, what would the partial fractions look like?
If the denominator is
\[(ax + b)(cx + d)^2\]
, in partial fractions would would be in the denominator of the $A$ term?
If the denominator is
\[(ax + b)(cx + d)^2\]
, in partial fractions would would be in the denominator of the $B$ term?
If the denominator is
\[(ax + b)(cx + d)^2\]
, in partial fractions would would be in the denominator of the $C$ term?
\[\frac{A}{(ax + b)} + \frac{B}{(cx + d)^2} + \frac{C}{(cx + d)}\]
In partial fractions, why is the denominator of the $B$ term $(cx + d)^2$ rather than just $(cx + d)$?
Otherwise when you add the fractions together they don’t reduce properly.
\[\frac{A}{(ax + b)} + \frac{B}{(cx + d)^2} + \frac{C}{(cx + d)}\]
What does this look like as one fraction?
\[\frac{A}{(ax + b)} + \frac{B}{(cx + d)^2} + \frac{C}{(cx + d)}\]
What does $A$ multiply in the numerator when this is written as one fraction?
\[\frac{A}{(ax + b)} + \frac{B}{(cx + d)^2} + \frac{C}{(cx + d)}\]
What does $B$ multiply in the numerator when this is written as one fraction?
\[\frac{A}{(ax + b)} + \frac{B}{(cx + d)^2} + \frac{C}{(cx + d)}\]
What does $C$ multiply in the numerator when this is written as one fraction?
\[\frac{A}{x + 2} + \frac{B}{(x-1)^2} + \frac{C}{(x-1)}\]
What is the numerator of this fraction when all the terms are added together?
\[\frac{x^2 + 8x + 30}{(x + 2)(x - 3)^2}\]
In partial fractions, what is $x^2 + 8x + 30$ equivalent to in terms of $A$, $B$ and $C$?
\[A(x - 3)^2 + B(x + 2) + C(x + 2)(x - 1)\]
What would be the ‘gotcha’ for substituting in $x = -2$?
You have to square $(x - 3)^2$.
\[x^2 - 8x + 30 \equiv A(x - 3)^2 + B(x + 2) + C(x + 2)(x - 3)\]
If you know $A = 2$ and $B = 3$, what two different techniques could you use here in order to find the value of $C$?
- Equating coefficients
- Substituting in a value of $x$ and seeing what value of $C$ makes it true.
2021-05-05
\[\frac{5x^2 + 5x + 8}{(x + 2)(x^2 + 5)}\]
How would you write this for a partial fractions question?
When do you use $Bx + C$ for a partial fractions question?
When there is a quadratic $x^2$ term under the fraction, like $(x^2 + 5)$ or $(x^2 - 6)$.
2022-01-20
What’s the quick way of getting to the numerator equivalence in partial fractions?
Multiplying both sides by the denominator.