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??
$$ \frac{A(x - 3) + B(x + 2)}{(x+2)(x-3)} $$
$$\frac{5x}{(x + 2)(x - 3)} = \frac{A(x - 3) + B(x + 2)}{(x+2)(x-3)}$$ How could you simplify this??
$$ 5x = A(x - 3) + B(x + 2) $$
$$5x = A(x - 3) + B(x + 2)$$ 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??
$$ B = 3 $$
If the denominator is $$(ax + b)(cx + d)^2$$, how many partial fractions would there be??
$$ 3 $$
If the denominator is $$(ax + b)(cx + d)^2$$, what would the partial fractions look like??
$$ \frac{A}{(ax + b)} + \frac{B}{(cx + d)^2} + \frac{C}{(cx + d)} $$
If the denominator is $$(ax + b)(cx + d)^2$$, in partial fractions would would be in the denominator of the $A$ term??
$$ (ax + b) $$
If the denominator is $$(ax + b)(cx + d)^2$$, in partial fractions would would be in the denominator of the $B$ term??
$$ (cx + d)^2 $$
If the denominator is $$(ax + b)(cx + d)^2$$, in partial fractions would would be in the denominator of the $C$ term??
$$ (cx + d) $$
$$\frac{A}{(ax + b)} + \frac{B}{(cx + d)^2} + \frac{C}{(cx + d)}$$ 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??
$$ \frac{A(cx + d)^2 + B(ax + b) + C(ax + b)(cx + d)}{(ax + b)(cx + d)^2} $$
$$\frac{A}{(ax + b)} + \frac{B}{(cx + d)^2} + \frac{C}{(cx + d)}$$ What does $A$ multiply in the numerator when this is written as one fraction??
$$ A(cx + d)^2 $$
$$\frac{A}{(ax + b)} + \frac{B}{(cx + d)^2} + \frac{C}{(cx + d)}$$ What does $B$ multiply in the numerator when this is written as one fraction??
$$ B(ax + b) $$
$$\frac{A}{(ax + b)} + \frac{B}{(cx + d)^2} + \frac{C}{(cx + d)}$$ What does $C$ multiply in the numerator when this is written as one fraction??
$$ C(ax + b)(cx + d) $$
$$\frac{A}{x + 2} + \frac{B}{(x-1)^2} + \frac{C}{(x-1)}$$ What is the numerator of this fraction when all the terms are added together??
$$ A(x - 1)^2 + B(x + 2) + C(x + 2)(x - 1) $$
$$\frac{x^2 + 8x + 30}{(x + 2)(x - 3)^2}$$ In partial fractions, what is $x^2 + 8x + 30$ equivalent to in terms of $A$, $B$ and $C$??
$$ x^2 - 8x + 30 \equiv A(x - 3)^2 + B(x + 2) + C(x + 2)(x - 1) $$
$$A(x - 3)^2 + B(x + 2) + C(x + 2)(x - 1)$$ 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??
$$ \frac{A}{x+2} + \frac{Bx + C}{x^2 + 5} $$
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.
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Metadata
date: 2021-01-27 14:33
tags:
- '@?further-maths'
- '@?algebra'
- '@?public'
title: Further Maths - Partial Fractions