Question
Write the partial fraction decomposition of the following expression.
(20x + 35)/(x + 4)2
▶️Answer/Explanation
Solution:
(20x + 35)/(x + 4)2
(20x + 35)/(x + 4)2 = [A/(x + 4)] + [B/(x + 4)2]
(20x + 35)/(x + 4)2 = [A(x + 4) + B]/ (x + 4)2
Now, equating the numerators,
20x + 35 = A(x + 4) + B
20x + 35 = Ax + 4A + B
20x + 35 = Ax + (4A + B)
By equating the coefficients,
A = 20
4A + B = 35
4(20) + B = 35
B = 35 – 80 = -45
Therefore, (20x + 35)/(x + 4)2 = [20/(x + 4)] – [45/(x + 4)2]
Question
Decompose the given expression into partial fractions.
(x2 + 1)/ (x3 + 3x2 + 3x + 2)
▶️Answer/Explanation
Solution:
(x2 + 1)/ (x3 + 3x2 + 3x + 2)
Using the factor theorem, x + 2 is a factor of x3 + 3x2 + 3x + 2.
Thus, x3 + 3x2 + 3x + 2 = (x + 2)(x2 + x + 1)
Now, the given expression can be written as:
(x2 + 1)/ (x3 + 3x2 + 3x + 2) = (x2 + 1)/ [(x + 2)(x2 + x + 1)]
By the method of decomposition,
(x2 + 1)/(x + 2)(x2 + x + 1) = [A/(x + 2)] + [(Bx + C)/(x2 + x + 1)]
(x2 + 1)/(x + 2)(x2 + x + 1) = [A(x2 + x + 1) + (Bx + C)(x + 2)]/ [(x + 2)(x2 + x + 1)]
= [(A + B)x2 + (A + 2B + C)x + A + 2C]/ [(x + 2)(x2 + x + 1)]
Equating the coefficients in the numerators of both LHS and RHS,
A + B = 1
A + 2B + C = 0
A + 2C = 1
Solving these equations,
A = 5/3, B = -2/3 and C = -1/3
(x2 + 1)/(x + 2)(x2 + x + 1) = [5/3(x + 2)] – [(2x + 1)/3(x2 + x + 1)]
Question
Express Following in Partial Fraction
▶️Answer/Explanation
Solution
We can decompose it as such:
And so we can find A and B, by simplifying the RHS and compare the coefficients.
Now we have the case where we can compare the coefficients, since RHS must be equivalent to LHS, the coefficients must be equal. In other words:
We can solve it using our GDC, linSolve() again, and get the answers:
So we can re-write all of these:
Since the IB only requires us to perform this for at most 2 decomposed fractions, we can stop here. But if there are more, you can just imagine adding a third term with C and it’ll be the same! Do note that it’ll be slightly different when there are non-linear terms in the denominator, but that would be beyond the scope of IB maths.
Question
Determine the partial fraction decomposition of each of the following expression.
\[\frac{{17x – 53}}{{{x^2} – 2x – 15}}\]
▶️Answer/Explanation
Solution
The first step is to determine the form of the partial fraction decomposition. However, in order to do that we first need to factor the denominator as much as possible. Doing this gives,
\[\frac{{17x – 53}}{{\left( {x – 5} \right)\left( {x + 3} \right)}}\]
Okay, we can now see that the partial fraction decomposition is,
\[\frac{{17x – 53}}{{{x^2} – 2x – 15}} = \frac{A}{{x – 5}} + \frac{B}{{x + 3}}\] Show Step 2
The LCD for this expression is \(\left( {x – 5} \right)\left( {x + 3} \right)\). Adding the two terms back up gives,
\[\frac{{17x – 53}}{{{x^2} – 2x – 15}} = \frac{{A\left( {x + 3} \right) + B\left( {x – 5} \right)}}{{\left( {x – 5} \right)\left( {x + 3} \right)}}\] Show Step 3
Setting the numerators equal gives,
\[17x – 53 = A\left( {x + 3} \right) + B\left( {x – 5} \right)\] Show Step 4
Now all we need to do is pick “good” values of \(x\) to determine the constants. Here is that work.
\[\begin{array}{l}{x = 5:}\\{x = – 3:}\end{array}\hspace{0.25in}\begin{aligned}32 & = 8A\\ – 104 & = – 8B\end{aligned}\hspace{0.25in}\to \hspace{0.25in}\begin{array}{l}{A = 4}\\{B = 13}\end{array}\] Show Step 5
The partial fraction decomposition is then,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{17x – 53}}{{{x^2} – 2x – 15}} = \frac{4}{{x – 5}} + \frac{{13}}{{x + 3}}}}\]