Corrected exercises: partial fraction decomposition (201-SN3-RE)

For any polynomials $A$ and $B$ with $B$ nonzero, there is a UNIQUE pair $(Q,R)$ such that $A(x)=B(x)\times Q(x)+R(x)$ with $\deg(R)<\deg(B)$. That last inequality, and it alone, tells you when to stop: you go down one degree at each step, and you put the pencil down as soon as the remainder falls strictly below the degree of the divisor. Dividing the equality by $B(x)$ gives the form that is useful here: $\frac{A(x)}{B(x)}=Q(x)+\frac{R(x)}{B(x)}$, where the second fraction now has a numerator smaller than its denominator.

Divide $2x+1$ by $x-3$. The leading term $2x$ comes from $2\times x$, so the quotient starts with 2; subtract $2(x-3)=2x-6$ and what is left is $1-(-6)=7$. Since $\deg(7)=0<\deg(x-3)=1$, the division is over: $2x+1=(x-3)\times 2+7$, so $\frac{2x+1}{x-3}=2+\frac{7}{x-3}$. This is the founding example of the chapter: on the right, everything can be read off, the constant 2 is the limiting value at infinity and the piece $\frac{7}{x-3}$ carries the pole on its own.

On the figure: $2x^4=2x^2\times(x^2-3)$ up to the first term, subtract $2x^4-6x^2$ and $6x^2+x-1$ is left, which explains the middle line. Then $6x^2=6\times(x^2-3)$ up to the first term, subtract $6x^2-18$ and $x+17$ is left. Since $\deg(x+17)=1<\deg(x^2-3)=2$, you stop. So $Q(x)=2x^2+6$, $R(x)=x+17$, and $\frac{2x^4+x-1}{x^2-3}=2x^2+6+\frac{x+17}{x^2-3}$. Watch out for the layout trap: the dividend has no $x^3$ term and no $x^2$ term, so leave those columns empty rather than shifting the terms.

For $\frac{x^2+3}{x+2}$: $x^2+3=(x+2)(x-2)+7$, so $\frac{x^2+3}{x+2}=x-2+\frac{7}{x+2}$. For $\frac{x^3}{2x-1}$, the leading coefficient of the divisor is not 1, which puts fractions in the quotient and makes many students give up: $x^3=(2x-1)\left(\frac12x^2+\frac14x+\frac18\right)+\frac18$, so $\frac{x^3}{2x-1}=\frac12x^2+\frac14x+\frac18+\frac{1}{8(2x-1)}$. Check by expanding: it takes ten seconds and catches every sign error.

None of the four remainders is zero: they are 7, $x+17$, 7 and $\frac18$. A zero remainder would mean that $B$ divides $A$, so that the rational function is in fact a polynomial, and the chapter would stop there: there would be nothing to decompose. The case $R\neq 0$ is exactly the one that makes the chapter necessary, because it is $\frac{R(x)}{B(x)}$, and it alone, that will be broken into partial fractions.

Exclude the roots of the denominator. $2x-5=0$ gives $x=\frac52$, so the domain is $\mathbb{R}\setminus\left\{\frac52\right\}$. $x^2+9$ never vanishes on $\mathbb{R}$, so the domain is all of $\mathbb{R}$: an irreducible denominator creates no excluded value, which is the first visible sign that it is irreducible. $2x^2-8x=2x(x-4)$ vanishes at 0 and at 4, so the domain is $\mathbb{R}\setminus\{0, 4\}$. Finally $(x-3)^2$ vanishes only at 3, domain $\mathbb{R}\setminus\{3\}$.

Already decomposed: $\frac{3}{2x-5}$, because the denominator has degree 1 and nothing more can be broken; $\frac{2}{x^2+x+1}$ and $\frac{2x-3}{x^2+x+1}$, because $\Delta=1-4=-3<0$ and the numerator has degree at most 1; $\frac{2x}{x^2+9}$, for the same reason with $\Delta=-36<0$; and $\frac{2}{(x-3)^2}$, because the numerator is constant and the denominator is a power of a linear factor.

$\frac{x+3}{x-3}$ is blocked by the degree: $\deg(A)=\deg(B)=1$, you must divide, and you get $1+\frac{6}{x-3}$. $\frac{x^3}{2x-1}$ is blocked the same way, with $\deg(A)=3>1$; the division was done in exercise 1. $\frac{1}{2x^2-8x}$ is blocked by the factorization: $\Delta=64>0$, the denominator breaks into $2x(x-4)$.

$\frac{1}{2x^2-8x}=\frac{1}{2x(x-4)}=\frac{A}{x}+\frac{B}{x-4}$. Over the common denominator $2x(x-4)$, the numerator becomes $2A(x-4)+2Bx$, which must equal 1 for every $x$. At $x=0$: $-8A=1$, so $A=-\frac18$. At $x=4$: $8B=1$, so $B=\frac18$. Hence $\frac{1}{2x^2-8x}=-\frac{1}{8x}+\frac{1}{8(x-4)}$. Check at $x=2$: on the left $\frac{1}{8-16}=-\frac18$, on the right $-\frac{1}{16}-\frac{1}{16}=-\frac18$.

The difference is not in the denominator but in the NUMERATOR. For $\frac{2}{(x-3)^2}$, the numerator is constant: there is nothing to share out, the fraction is a partial fraction on its own. For $\frac{3x+1}{(x+1)^2}$, the numerator has degree 1 and can be split into a part that cancels a factor $(x+1)$ and a part that does not, hence two terms. This is the general rule for a repeated factor, treated in exercise 5: a square in the denominator opens TWO terms, and the degree of the numerator decides whether the first one is empty or not.

Look for two numbers with sum 5 and product 6: they are 2 and 3, so $x^2-5x+6=(x-2)(x-3)$. The domain is $\mathbb{R}\setminus\{2, 3\}$. You can also go through the discriminant, $\Delta=25-24=1$, roots $\frac{5\pm1}{2}$, but sum and product is twice as fast on a monic quadratic with integer roots.

The figure shows two vertical asymptotes, at $x=2$ and $x=3$, drawn as dashed lines. Each root of the denominator makes it vanish without making the numerator vanish, so the fraction tends to infinity in absolute value: one root, one asymptote. This is the graphical translation of the thread of the set, THE DENOMINATOR IS IN CHARGE. Notice the middle branch, between 2 and 3: it stays below $-12$, because $(x-2)(x-3)$ is negative there with absolute value at most $\frac14$.

Set $\frac{3}{(x-2)(x-3)}=\frac{A}{x-2}+\frac{B}{x-3}$. Over a common denominator: $\frac{A(x-3)+B(x-2)}{(x-2)(x-3)}=\frac{(A+B)x-3A-2B}{(x-2)(x-3)}$. The two fractions have the same denominator and are equal for every $x$ in the domain, so the numerators are the same polynomial: $A+B=0$ and $-3A-2B=3$. From the first, $A=-B$, which you substitute: $3B-2B=3$, so $B=3$ and $A=-3$. Conclusion: $\frac{3}{x^2-5x+6}=\frac{-3}{x-2}+\frac{3}{x-3}$.

Same form, different numerator: $A(x-3)+B(x-2)=3x-7$, so $A+B=3$ and $-3A-2B=-7$. Multiplying the first by 2 and adding: $-A=-7+6=-1$, so $A=1$ and $B=2$. Hence $\frac{3x-7}{x^2-5x+6}=\frac{1}{x-2}+\frac{2}{x-3}$. Remember: the FORM of the decomposition did not move one bit between c) and d), only the coefficients changed. The numerator never decides the form.

At $x=0$ for the first: on the left $\frac{3}{6}=\frac12$, on the right $\frac{-3}{-2}+\frac{3}{-3}=\frac32-1=\frac12$. For the second: on the left $\frac{-7}{6}$, on the right $\frac{1}{-2}+\frac{2}{-3}=-\frac12-\frac23=-\frac76$. This check is excellent but it is not a proof: two functions can agree at one point without being equal. It becomes a proof if you do it at as many points as there are unknowns, here two, because two polynomials of degree at most 1 that agree at two points are equal.

Over a common denominator, the numerator on the right is $A(x+2)(x-3)+B(x-1)(x-3)+C(x-1)(x+2)$. Expanding and collecting: the coefficient of $x^2$ is $A+B+C$, that of $x$ is $-A-4B+C$, and the constant term is $-6A+3B-2C$. Since this polynomial must be the constant 3, you get $A+B+C=0$, $-A-4B+C=0$ and $-6A+3B-2C=3$. Three equations, three unknowns, and plenty of chances to get a sign wrong.

Multiply everything by $(x-1)$: on the left $\frac{3}{(x+2)(x-3)}$ remains, on the right $A+\frac{B(x-1)}{x+2}+\frac{C(x-1)}{x-3}$. The equality holds for every $x$ other than 1, $-2$ and 3, hence also at $x=1$ by continuity. At $x=1$, the factors $(x-1)$ in the terms in $B$ and $C$ vanish and those two terms disappear: what is left is $A=\frac{3}{(1+2)(1-3)}=\frac{3}{-6}=-\frac12$.

The same move with $(x+2)$ and $x=-2$ gives $B=\frac{3}{(-2-1)(-2-3)}=\frac{3}{15}=\frac15$. With $(x-3)$ and $x=3$: $C=\frac{3}{(3-1)(3+2)}=\frac{3}{10}$. In practice you rewrite nothing: you COVER the chosen factor in the original denominator with your finger and evaluate what is left at the root of that factor. Hence the name. Conclusion: $\frac{3}{(x-1)(x+2)(x-3)}=-\frac{1}{2(x-1)}+\frac{1}{5(x+2)}+\frac{3}{10(x-3)}$.

$A+B+C=-\frac12+\frac15+\frac3{10}=\frac{-5+2+3}{10}=0$. Next $-A-4B+C=\frac12-\frac45+\frac3{10}=\frac{5-8+3}{10}=0$. Finally $-6A+3B-2C=3+\frac35-\frac35=3$. All three equations hold: the shortcut gives the exact solution of the system, not an approximation. A limitation to know: it applies as is only to SIMPLE poles, and for a repeated factor it only delivers the coefficient of the highest power.

$\frac{1}{k(k+1)}$ has two simple poles, 0 and $-1$: by cover-up, the coefficient of $\frac1k$ is $\frac{1}{0+1}=1$ and that of $\frac{1}{k+1}$ is $\frac{1}{-1}=-1$, so $\frac{1}{k(k+1)}=\frac1k-\frac1{k+1}$. The sum telescopes: $\sum_{k=1}^{n}\left(\frac1k-\frac1{k+1}\right)=1-\frac1{n+1}=\frac{n}{n+1}$, because each $-\frac1{k+1}$ cancels the $\frac{1}{k+1}$ of the next term. The limit is 1. This is the textbook example of a telescoping series, and it shows that the decomposition is useful well beyond antiderivatives.

If $\frac{3x+1}{(x+1)^2}=\frac{A}{(x+1)^2}$, then multiplying by $(x+1)^2$ gives $3x+1=A$ for every $x$ in the domain, that is, a polynomial of degree 1 equal to a constant: impossible. The reasoning is general and gives the rule: a factor $(x-a)^m$ opens $m$ terms, $\frac{A_1}{x-a}+\dots+\frac{A_m}{(x-a)^m}$, and it is the degree of the numerator that decides which ones are nonzero. With a constant numerator, only the last one survives, which explains the case $\frac{2}{(x-3)^2}$ in exercise 2.

$3x+1=A(x+1)+B$. At $x=-1$: $-3+1=-2=B$. The coefficient of $x$ gives $A=3$. So $\frac{3x+1}{(x+1)^2}=\frac{3}{x+1}-\frac{2}{(x+1)^2}$. Another route, often faster: write $3x+1=3(x+1)-2$ then divide term by term by $(x+1)^2$. It is the same idea as long division, applied to the numerator rather than to the whole fraction.

Recognize $x^3+3x^2+3x+1=(x+1)^3$, a special product of degree 3. The substitution $u=x+1$, so $x=u-1$, turns the numerator into $(u-1)^2+(u-1)+1=u^2-2u+1+u-1+1=u^2-u+1$. Hence $\frac{x^2+x+1}{(x+1)^3}=\frac{u^2-u+1}{u^3}=\frac1u-\frac1{u^2}+\frac1{u^3}$, that is, $\frac{1}{x+1}-\frac{1}{(x+1)^2}+\frac{1}{(x+1)^3}$. The substitution avoids the system entirely: as soon as there is only one repeated pole, it is the method to use.

$x^2+4x+4=(x+2)^2$, so the form is $\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{(x+2)^2}$. Covering $(x-1)$ at $x=1$ gives $A=\frac{2+1}{(1+2)^2}=\frac39=\frac13$. Covering $(x+2)^2$ at $x=-2$ gives $C=\frac{2(-2)+1}{-2-1}=\frac{-3}{-3}=1$. For $B$, take a simple value, $x=0$: on the left $\frac{1}{(-1)(4)}=-\frac14$; on the right $-\frac13+\frac B2+\frac14$. Hence $\frac B2=-\frac14+\frac13-\frac14=-\frac16$ and $B=-\frac13$. Conclusion: $\frac{1}{3(x-1)}-\frac{1}{3(x+2)}+\frac{1}{(x+2)^2}$.

Multiplying by $(x+2)$ alone does not clear the denominator: an $(x+2)$ remains under the term in $C$, and the expression blows up at $x=-2$ instead of simplifying. Multiplying by $(x+2)^2$ clears everything, but then the term in $B$ is multiplied by $(x+2)$ and vanishes at $x=-2$: you recover $C$, not $B$. In other words, the cover-up method always reaches the HIGHEST power of a factor, never the intermediate powers. For those you need a convenient value, a derivative, or equating coefficients.

Expand: $(x-1)(x^2+x+1)=x^3+x^2+x-x^2-x-1=x^3-1$. For $x^2+x+1$, the discriminant is $\Delta=1-4=-3<0$, so there is no real root: the quadratic is irreducible over $\mathbb{R}$. Another reading, useful in an exam: $x^2+x+1=\left(x+\frac12\right)^2+\frac34$, the sum of a square and a strictly positive number, so it is always at least $\frac34$.

The principle is a count of unknowns. A denominator of degree $n$ requires exactly $n$ coefficients to determine, because the generic numerator of the original fraction has $n$ free coefficients. Here the denominator $(x-1)(x^2+x+1)$ has degree 3: three unknowns are needed. The linear factor brings one, $A$; the quadratic factor must therefore bring two, hence a numerator $Bx+C$. With $\frac{B}{x^2+x+1}$ there would be only two in total, and the system would in general have no solution.

Covering $(x-1)$ at $x=1$ gives $A=\frac{3+1}{1+1+1}=\frac43$. Next write $\frac{3x^2+1}{(x-1)(x^2+x+1)}=\frac{4}{3(x-1)}+\frac{Bx+C}{x^2+x+1}$ and choose simple values. At $x=0$: on the left $\frac{1}{(-1)(1)}=-1$, on the right $-\frac43+C$, so $C=\frac13$. At $x=-1$: on the left $\frac{4}{(-2)(1)}=-2$, on the right $-\frac23+(-B+C)$, so $-B=-2+\frac23-\frac13=-\frac53$ and $B=\frac53$. Conclusion: $\frac{4}{3(x-1)}+\frac{5x+1}{3(x^2+x+1)}$.

At $x=2$: on the left $\frac{3\times4+1}{(1)(4+2+1)}=\frac{13}{7}$. On the right $\frac{4}{3}+\frac{11}{3\times7}=\frac{4}{3}+\frac{11}{21}=\frac{28+11}{21}=\frac{39}{21}=\frac{13}{7}$. The two agree. Since the decomposition has three unknowns, a check at three distinct points would be a complete proof; in practice one well-chosen point is enough to catch the vast majority of sign errors.

$x^2+4$ has discriminant $-16<0$ and $x^2+x+1$ has discriminant $-3<0$: both factors are irreducible, neither breaks. The form is therefore $\frac{Ax+B}{x^2+4}+\frac{Cx+D}{x^2+x+1}$, with four unknowns, which is indeed the degree 4 of the denominator. Note that the cover-up method is useless here, for lack of a real root: you go through equating coefficients or convenient values.