In YÖS exams a typical problem gives you an identity such as $$\frac{2x-1}{x^2-3x-4}=\frac{A}{x-4}+\frac{B}{x+1}$$ and asks for the value of $A+B$, $A-2B$, $A\cdot B$ and so on. This technique is called partial fraction decomposition: a complicated rational fraction is written as a sum of simple fractions whose denominators are the factors of the original denominator. It is the reverse of adding fractions with a common denominator. To solve such problems you need two skills: (1) factoring the denominator into linear factors, and (2) finding the unknown constants $A$ and $B$. Both are covered step by step below.
If $x_1$ and $x_2$ are the roots of $ax^2+bx+c=0$, then $$ax^2+bx+c=a(x-x_1)(x-x_2).$$ The roots can be found by the discriminant: $x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$, or (for $a=1$) by Vieta's theorem: $$x_1+x_2=-b,\qquad x_1\cdot x_2=c.$$ Example: for $x^2-3x-4$ we need two numbers with sum $3$ and product $-4$; these are $4$ and $-1$, so $x^2-3x-4=(x-4)(x+1)$. A special case worth memorizing is the difference of squares: $$a^2-b^2=(a-b)(a+b),$$ for example $x^2-4=(x-2)(x+2)$.
Definition. If the denominator factors into *distinct* linear factors, the fraction decomposes into a sum of simple fractions — one for each factor, with an unknown constant in each numerator: $$\frac{px+q}{(x-x_1)(x-x_2)}=\frac{A}{x-x_1}+\frac{B}{x-x_2}.$$ This is an identity: it must hold for *every* admissible value of $x$, which is exactly what allows us to determine $A$ and $B$. Multiplying both sides by the denominator gives the key working equation: $$px+q=A(x-x_2)+B(x-x_1).$$ All methods for finding $A$ and $B$ start from this equation.
Since the equation $px+q=A(x-x_2)+B(x-x_1)$ holds for every $x$, we may substitute the *most convenient* values — the roots of the denominator. Substituting $x=x_1$ kills the term with $B$ (because $x_1-x_1=0$) and leaves a one-step equation for $A$; substituting $x=x_2$ likewise gives $B$ immediately. This is the fastest method and should be your default on the exam. It is also called the cover-up method: to find $A$, cover the factor $(x-x_1)$ in the original fraction and evaluate what remains at $x=x_1$: $$A=\frac{px_1+q}{x_1-x_2},\qquad B=\frac{px_2+q}{x_2-x_1}.$$ No system of equations is needed.
Expand the right-hand side of $px+q=A(x-x_2)+B(x-x_1)$ and collect powers of $x$: $$px+q=(A+B)\,x-(Ax_2+Bx_1).$$ Two polynomials are identically equal only if the coefficients of equal powers match. This gives a system: $$A+B=p,\qquad -(Ax_2+Bx_1)=q.$$ Solve it for $A$ and $B$. This method is slightly longer than substitution, but it is universal and works even when the question asks directly about a *combination* like $A+B$ — often the first equation of the system already contains the answer! Tip: use coefficient comparison as a quick check of the values found by Method 1.
Definition. If the denominator contains a *repeated* linear factor, the decomposition must include every power of that factor, from the first up to the multiplicity: $$\frac{px+q}{(x+a)^2}=\frac{A}{x+a}+\frac{B}{(x+a)^2}.$$ A common mistake is to write only $\frac{B}{(x+a)^2}$ — that is not enough! Multiplying by $(x+a)^2$ gives $$px+q=A(x+a)+B.$$ How to find the constants: substitute the root $x=-a$ to get $B=q-pa$ at once; then compare the coefficients of $x$ to get $A=p$. So for a repeated root the two methods are naturally *combined*: substitution gives $B$, coefficient comparison gives $A$.
If the denominator is, say, $2x^2-x-1$, first factor it. Its roots are $x=1$ and $x=-\frac{1}{2}$, so $2x^2-x-1=2(x-1)\left(x+\frac{1}{2}\right)=(x-1)(2x+1)$. In YÖS problems the given decomposition usually keeps the factor as written, e.g. $$\frac{2x+3}{2x^2-x-1}=\frac{A}{x-1}+\frac{B}{2x+1}.$$ The technique does not change: multiply by $(x-1)(2x+1)$ to get $2x+3=A(2x+1)+B(x-1)$, then substitute the roots $x=1$ and $x=-\frac{1}{2}$. Two warnings: (1) the root of $2x+1$ is $x=-\frac{1}{2}$, not $-1$; (2) the quick shortcut $A+B=p$ no longer holds, because comparing coefficients of $x$ now gives $2A+B=p$.
If both denominators on the right are monic ($x$ has coefficient $1$) and the factors are distinct, $$\frac{px+q}{(x-x_1)(x-x_2)}=\frac{A}{x-x_1}+\frac{B}{x-x_2}\ \Rightarrow\ A+B=p,$$ i.e. $A+B$ equals the coefficient of $x$ in the numerator — instantly, with no computation. Check: $\frac{2x-1}{(x-4)(x+1)}$ gives $A+B=2$. Caveats: (1) if a factor is $2x+1$ (not monic), the rule becomes $2A+B=p$; (2) for a repeated root $\frac{px+q}{(x+a)^2}=\frac{A}{x+a}+\frac{B}{(x+a)^2}$ the rule is $A=p$, and $A+B$ requires finding $B$ too; (3) if the question asks for $A-2B$ or $A\cdot B$, you must find both constants anyway. Use the shortcut to save time or to verify your answer.
Problem. Given $$\frac{2x-1}{x^2-3x-4}=\frac{A}{x-4}+\frac{B}{x+1},$$ find $A+B$.
Step 1. Factor the denominator. By Vieta: sum of roots $3$, product $-4$, so the roots are $4$ and $-1$: $x^2-3x-4=(x-4)(x+1)$.
Step 2. Multiply both sides by $(x-4)(x+1)$: $$2x-1=A(x+1)+B(x-4).$$
Step 3. Substitute the roots. At $x=4$: $\ 8-1=A\cdot 5$, so $A=\frac{7}{5}$. At $x=-1$: $\ -2-1=B\cdot(-5)$, so $B=\frac{3}{5}$.
Step 4. Answer. $A+B=\frac{7}{5}+\frac{3}{5}=2$.
Check via the shortcut: both factors are monic, so $A+B$ must equal the coefficient of $x$ in the numerator, which is $2$. It matches.
Problem. Given $$\frac{5x-2}{x^2-4}=\frac{A}{x-2}+\frac{B}{x+2},$$ find $A-2B$.
Step 1. Factor: difference of squares, $x^2-4=(x-2)(x+2)$.
Step 2. Working equation: $$5x-2=A(x+2)+B(x-2).$$
Step 3. Compare coefficients. Expand: $5x-2=(A+B)x+(2A-2B)$. Matching powers of $x$: $$A+B=5,\qquad 2A-2B=-2\ \Rightarrow\ A-B=-1.$$ Adding the equations: $2A=4$, so $A=2$ and $B=3$.
Step 4. Answer. $A-2B=2-6=-4$.
Check by substitution: at $x=2$: $10-2=4A$, so $A=2$; at $x=-2$: $-12=-4B$, so $B=3$. Same values.
Problem. Given $$\frac{x+3}{(x+2)^2}=\frac{A}{x+2}+\frac{B}{(x+2)^2},$$ find $A+B$.
Note the setup: because the root $x=-2$ is repeated, both powers of $(x+2)$ appear in the decomposition.
Step 1. Multiply both sides by $(x+2)^2$: $$x+3=A(x+2)+B.$$
Step 2. Substitute the root $x=-2$: $\ -2+3=A\cdot 0+B$, so $B=1$.
Step 3. Compare coefficients of $x$: on the left the coefficient is $1$, on the right it is $A$, so $A=1$. (Alternatively substitute any other value, e.g. $x=0$: $3=2A+B=2A+1$, again $A=1$.)
Step 4. Answer. $A+B=1+1=2$.
Note that here $A$ (not $A+B$) equals the coefficient of $x$ in the numerator — the monic-shortcut for distinct factors does not apply to repeated roots.
Problem. Given $$\frac{2x+3}{2x^2-x-1}=\frac{A}{x-1}+\frac{B}{2x+1},$$ find $A$ and $B$.
Step 1. Factor the denominator. Roots of $2x^2-x-1=0$: discriminant $D=1+8=9$, so $x=\frac{1\pm 3}{4}$, i.e. $x=1$ and $x=-\frac{1}{2}$. Hence $2x^2-x-1=(x-1)(2x+1)$ (expand to verify).
Step 2. Working equation: $$2x+3=A(2x+1)+B(x-1).$$
Step 3. Substitute the roots. At $x=1$: $\ 5=3A$, so $A=\frac{5}{3}$. At $x=-\frac{1}{2}$: $\ 2=B\left(-\frac{3}{2}\right)$, so $B=-\frac{4}{3}$.
Step 4. Check by coefficients: $2A+B=\frac{10}{3}-\frac{4}{3}=2$ (coefficient of $x$) and $A-B=\frac{5}{3}+\frac{4}{3}=3$ (constant term). Both match. Note that here $2A+B=2$, not $A+B$ — the non-monic factor changes the shortcut.
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