Polynomials and Partial Fractions

Secondary 3 and 4 Additional Mathematics · Study guide

Use division, remainder and factor theorems to solve cubic equations, then decompose proper and improper fractions with all denominator restrictions.

Concept → worked examples → practice29 min read · practice linked below

One division identity connects the chapter

A polynomial has non-negative integer powers of $x$ . Its degree is its highest power with a nonzero coefficient. Division writes a polynomial as

P(x)=D(x)Q(x)+R(x),\qquad \deg R<\deg D,

where $D$ is the nonzero divisor, $Q$ the quotient and $R$ the remainder; the remainder may be zero. The degree rule is what tells you whether to use a constant or a linear remainder. An identity holds for every $x$ , so matching coefficients is valid; an equation may hold only at particular roots.

A linear divisor leaves a constant remainder. Substituting its root removes the whole quotient term.

Division, remainders and unknown coefficients

In long division, divide the leading terms to obtain the next quotient term, multiply the whole divisor by it, and subtract. Repeat until the remaining degree is smaller than the divisor’s. Insert zero coefficients for missing powers so the columns stay aligned.

The remainder on division by $x-a$ is $P(a)$ . For $mx+n$ , with $m\ne0$ , substitute its root $-n/m$ , not $-n$ . The factor theorem says $x-a$ is a factor exactly when $P(a)=0$ .

Example 1: a non-monic divisor

Divide $2x^3+3x^2-x+5$ by $2x-1$ , and check the remainder.

The leading quotient term is $x^2$ . Subtract $2x^3-x^2$ , leaving $4x^2-x+5$ .
The next term is $2x$ . Subtract $4x^2-2x$ , leaving $x+5$ .
The final term is $1/2$ . Subtract $x-1/2$ , leaving $11/2$ .

2x^3+3x^2-x+5=(2x-1)\left(x^2+2x+\frac12\right)+\frac{11}2.

Thus $Q=x^2+2x+1/2$ , $R=11/2$ . The remainder theorem checks this at the divisor’s root: $P(1/2)=1/4+3/4-1/2+5=11/2$ .

Example 2: use two conditions, then solve a cubic

$P(x)=x^3+ax^2+bx+6$ has factor $x-1$ and remainder 12 on division by $x+1$ . Find $a,b$ , then solve $P(x)=0$ .

P(1)=0\Rightarrow a+b=-7,\qquad P(-1)=12\Rightarrow a-b=7.

Adding gives $a=0$ , then $b=-7$ . Division by the known factor gives

P(x)=x^3-7x+6=(x-1)(x^2+x-6)=(x-1)(x+3)(x-2).

Hence $x=1,-3,2$ . Expanding the factors recovers $x^3-7x+6$ . A factor is an expression such as $x-1$ ; a root is the value 1.

Example 3: a quadratic divisor needs a linear remainder

A polynomial has $P(2)=7$ and $P(-1)=-2$ . Find the remainder when it is divided by $(x-2)(x+1)$ .

Write $P(x)=(x-2)(x+1)Q(x)+Ax+B$ . Substituting the two divisor roots gives

2A+B=7,\qquad -A+B=-2.

Subtracting yields $3A=9$ , so $A=3,B=1$ . The remainder is $3x+1$ . It takes the same values as $P$ at both roots, as required. A constant remainder could not fit these two different values. If the divisor has no real roots, use division or compare coefficients instead of inventing real substitution values.

Cubic factors and repeated factors

Find one root by a given condition or a sensible trial, use the factor theorem, divide to obtain a quadratic, and solve that quadratic. For integer coefficients, rational-root trials $p/q$ may be taken with $p$ dividing the constant term and $q$ dividing the leading coefficient. This gives candidates, not guaranteed roots.

a^3+b^3=(a+b)(a^2-ab+b^2),\qquad a^3-b^3=(a-b)(a^2+ab+b^2).

These identities follow by expansion; the cross terms cancel. A real cubic can have one, two or three distinct real roots, because roots may repeat or the remaining quadratic may have no real roots.

Example 4: a cubic need not have three real roots

Solve $x^3+8=0$ over the real numbers.

x^3+8=(x+2)(x^2-2x+4).

The quadratic factor is $(x-1)^2+3>0$ for every real $x$ ; equivalently its discriminant is $-12<0$ . It cannot equal zero. Therefore $x=-2$ is the only real solution. Checking gives $(-2)^3+8=0$ .

Example 5: a repeated factor needs more than one root check

Find $a,b$ if $x^3+ax^2+bx+2$ has factor $(x-1)^2$ .

The quotient must be a monic linear polynomial, $x+c$ . Comparing constant terms gives $c=2$ . Expand the identity:

(x-1)^2(x+2)=(x^2-2x+1)(x+2)=x^3-3x+2.

Thus $a=0,b=-3$ . Merely using $P(1)=0$ gives only $a+b=-3$ , which does not prove the factor repeats. The displayed factorisation does prove it.

Explore: the remainder is a graph height

For $P(x)=x^3-x^2-4x+4$ , divide by $x-a$ . The point on the graph at $x=a$ has height $P(a)$ , exactly the remainder. Move the point onto the x-axis to find a factor.

Divisor root a:1

A zero remainder means the division has no leftover term. A nonzero height means the proposed divisor is not a factor.

a = 1. Quotient: x² − 4. Remainder: 0. P(x) = (x − 1)(x² − 4). Therefore x − 1 is a factor.

Partial fractions: choose the complete form

Partial fractions reverse the process of combining fractions. First factorise the denominator and retain its exclusions. If the numerator’s degree is at least the denominator’s degree, divide first; decompose only the proper fraction left over.

Numerators and denominator factors
Denominator type	Required form
Distinct linear factors $(x-a)(x-b)$	$\dfrac A{x-a}+\dfrac B{x-b}$
Repeated linear factors $(x-a)^2(x-b)$	$\dfrac A{x-a}+\dfrac B{(x-a)^2}+\dfrac C{x-b}$
Linear factor and positive quadratic $(x-a)(x^2+c^2)$ , $c>0$	$\dfrac A{x-a}+\dfrac{Bx+C}{x^2+c^2}$

Each numerator has degree smaller than its associated factor. Repeated factors require every power up to the repeated power. A general quadratic numerator is linear, so using just a constant may make a valid decomposition impossible.

Multiply through by the common denominator to obtain a polynomial identity. That identity can be evaluated even at a former denominator root; the original rational expression still excludes that input. Use convenient substitutions to find constants, then compare coefficients for any remaining constants.

Example 6: two distinct linear factors

Decompose $\dfrac{2x+3}{(x-1)(x+2)}$ , where $x\ne1,-2$ .

\frac{2x+3}{(x-1)(x+2)}=\frac A{x-1}+\frac B{x+2}\quad\Longrightarrow\quad 2x+3=A(x+2)+B(x-1).

Set $x=1$ : $5=3A$ , so $A=5/3$ . Set $x=-2$ : $-1=-3B$ , so $B=1/3$ . Therefore

\boxed{\frac5{3(x-1)}+\frac1{3(x+2)}},\qquad x\ne1,-2.

Recombining gives numerator $[5(x+2)+(x-1)]/3=2x+3$ .

Example 7: a repeated linear factor

Decompose $\dfrac{x^2+7x+1}{(x-1)^2(x+2)}$ , where $x\ne1,-2$ .

\frac{x^2+7x+1}{(x-1)^2(x+2)}=\frac A{x-1}+\frac B{(x-1)^2}+\frac C{x+2}.

x^2+7x+1=A(x-1)(x+2)+B(x+2)+C(x-1)^2.

At $x=1$ , $9=3B$ , so $B=3$ . At $x=-2$ , $-9=9C$ , so $C=-1$ . Comparing coefficients of $x^2$ , $A+C=1$ , giving $A=2$ . Hence

\boxed{\frac2{x-1}+\frac3{(x-1)^2}-\frac1{x+2}},\qquad x\ne1,-2.

The recombined numerator is $2(x^2+x-2)+3x+6-(x^2-2x+1)=x^2+7x+1$ .

Example 8: a quadratic factor that stays whole

Decompose $\dfrac{3x^2+x-2}{(x-1)(x^2+1)}$ for real $x\ne1$ .

Since $x^2+1>0$ , it has no real linear factors. Write

\frac A{x-1}+\frac{Bx+C}{x^2+1}.

The cleared identity is $3x^2+x-2=A(x^2+1)+(Bx+C)(x-1)$ . At $x=1$ , $2=2A$ , so $A=1$ . Comparing $x^2$ coefficients gives $A+B=3$ , hence $B=2$ . Constants give $A-C=-2$ , hence $C=3$ . The x-coefficient also checks: $C-B=1$ .

\boxed{\frac1{x-1}+\frac{2x+3}{x^2+1}},\qquad x\ne1.

Example 9: divide an improper fraction first

Decompose $\dfrac{x^2+1}{(x-1)(x+2)}$ , where $x\ne1,-2$ .

The numerator and denominator both have degree 2. Since $(x-1)(x+2)=x^2+x-2$ , division gives

\frac{x^2+1}{x^2+x-2}=1+\frac{-x+3}{(x-1)(x+2)}.

For the proper part, $-x+3=A(x+2)+B(x-1)$ . At $x=1$ , $A=2/3$ ; at $x=-2$ , $B=-5/3$ . Therefore

\boxed{1+\frac2{3(x-1)}-\frac5{3(x+2)}},\qquad x\ne1,-2.

The polynomial part 1 is essential: the original fraction approaches 1 for large $|x|$ , whereas a sum of these proper fractions alone approaches zero.

Common mistakes and checks

Using the wrong substitution in a non-monic divisor. Solve $mx+n=0$ before evaluating $P$ .
Subtracting only the leading term in division. Subtract every term of the divisor–quotient product with brackets.
Assuming a quadratic divisor has a constant remainder. Allow $Ax+B$ , even if the answer later has $A=0$ .
Calling one zero value proof of a repeated factor. Divide again or establish an identity with the squared factor.
Missing the first power of a repeated denominator factor. Include both $A/(x-a)$ and $B/(x-a)^2$ .
Using a constant over a quadratic factor. Begin with $Bx+C$ .
Losing original exclusions after cancellation. The decomposition has the same permitted inputs as the original fraction.
Skipping a final identity check. Multiply the result by the common denominator and expand.

Practise this chapter

This guide supports both Secondary 3 learning and Secondary 4 revision. Choose the question bank for your level: Secondary 3: polynomials and partial fractions or Secondary 4: polynomials and partial fractions. Attempt a question before reading its worked solution.