P2 2.1 - Factor and remainder theorems | Edexcel A-Level International Mathematics

Find a remainder by substitution, recognise linear factors, factorise cubics and determine unknown coefficients. These theorems come from the division identity, so the value you substitute and the equation you form both have a reason.

Remainders at roots of a linear divisor

Can you find a remainder without doing the division? If $f(x)=(x-2)q(x)+R$ , putting $x=2$ makes the whole product disappear, leaving $f(2)=R$ .

For instance, $f(x)=x^3-2x+4$ gives $f(2)=8-4+4=8$ . That single evaluation finds the remainder on division by $x-2$ , although it does not find the quotient.

Remainder theorem

For division of a polynomial $f(x)$ by $ax+b$ , where $a\ne0$ , the remainder is

R=f\left(-\frac ba\right).

This follows by substituting the zero of the divisor into $f(x)=(ax+b)q(x)+R$ .

Not in the formula booklet: learn it.

Solve “divisor equals zero” before substituting. This handles both signs and non-monic divisors.

Worked example 1: a whole-number input

Question. Find the remainder when $f(x)=x^3-2x+4$ is divided by $x-2$ .

Plan. Set $x-2=0$ , then evaluate $f(2)$ .

R=f(2)=2^3-2(2)+4=8.

Answer. The remainder is $8$ .

Check. Independent division gives $f(x)=(x-2)(x^2+2x+2)+8$ , whose expansion is $x^3-2x+4$ .

Change only the divisor, to $2x-1$ . Its zero is now a fraction; the theorem does not require an integer input.

Worked example 2: a non-monic divisor

Question. Find the remainder when the same $f(x)=x^3-2x+4$ is divided by $2x-1$ .

Plan. Solve $2x-1=0$ to obtain $x=1/2$ , then evaluate exactly.

R=f\left(\frac12\right)=\frac18-1+4=\frac{25}{8}.

Answer. The remainder is $25/8$ .

Check. Division gives quotient $\frac12x^2+\frac14x-\frac78$ . Expanding its product with $2x-1$ gives $x^3-2x+\frac78$ ; adding $25/8$ restores the constant $4$ .

The coefficient $2$ changes the zero of $2x-1$ ; it is not a reason to divide the resulting remainder by $2$ .

Watch out

For division by $2x-1$ , use $f(1/2)$ , not $f(1)$ or $f(-1/2)$ . For division by $x+3$ , use $f(-3)$ .

Find the divisor's zero first in this next calculation.

The factor theorem

A linear divisor is a factor exactly when it leaves remainder zero.

If $f(c)=0$ , division by $x-c$ gives $f(x)=(x-c)q(x)$ . Conversely, a factor $x-c$ makes $f(c)=0$ . This gives a test in either direction.

Factor theorem

For a polynomial $f(x)$ ,

f(c)=0\quad\Longleftrightarrow\quad x-c\text{ is a factor of }f(x).

More generally, $ax+b$ is a factor exactly when $f(-b/a)=0$ , where $a\ne0$ .

Not in the formula booklet: learn it.

A root of $f(x)=0$ is an input giving output zero. The root is a number; the associated factor is a linear expression.

Worked example 3: testing a monic factor

Question. Show that $x-1$ is a factor of $f(x)=2x^3-3x^2-5x+6$ .

Plan. Evaluate at the zero of $x-1$ , then use the factor theorem.

f(1)=2-3-5+6=0.

Answer. Since $f(1)=0$ , $x-1$ is a factor.

Check. Multiplication verifies $f(x)=(x-1)(2x^2-x-6)$ .

Keep the polynomial and change the proposed factor to $2x+3$ . Now its zero is negative and fractional.

Worked example 4: testing a non-monic factor

Question. Show that $2x+3$ is a factor of $f(x)=2x^3-3x^2-5x+6$ .

Plan. Solve $2x+3=0$ , then substitute $x=-3/2$ with brackets around negative inputs.

\begin{aligned} f(-3/2)&=2(-3/2)^3-3(-3/2)^2-5(-3/2)+6\\ &=-\frac{27}{4}-\frac{27}{4}+\frac{30}{4}+\frac{24}{4}=0 \end{aligned}

Answer. Since $f(-3/2)=0$ , $2x+3$ is a factor.

Check. Expanding $(2x+3)(x^2-3x+2)$ gives $2x^3-3x^2-5x+6$ .

Show the zero value explicitly before drawing the factor conclusion.

Exam tip

“ $f(c)=0$ , therefore $x-c$ is a factor” contains both the calculation and its meaning. Merely listing $f(c)$ without showing that it equals zero leaves the argument unfinished.

Try a new polynomial and a fractional root.

Factorising cubics

One known linear factor reduces a cubic factorisation to the quadratic factorisation you already know.

If no factor is given and the coefficients are integers, small positive and negative integer divisors of the constant are useful candidates for integer roots. For a non-monic polynomial, a rational root may be needed. A candidate becomes a known root only after evaluation gives zero.

You already know

Division by a linear factor gives a quadratic quotient with remainder zero. Factorise that quadratic, then include the original linear factor in the final product.

Do not stop after finding the quadratic: “fully factorise” means continue wherever real linear factors can be obtained by the taught algebra.

Worked example 5: reducing to a quadratic

Question. Given that $x-1$ is a factor, fully factorise $x^3-3x^2-x+3$ using algebra.

Plan. Divide by $x-1$ , then factorise the quadratic quotient.

\begin{aligned} x^3-3x^2-x+3&=(x-1)(x^2-2x-3) &&\text{division}\\ &=(x-1)(x-3)(x+1) &&\text{product }-3,\ \text{sum }-2 \end{aligned}

Answer. $(x-1)(x-3)(x+1)$ .

Check. Expanding the last two factors gives $x^2-2x-3$ ; multiplying by $x-1$ restores all four original coefficients.

Change only the known linear factor from $x-1$ to $2x-1$ , leaving the quadratic quotient unchanged. The new cubic therefore has a different leading coefficient.

Worked example 6: retaining a non-monic factor

Question. Given that $2x-1$ is a factor, fully factorise $2x^3-5x^2-4x+3$ using algebra.

Plan. Divide by the stated factor, keeping its leading coefficient $2$ .

\begin{aligned} 2x^3-5x^2-4x+3&=(2x-1)(x^2-2x-3)\\ &=(2x-1)(x-3)(x+1) \end{aligned}

The successive quotient terms are $x^2,-2x,-3$ ; subtracting their products leaves zero.

Answer. $(2x-1)(x-3)(x+1)$ .

Check. At $x=0$ , the product is $(-1)(-3)(1)=3$ . Its leading term is $2x^3$ , and full expansion gives $2x^3-5x^2-4x+3$ .

A list of calculator roots alone does not show algebraic factorisation when the question requires algebra.

Watch out

The root $x=1/2$ corresponds to $2x-1$ , but replacing $2x-1$ by $x-1/2$ without an extra factor $2$ halves the polynomial. Check the leading coefficient as well as the roots.

Use the given factor to choose the division in this new cubic.

Finding unknown coefficients

Factor and remainder information become equations in unknown coefficients.

Write the condition before simplifying: a factor gives $f(c)=0$ ; a stated remainder $R$ gives $f(c)=R$ . Two independent conditions can determine two unknowns.

Worked example 7: one factor and one remainder

Question. Let $f(x)=x^3+ax^2-4x+b$ . Given that $x-2$ is a factor and division by $x+1$ leaves remainder $-3$ , find $a,b$ .

Plan. Form equations from $f(2)=0$ and $f(-1)=-3$ , then eliminate one unknown.

\begin{aligned} f(2)=0&\Rightarrow8+4a-8+b=0\\ &\Rightarrow4a+b=0\\ f(-1)=-3&\Rightarrow-1+a+4+b=-3\\ &\Rightarrow a+b=-6 \end{aligned}

Subtracting the second equation from the first gives $3a=6$ . Thus $a=2$ and $b=-8$ .

Answer. $a=2, b=-8$ .

Check. Independently, $f(x)=(x-2)(x+2)^2$ , so $f(2)=0$ , while $f(-1)=(-3)(1)=-3$ .

Change just the stated remainder from $-3$ to $12$ . The factor condition stays the same, but the second equation changes.

Worked example 8: changing the remainder condition

Question. Let $f(x)=x^3+ax^2-4x+b$ . Given that $x-2$ is a factor and division by $x+1$ leaves remainder $12$ , find $a,b$ .

Plan. Keep $4a+b=0$ , then use $f(-1)=12$ .

\begin{aligned} -1+a+4+b=12&\Rightarrow a+b=9\\ (4a+b)-(a+b)=0-9&\Rightarrow3a=-9\\ a&=-3,\qquad b=12 \end{aligned}

Answer. $a=-3, b=12$ .

Check. Direct substitution gives $f(2)=8-12-8+12=0$ and $f(-1)=-1-3+4+12=12$ , satisfying both original conditions.

If a divisor is $2x-1$ , the corresponding input would be $1/2$ ; the same equation-building method applies.

Watch out

A remainder is not automatically zero. Setting $f(-1)=0$ in these examples would replace the given information by a different question. Write the stated remainder on the right-hand side first.

Combine the two conditions for this fresh polynomial.

Quick Recap

Find a remainder by evaluating at the zero of the linear divisor.
A zero remainder identifies a factor.
Divide out a known factor, then factorise the quadratic quotient.
Translate each piece of factor or remainder information into its own equation.

The opening shortcut works because the divisor's zero makes the quotient term disappear.

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P2 2.1 - Factor and remainder theorems

Remainders at roots of a linear divisor

The factor theorem

Factorising cubics

Finding unknown coefficients

Quick Recap

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