3B - Remainders and factors

Can one substitution replace a whole polynomial division? You will use the remainder theorem to find what is left over, recognise linear factors and turn factor or remainder information into equations for unknown coefficients. The link comes directly from the division identity.

Finding a remainder by substitution

How could you find the remainder when $x^3-4x^2+x+8$ is divided by $x-3$ , without finding the quotient?

Call the polynomial $f(x)$ . The division identity says

f(x)=(x-3)q(x)+r.

At $x=3$ , the entire quotient term disappears: $f(3)=0q(3)+r=r$ . The value that makes the divisor zero gives the remainder directly.

Remainder theorem

For a polynomial $f(x)$ , the remainder on division by $x-a$ is $f(a)$ .

More generally, on division by $ax+b$ , where $a\ne0$ , the remainder is

r=f\!\left(-\frac{b}{a}\right).

Not on the formulae sheet: learn it.

Always solve “divisor equals zero” first. This handles both signs and non-monic divisors without a separate rule to memorise.

Worked example 1: a monic divisor

Question. Find the remainder when $f(x)=x^3-4x^2+x+8$ is divided by $x-3$ .

Plan. Set $x-3=0$ , then evaluate $f(3)$ .

\begin{aligned} f(3)&=3^3-4(3^2)+3+8 &&\text{substitute into every term}\\ &=27-36+3+8\\ &=2 \end{aligned}

Answer. The remainder is $2$ .

Check. Independent division gives quotient $x^2-x-2$ . Reconstruction confirms

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

Keep the same polynomial, but change the divisor to $2x-3$ . Its zero changes from $3$ to $3/2$ ; you must change the substitution accordingly.

Worked example 2: a non-monic divisor

Question. Find the remainder when $f(x)=x^3-4x^2+x+8$ is divided by $2x-3$ .

Plan. Solve $2x-3=0$ , then substitute $x=3/2$ exactly.

\begin{aligned} f\!\left(\frac32\right) &=\frac{27}{8}-4\left(\frac94\right)+\frac32+8\\ &=\frac{27-72+12+64}{8} &&\text{use a common denominator}\\ &=\frac{31}{8} \end{aligned}

Answer. The remainder is $31/8$ .

Check. Dividing instead gives

f(x)=(2x-3)\left(\frac{x^2}{2}-\frac{5x}{4}-\frac{11}{8}\right)+\frac{31}{8}.

Expanding gives $x^3-4x^2+x+33/8+31/8=f(x)$ , confirming the remainder.

The non-monic divisor does not require dividing the remainder by its leading coefficient.

Watch out

For $2x-3$ , substitute $3/2$ , not $3$ , and report $f(3/2)$ , not $f(3/2)/2$ . The divisor's whole product is already zero at $3/2$ .

Use that same decision for a divisor with a plus sign.

Recognising factors

A zero remainder tells you that a linear divisor is a factor of the polynomial.

If $f(x)=(x-a)q(x)+r$ , substitution gives $f(a)=r$ . Thus $f(a)=0$ makes the division exact. Conversely, if $x-a$ is a factor, substituting $a$ into that product gives zero.

Factor theorem

For a polynomial $f(x)$ ,

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

The double arrow means that the statement works in both directions. The number $a$ is a root of $f(x)=0$ ; the expression $x-a$ is a factor of $f(x)$ .

Not on the formulae sheet: learn it.

For a non-monic linear expression, apply the same reasoning at its zero. For instance, $2x-1$ is a factor precisely when $f(1/2)=0$ .

Worked example 3: showing a factor

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

Plan. Test $x=-1$ , because $-1+1=0$ .

\begin{aligned} f(-1)&=(-1)^3+2(-1)^2-5(-1)-6\\ &=-1+2+5-6\\ &=0 \end{aligned}

Answer. Since $f(-1)=0$ , the factor theorem shows that $x+1$ is a factor.

Check. Multiplication gives $(x+1)(x^2+x-6)=x^3+2x^2-5x-6$ , so the division really is exact.

Change only the proposed factor's sign. Testing $x-1$ now requires $1$ , and the same polynomial need not give zero.

Worked example 4: rejecting a proposed factor

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

Plan. Evaluate $f(1)$ and compare the result with zero.

f(1)=1+2-5-6=-8.

Answer. $x-1$ is not a factor, because $f(1)=-8\ne0$ .

Check. Division gives

f(x)=(x-1)(x^2+3x-2)-8.

Expanding confirms the nonzero remainder $-8$ .

A “show that” answer needs the substitution and its result, followed by the factor conclusion. Just naming a root does not demonstrate the calculation.

Watch out

The root $-1$ corresponds to the factor $x+1$ . Writing “ $-1$ is a factor” confuses a number solving the equation with a polynomial dividing the expression.

Check the two proposed factors separately.

Finding unknown coefficients

Factor and remainder information can give equations for coefficients whose values you do not yet know.

Write the required substitution before doing any algebra. “ $x-a$ is a factor” gives $f(a)=0$ ; “remainder $r$ on division by $x-a$ ” gives $f(a)=r$ . Two unknown coefficients usually need two independent conditions.

Worked example 5: one unknown coefficient

Question. Given that $x-2$ is a factor of $f(x)=x^3+kx^2-4x+2$ , find $k$ .

Plan. Set $f(2)=0$ and solve the resulting linear equation.

\begin{aligned} 2^3+k(2^2)-4(2)+2&=0 &&\text{factor condition}\\ 8+4k-8+2&=0\\ 4k+2&=0\\ k&=-\frac12 \end{aligned}

Answer. $k=-1/2$ .

Check. With this value, division gives

x^3-\frac12x^2-4x+2=(x-2)\left(x^2+\frac32x-1\right),

which expands correctly and has zero remainder.

Now allow two coefficients to be unknown. A second, different remainder condition supplies the second equation needed to determine them.

Worked example 6: two conditions

Question. The polynomial $f(x)=x^3+px^2+qx+6$ has factor $x-1$ . Its remainder on division by $x+1$ is $8$ . Find $p$ and $q$ .

Plan. Translate the two conditions into equations, then eliminate one unknown.

\begin{aligned} f(1)=0&\quad\Rightarrow\quad1+p+q+6=0\\ &\quad\Rightarrow\quad p+q=-7\\[2pt] f(-1)=8&\quad\Rightarrow\quad-1+p-q+6=8\\ &\quad\Rightarrow\quad p-q=3 \end{aligned}

Add the equations to cancel $q$ :

2p=-4\quad\Rightarrow\quad p=-2.

Then $-2+q=-7$ , so $q=-5$ .

Answer. $p=-2$ and $q=-5$ .

Check. The completed polynomial is $x^3-2x^2-5x+6$ . Direct substitution gives $f(1)=1-2-5+6=0$ and $f(-1)=-1-2+5+6=8$ . Both original conditions hold.

This answers the opening question: substitution finds a remainder without the quotient, and can also reveal information about the polynomial itself.

Problem-solving

Label each equation by its condition before solving. A factor condition has right-hand side zero; a stated remainder condition has that remainder on the right-hand side.

Use both conditions, then check them in the completed polynomial.

Quick Recap

Make the linear divisor zero, then substitute that value into the polynomial.
The resulting value is the remainder; zero means the divisor is a factor.
A root $a$ corresponds to a factor $x-a$ .
Use independent factor or remainder conditions to find unknown coefficients.

For the opening polynomial divided by $x-3$ , the single calculation $f(3)=2$ gives the remainder.

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3B - Remainders and factors

Finding a remainder by substitution

Recognising factors

Finding unknown coefficients

Quick Recap

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