2.1, 3.1 - Factor and remainder theorems

Remainder theorem

When a polynomial $f(x)$ is divided by $x-a$, the remainder is $f(a)$. More generally, division by $px+q$, where $p\ne0$, leaves remainder

Not in the formula booklet: learn it.

Factor theorem

For a polynomial $f$,

Equivalently, $px+q$ is a factor exactly when $f(-q/p)=0$, for $p\ne0$. Not in the formula booklet: learn it.

Worked example 1: a monic linear divisor

Question. Find the remainder when $f(x)=2x^3+3x^2-5x+7$ is divided by $x-2$.

Plan. Set the divisor to zero and evaluate the polynomial there.

Answer. Remainder $25$.

Check. Division gives $f(x)=(x-2)(2x^2+7x+9)+25$; expansion recovers every coefficient.

Worked example 2: a non-monic linear divisor

Question. Find the remainder when the same $f(x)$ is divided by $2x-4$.

Plan. Solve $2x-4=0$ before substituting; do not use the constant term alone.

Answer. Remainder $25$, answering the opening question.

Check. $f(x)=(2x-4)(x^2+\tfrac72x+\tfrac92)+25$. The divisor doubled and the quotient halved, so their product is unchanged.

Watch out

For $3x+2$, substitute $x=-2/3$, not $-2$. Always solve “divisor = zero”; no extra factor of $3$ multiplies the remainder.

Worked example 3: one unknown coefficient

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

Plan. Substitute the factor's zero and equate the result to zero.

Answer. $a=-2$.

Check. With this value, $f(1)=1-2-5+6=0$, and division gives $(x-1)(x^2-x-6)$.

Worked example 4: two unknown coefficients

Question. The polynomial $f(x)=x^3+ax^2+bx+6$ has factor $x-1$ and leaves remainder $12$ when divided by $x+2$. Find $a$ and $b$.

Plan. Use the two divisor zeros and their stated remainders to form simultaneous linear equations, then eliminate one unknown.

Adding the equations gives $3a=0$, hence $a=0$. Substituting into $a+b=-7$ gives $b=-7$.

Answer. $a=0$, $b=-7$.

Check. The polynomial is $x^3-7x+6$. Its values are $f(1)=1-7+6=0$ and $f(-2)=-8+14+6=12$, confirming both stated conditions.