Find the remainder when $4x^3 - 2x^2 + 3x - 5$ is divided by $2x + 1$.

The remainder is $-\frac{15}{2}$.

$P(x) = x^3 + ax^2 + bx + 6$ has $x + 2$ as a factor and leaves a remainder of $12$ when divided by $x - 1$. Find $a$ and $b$, then factor $P(x)$ fully.

$a = 2$, $b = 3$, and $P(x) = (x + 2)(x^2 + 3)$.

Grade 12 · MHF4U · Advanced Functions

MHF4U: Remainder and Factor Theorem

Dividing $P(x)$ by $x - a$ leaves a remainder, and the remainder theorem says you can get it by substitution: it is $P(a)$ . The factor theorem is the special case where that remainder is zero.

How It Works

Remainder theorem: when $P(x)$ is divided by $x - a$ , the remainder is $P(a)$ . It comes straight from the division statement $P(x) = (x - a)Q(x) + R$ : substitute $x = a$ and the first term dies, leaving $P(a) = R$ . For a divisor $ax - b$ , substitute the zero of the divisor, $x = \frac{b}{a}$ .

Factor theorem: $x - a$ is a factor of $P(x)$ exactly when $P(a) = 0$ . Both directions matter. If $P(a) = 0$ then $x - a$ divides evenly, and if $x - a$ is a factor then $P(a)$ must be $0$ .

The theorems do three jobs on a test. Finding a remainder without dividing: substitute. Testing a factor: substitute and look for zero. Finding an unknown coefficient: substitute, set the result equal to the given remainder (or to $0$ for a factor), and solve.

Finding a factor to start with is the rational zero search. For integer coefficients, any rational zero is $\pm\frac{p}{q}$ with $p$ a factor of the constant term and $q$ a factor of the leading coefficient. Try the small ones first, and once $P(a) = 0$ , divide by $x - a$ to get a lower-degree quotient.

Two unknowns need two facts, which give two equations. Solve them together. And once you find a factor, keep going until the polynomial is fully factored, which is usually what the marks are for.

5 Worked Examples

Easy first, test level last. Try each one on paper, then open the solution.

Easy
Find the remainder when $x^3 - 2x^2 + 5x - 1$ is divided by $x - 2$ .
Show solution
1. The divisor is zero at $x = 2$ , so the remainder is $P(2)$ .
2. $P(2) = 8 - 8 + 10 - 1 = 9$ .
Answer: The remainder is $9$ .
Easy
Is $x + 3$ a factor of $P(x) = x^3 + 2x^2 - 5x - 6$ ? If so, factor $P(x)$ fully.
Show solution
1. $x + 3 = x - (-3)$ , so test $P(-3) = -27 + 18 + 15 - 6 = 0$ . Yes, it is a factor.
2. Divide by $x + 3$ (synthetic division with $-3$ ): the quotient is $x^2 - x - 2$ .
3. Factor the quotient: $x^2 - x - 2 = (x - 2)(x + 1)$ .
Answer: $P(x) = (x + 3)(x + 1)(x - 2)$
Medium
Find $k$ so that $x - 1$ is a factor of $2x^3 + kx^2 - 7x + 2$ .
Show solution
1. By the factor theorem, $P(1) = 0$ .
2. $P(1) = 2 + k - 7 + 2 = k - 3$ .
3. $k - 3 = 0$ gives $k = 3$ .
Answer: $k = 3$
Medium
Find the remainder when $4x^3 - 2x^2 + 3x - 5$ is divided by $2x + 1$ .
Show solution
1. The divisor is zero at $x = -\frac{1}{2}$ , so the remainder is $P(-\frac{1}{2})$ .
2. $P(-\frac{1}{2}) = 4(-\frac{1}{8}) - 2(\frac{1}{4}) + 3(-\frac{1}{2}) - 5 = -\frac{1}{2} - \frac{1}{2} - \frac{3}{2} - 5$ .
Answer: The remainder is $-\frac{15}{2}$ .
Test level
$P(x) = x^3 + ax^2 + bx + 6$ has $x + 2$ as a factor and leaves a remainder of $12$ when divided by $x - 1$ . Find $a$ and $b$ , then factor $P(x)$ fully.
Show solution
1. Factor: $P(-2) = -8 + 4a - 2b + 6 = 0$ , so $4a - 2b = 2$ , or $2a - b = 1$ .
2. Remainder: $P(1) = 1 + a + b + 6 = 12$ , so $a + b = 5$ .
3. Add the two equations: $3a = 6$ , so $a = 2$ and $b = 3$ .
4. $P(x) = x^3 + 2x^2 + 3x + 6$ . Group: $x^2(x + 2) + 3(x + 2) = (x + 2)(x^2 + 3)$ . The quadratic has no real zeros, so stop.
Answer: $a = 2$ , $b = 3$ , and $P(x) = (x + 2)(x^2 + 3)$ .

Learn It on Martian Lab

Martian Lab teaches this in the Factor and Remainder Theorem lesson of its MHF4U Polynomial Equations unit. Watch it, then take the Polynomial Equations practice tests to see if it stuck.

Watch the Factor and Remainder Theorem lesson free

Keep Going

Free Polynomial Equations practice test, with solutions

More Polynomial Equations topics: Long and Synthetic Division · Solving Polynomial Equations · Polynomial Inequalities · Sum and Difference of Cubes

More MHF4U topics: Vertical and Horizontal Asymptotes · Proving Trig Identities · Solving Trig Equations · Laws of Logarithms · Solving Logarithmic Equations · End Behaviour and Degree · Finding the Equation of a Polynomial · Even and Odd Functions · Holes in Rational Functions · Graphing Rational Functions · Rational Equations and Inequalities · Radian Measure · Unit Circle and Special Angles · Compound Angle Formulas · Double Angle Formulas · Graphing Sine and Cosine with Transformations · Graphs of Tan, Sec, Csc and Cot · Sinusoidal Word Problems · Solving Exponential Equations · Graphing Logarithmic Functions · Exponential and Log Word Problems · Sums, Products and Quotients of Functions · Composition of Functions · Average vs Instantaneous Rate of Change

Every MHF4U unit and a free sample test

Start Practicing

No credit card required to start.

Get started