Training Quadratic and Polynomial Functions Remainder and Factor Theorems
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Remainder and Factor Theorems

22 min Quadratic and Polynomial Functions

Dividing Polynomials

The Remainder Theorem and Factor Theorem are elegant connections between polynomial division and polynomial evaluation. The Remainder Theorem says that the remainder when you divide a polynomial p(x) by (x − c) is simply p(c) — you can evaluate instead of dividing.

The Factor Theorem is a direct consequence: if p(c) = 0, then (x − c) is a factor of p(x). This gives you a fast way to check potential factors and to find roots of higher-degree polynomials.

This lesson covers synthetic division as an efficient alternative to long division, and shows how the two theorems work together to factor polynomials completely.

Synthetic Division

Method

A shortcut for dividing by $(x - c)$: write coefficients, bring down, multiply by $c$, add.

Example 1

Divide $2x^3 - 3x^2 + x - 5$ by $(x - 2)$.

Synthetic division with $c = 2$:

$2$ |$2$$-3$$1$$-5$
$4$$2$$6$
$2$$1$$3$$1$

Quotient: $2x^2 + x + 3$, Remainder: $1$.

Remainder Theorem

Theorem

When $P(x)$ is divided by $(x - c)$, the remainder equals $P(c)$.

Example 2

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

$P(-1) = -1 + 2 + 4 = 5$. Remainder $= 5$.

Factor Theorem

Theorem

$(x - c)$ is a factor of $P(x)$ if and only if $P(c) = 0$.

Example 3

Is $(x + 1)$ a factor of $x^3 + 2x^2 - x - 2$?

$P(-1) = -1 + 2 + 1 - 2 = 0$. Yes, $(x + 1)$ is a factor.

Rational Zero Theorem

Theorem

If $P(x) = a_n x^n + \cdots + a_0$ has rational zero $\dfrac{p}{q}$ in lowest terms, then $p$ divides $a_0$ and $q$ divides $a_n$.

Example 4

Find all rational zeros of $P(x) = 2x^3 - x^2 - 7x + 6$.

Possible: $\pm\{1, 2, 3, 6, \frac{1}{2}, \frac{3}{2}\}$.

$P(1) = 2 - 1 - 7 + 6 = 0$ ✓. So $(x - 1)$ is a factor.

Divide: $2x^3 - x^2 - 7x + 6 = (x - 1)(2x^2 + x - 6) = (x-1)(2x-3)(x+2)$.

Zeros: $x = 1, \dfrac{3}{2}, -2$.

Example 5

Use synthetic division to divide $x^4 - 3x^3 + 2x - 5$ by $(x - 1)$.

Coefficients (include $0x^2$): $1, -3, 0, 2, -5$; $c = 1$.

$1$ |$1$$-3$$0$$2$$-5$
$1$$-2$$-2$$0$
$1$$-2$$-2$$0$$-5$

Quotient: $x^3 - 2x^2 - 2x$, Remainder: $-5$.

Practice Problems

1. Divide $x^3 + 2x^2 - 5x - 6$ by $(x - 2)$
2. Remainder when $x^3 - 4x + 3$ is divided by $(x - 1)$
3. Is $(x - 3)$ a factor of $x^3 - 27$?
4. List possible rational zeros of $3x^3 + x^2 - 12x - 4$
5. Divide $2x^3 - 5x^2 + 3x + 1$ by $(x - 1)$
6. Find $P(2)$ for $P(x) = x^4 - 3x^2 + 2$
7. Is $(x + 2)$ a factor of $x^3 + 3x^2 + 2x$?
8. Factor $x^3 - 6x^2 + 11x - 6$ completely (test $x = 1,2,3$)
9. Remainder when $2x^4 - x$ is divided by $(x + 1)$
10. Find all zeros of $x^3 - x^2 - 4x + 4$
Show Answer Key

1. $x^2 + 4x + 3$, R $= 0$

2. $P(1) = 1 - 4 + 3 = 0$

3. $P(3) = 27 - 27 = 0$. Yes.

4. $\pm\{1, 2, 4, \frac{1}{3}, \frac{2}{3}, \frac{4}{3}\}$

5. $2x^2 - 3x$, R $= 1$

6. $16 - 12 + 2 = 6$

7. $P(-2) = -8 + 12 - 4 = 0$. Yes.

8. $(x-1)(x-2)(x-3)$

9. $P(-1) = 2(1) - (-1) = 3$

10. $x^2(x-1) - 4(x-1) = (x-1)(x-2)(x+2)$; $x = 1, 2, -2$

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