Remainder theorem and factor theorem — proof and applications

The Remainder Theorem

Derivation from First Principles

Let's prove this step-by-step, understanding why each step works.

Starting Point: When we divide any polynomial $p(x)$ by $(x - a)$ , we can write:

$p(x) = (x - a) \cdot q(x) + r$

Where:

$q(x)$ is the quotient polynomial
$r$ is the remainder (which must be a constant because we're dividing by a linear term)

Why is the remainder a constant? The degree of the remainder must be less than the degree of the divisor. Since $(x - a)$ has degree 1, the remainder must have degree 0 (a constant) or be zero.

The Key Insight: This equation must be true for all values of x. So let's substitute $x = a$ :

$p(a) = (a - a) \cdot q(a) + r$

Why this step? We chose $x = a$ specifically to make the $(x - a)$ term vanish!

$p(a) = 0 \cdot q(a) + r$

$p(a) = r$

Conclusion: The remainder when $p(x)$ is divided by $(x - a)$ is exactly $p(a)$ . ∎

Figure — Remainder theorem and factor theorem — proof and applications

The Factor Theorem

Derivation from Remainder Theorem

The Factor Theorem is a special case of the Remainder Theorem.

From Remainder Theorem: $p(x) = (x - a) \cdot q(x) + p(a)$

Forward direction (If $(x-a)$ is a factor, then $p(a) = 0$ ):

If $(x - a)$ is a factor, then $p(x)$ divides evenly with no remainder
This means: $p(x) = (x - a) \cdot q(x) + 0$
Substituting $x = a$ : $p(a) = 0 + 0 = 0$ ✓

Reverse direction (If $p(a) = 0$ , then $(x-a)$ is a factor):

If $p(a) = 0$ , then from Remainder Theorem: $r = p(a) = 0$
So: $p(x) = (x - a) \cdot q(x) + 0 = (x - a) \cdot q(x)$
This means $(x - a)$ divides $p(x)$ exactly, so it's a factor ✓

Formula Factor Theorem Formula

$(x - a) \text{ is a factor of } p(x) \iff p(a) = 0$ \begin{aligned} &\text{Worked Examples} \[6pt] &\textbf{Example 1: Finding the Remainder (Basic)} \ &\text{Find the remainder when } p(x) = x^3 - 4x^2 + 6x - 5 \text{ is divided by } (x - 2). \[6pt] &\textbf{Solution:} \ &\textbf{Step 1: } \text{Identify } a \text{ from the divisor } (x - a). \ &\quad \text{Here, } (x - 2), \text{ so } a = 2 \ &\quad \textbf{Why this step? } \text{We need to evaluate } p(a). \[6pt] &\textbf{Step 2: } \text{Calculate } p(2). \end{aligned} $p(2) = (2)^3 - 4(2)^2 + 6(2) - 5$ $= 8 - 16 + 12 - 5$ $= -1$

Why this step? By Remainder Theorem, $p(a)$ is the remainder.

Answer: Remainder = -1

Verification: If we did long division, we'd get $q(x) = x^2 - 2x + 2$ and remainder $-1$ .

Worked example Example 2: Finding the Remainder (General Form)

Find the remainder when $p(x) = 2x^3 + 3x^2 - 5x + 1$ is divided by $(2x - 1)$ . Solution: Step 1: Rewrite divisor in the form $(x - a)$ or find where it equals zero.

$2x - 1 = 0 \implies x = \frac{1}{2}$
Why? For the general form $(ax - b)$ , we evaluate at $x = \frac{b}{a}$ . Step 2: Calculate $p\left(\frac{1}{2}\right)$ . $p\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right)^3 + 3\left(\frac{1}{2}\right)^2 - 5\left(\frac{1}{2}\right) + 1$ $= 2 \cdot \frac{1}{8} + 3 \cdot \frac{1}{4} - \frac{5}{2} + 1$ $= \frac{1}{4} + \frac{3}{4} - \frac{5}{2} + 1$ $= 1 - \frac{5}{2} + 1 = 2 - 2.5 = -\frac{1}{2}$ Answer: Remainder = -1/2

Worked example Example 3: Using Factor Theorem to Check Factors

Is $(x - 3)$ a factor of $p(x) = x^3 - 7x + 6$ ? Solution: Step 1: Apply Factor Theorem — check if $p(3) = 0$ . $p(3) = (3)^3 - 7(3) + 6$ $= 27 - 21 + 6$ $= 12 \neq 0$

Why this step? If p(3) = 0, then (x-3) would be a factor. Answer: Since $p(3) \neq 0$ , (x - 3) is not a factor.

[!example] Example 4: Finding Unknown Coefficients If (x - 2) is a factor of p(x) = x^3 - 4x^2 + ax +6, find a.

Solution: Step 1: Since (x - 2) is a factor, by Factor Theorem: p(2) = 0.

Why? Factor Theorem guarantees this condition.

Step 2: Substitute x = 2 and set equal to zero. $p(2) = (2)^3 - 4(2)^2 + a(2) + 6 = 0$ $8 - 16 + 2a + 6 = 0$ $-2 + 2a = 0$ $2a = 2$ $a = 1$ Answer: $a = 1$

Worked example Example 5: Complete Factorization

Factor completely: $p(x) = x^3 - 6x^2 + 11x - 6$ .

Solution:

Step 1: Use Factor Theorem to find one factor by testing small integers.

Try $x = 1$ : $p(1) = 1 - 6 + 11 - 6 = 0$ ✓
Why test integers? By Rational Root Theorem, if there's a rational root, it divides the constant term (6).

Step 2: Since $p(1) = 0$ , $(x - 1)$ is a factor. Divide $p(x)$ by $(x - 1)$ . Using synthetic division or long division: $p(x) = (x - 1)(x^2 - 5x + 6)$

Why this step? We need to find q(x) in p(x) = (x-1) \cdot q(x).

Step 3: Factor the quadratic x^2 - 5x + 6. $x^2 - 5x + 6 = (x - 2)(x - 3)$

Final Answer: $p(x) = (x - 1)(x - 2)(x - 3)$

Verification: Roots are $x = 1, 2, 3$ . Check: $p(1) = p(2) = p(3) = 0$ ✓

Common Mistakes

Recall Explain to a 12-year-old

Imagine you have a big number like 17, and you want to divide it by 5. You get3 with a remainder of 2, right?

Polynomials work the same way! If you have $x^3 - 4x^2 + 6x - 5$ and divide it by $(x - 2)$ , you get some quotient and a remainder.

But here's the magic trick: you don't actually need to do the long division. Just plug in the number2 into your polynomial (because $x - 2 = 0$ when $x = 2$ ), and whatever answer you get — that's your remainder! It's like a shortcut calculator built right into the math.

And there's a bonus: if your remainder is exactly zero, that means $(x - 2)$ fits perfectly into your polynomial — it's a factor, like how 3 is a factor of 12because it divides evenly.

Key Applications

Quick remainder calculation without polynomial long division
Finding factors of polynomials by testing potential roots
Solving polynomial equations by factoring
Determining unknown coefficients in polynomials
Checking divisibility of one polynomial by another

Connections

Polynomial Long Division — the "slow way" that these theorems shortcut
Rational Root Theorem — helps identify which values of $a$ to test
Fundamental Theorem of Algebra — guarantees existence of roots (and therefore factors)
Synthetic Division — efficient computation technique using Remainder Theorem
Factoring Polynomials — Factor Theorem is the main tool
Solving Polynomial Equations — finding roots = finding factors

Flashcards

#flashcards/maths

What does the Remainder Theorem state? :: When polynomial $p(x)$ is divided by $(x - a)$ , the remainder is $p(a)$ .

State the Factor Theorem

(x - a)

is a factor of

p(x)

if and only if

p(a) = 0

When dividing

p(x)

(2x - 3)

, what value do you evaluate to find the remainder?

p\left(\frac{3}{2}\right)

, because

2x - 3 = 0

when

x = \frac{3}{2}

p(5) = 0

, what can you conclude?

(x - 5)

is a factor of

p(x)

(x + 4)

is a factor of

p(x)

, what is

p(-4)

p(-4) = 0

, because

(x + 4) = (x - (-4))

is a factor.

Why must the remainder be a constant when dividing by a linear polynomial?

The degree of the remainder must be less than the degree of the divisor. Since a linear polynomial has degree 1, the remainder has degree 0 (constant) or is zero.

Derive the Remainder Theorem in 3 steps :: (1) Write $p(x) = (x-a) \cdot q(x) + r$ ; (2) Substitute $x = a$ to get $p(a) = 0 \cdot q(a) + r$ ; (3) Therefore $r = p(a)$ .

How are the Remainder Theorem and Factor Theorem related?

Factor Theorem is a special case of Remainder Theorem when the remainder equals zero.

If you want to check if

(x - 7)

is a factor of

p(x) = x^3 - 9x + 2

, what single calculation do you perform?

Calculate

p(7)

. If it equals0, then

(x-7)

is a factor; otherwise it's not.

What does it mean for

(x - a)

to be a factor of

p(x)

p(x)

can be written as

(x - a) \cdot q(x)

for some polynomial

q(x)

, with zero remainder.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Remainder Theorem aur Factor Theorem algebra ke do bahut powerful tools hain jo polynomial division ko simple bana dete hain. Socho agar tumhe koi bada polynomial divide karna hai $(x - 3)$ se — toh normally tum long division karoge, bohot time lagega. Lekin Remainder Theorem kehta hai: "Arre, why waste time? Bas polynomial me $x = 3$ daal do, jo answerayega wohi tumara remainder hai!" Matlab $p(3)$ calculate karo, done. Ye shortcut kitna useful hai imagine karo — exams me time bachega!

Factor Theorem ek step age jata hai. Ye bata hai ki agar tumhara remainder zero nikla (matlab $p(3) = 0$ ), toh $(x - 3)$ tumhare polynomial ka proper factor hai. Iska matlab agar tum polynomial ko factorize karna chahte ho, toh bas different values test karte jao — jahan $p(a) = 0$ mile, wahan $(x - a)$ factor hai. Ye technique polynomial equations solve karne me kafi kaam ati hai, especially jab cubic ya higher degree polynomials ho.

Ye dono theorems algebraic problem-solving me foundation jaisi hain. Inhe samajhne se tumhe polynomials pe complete control mil jata hai — chahe roots dhoondhni ho, factorization karni ho, ya koi unknown coefficient nikalni ho. Real-world me jahan bhi polynomial expressions use hoti hain (physics equations, engineering formulas, economics models), ye theorems wahan kaam aate hain. Practice karte raho different types ke questions, aur ye concepts apne-ap natural lagenge!

Remainder theorem and factor theorem — proof and applications

Core Intuition

The Remainder Theorem

Derivation from First Principles

The Factor Theorem

Derivation from Remainder Theorem

Common Mistakes

Key Applications

Connections

Flashcards

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Algebra — Introduction & Intermediate

Connections