01 / 21
R \(p(a)\)

Applying the
Remainder Theorem

Learn to evaluate polynomials instantly using division and substitution, and discover when \( (x - a) \) is a factor.

Algebra 2

What is Polynomial Division?

Just like dividing whole numbers, dividing a polynomial \( p(x) \) by a divisor \( d(x) \) gives a quotient \( q(x) \) and a remainder \( r \):

When the divisor is linear, like \( (x - a) \), the remainder \( r \) is always a constant: just a number, not another polynomial.

Division Meets Evaluation

Consider dividing \( p(x) \) by \( (x - a) \). The division identity says:

Now substitute \( x = a \):

The remainder equals \( p(a) \). That is the entire insight behind the Remainder Theorem.

Why Does This Matter?

Fast Evaluation
Find \( p(a) \) using synthetic division instead of plugging in and computing large powers.
Factor Testing
If \( p(a) = 0 \), then \( (x - a) \) is a factor: the polynomial crosses the \( x \)-axis at \( x = a \).
Root Finding
Systematically test possible roots using the Rational Root Theorem, then confirm with the Remainder Theorem.

The Remainder Theorem

\( p(x) \div (x - a) \;\rightarrow\; \text{remainder} = p(a) \)
Factor Theorem (Corollary)
\( (x - a) \) is a factor of \( p(x) \iff p(a) = 0 \)

The Factor Theorem is the special case: remainder \( = 0 \) means \( (x - a) \) divides \( p(x) \) evenly.

Visualizing \( p(a) \) on the Graph

x y -1 1 3 2 \(p(2) = -3\) \(p(-1)=0\) \(p(1)=0\) \(p(3)=0\)

Remainder vs. Factor Theorem

Remainder Theorem Factor Theorem
Statement Remainder of \( p(x) \div (x-a) \) is \( p(a) \) \( (x-a) \) is a factor iff \( p(a) = 0 \)
Remainder Can be any value Must be exactly \( 0 \)
Tells you The value of \( p(a) \) Whether \( a \) is a root
Relationship General case Special case of Remainder Theorem

How to Apply the Theorem

1
Identify the divisor. Write it in the form \( (x - a) \). Be careful with signs: \( (x + 3) = (x - (-3)) \), so \( a = -3 \).
2
Substitute \( a \) into \( p(x) \). Replace every \( x \) with the value \( a \) and simplify.
3
Interpret the result. The number you get is the remainder. If it equals \( 0 \), then \( (x - a) \) is a factor.
4
Optional: verify with synthetic division. Use \( a \) as the synthetic divisor. The last entry should match \( p(a) \).

Example 1: Finding the Remainder

Find the remainder when \( p(x) = x^3 - 4x^2 + 5x - 2 \) is divided by \( (x - 2) \).

1
Divisor is \( (x - 2) \), so \( a = 2 \).
2
Evaluate: \( p(2) = (2)^3 - 4(2)^2 + 5(2) - 2 \)
3
\( = 8 - 16 + 10 - 2 = \) \( 0 \)

Example 2: Nonzero Remainder

Find the remainder when \( p(x) = 2x^3 + x^2 - 7x + 3 \) is divided by \( (x + 2) \).

1
Divisor is \( (x + 2) = (x - (-2)) \), so \( a = -2 \).
2
\( p(-2) = 2(-2)^3 + (-2)^2 - 7(-2) + 3 \)
3
\( = 2(-8) + 4 + 14 + 3 = -16 + 4 + 14 + 3 = \) \( 5 \)

Example 3: Synthetic Division Check

Verify Example 1 using synthetic division with \( a = 2 \).

2 1 \(-4\) 5 \(-2\) 2 \(-4\) 2 1 \(-2\) 1 0 \(x^2\) \(x\) const remainder \(p(2) = 0\) ✓

Example 4: Higher Degree Polynomial

Is \( (x - 1) \) a factor of \( p(x) = x^4 - 3x^3 + 3x^2 - 3x + 2 \)?

1
Divisor is \( (x - 1) \), so \( a = 1 \).
2
\( p(1) = (1)^4 - 3(1)^3 + 3(1)^2 - 3(1) + 2 \)
3
\( = 1 - 3 + 3 - 3 + 2 = \) \( 0 \)

Watch the Sign: \( (x + a) \) vs. \( (x - a) \)

The divisor must be written as \( (x - a) \). If you see a plus sign, rewrite it first:

\( (x - 5) \)
Already in the right form. Use \( a = 5 \). Evaluate \( p(5) \).
\( (x + 3) \)
Rewrite as \( (x - (-3)) \). Use \( a = -3 \). Evaluate \( p(-3) \).

Roots, Zeros, and Factors

The Remainder Theorem ties three ideas together. They are different names for the same relationship:

\(p(a) = 0\)
\((x - a)\) is a factor
\(x = a\) is a root
 all equivalent

Common Mistakes to Avoid

✕ Wrong Sign for \( a \)
Dividing by \( (x + 3) \) and plugging in \( a = 3 \) instead of \( a = -3 \). Always rewrite as \( (x - a) \) first.
✕ Arithmetic Errors
Forgetting that \( (-2)^3 = -8 \), not \( 8 \). Negative exponents trip up many students: double-check signs.
✕ Missing Terms
If \( p(x) = x^3 + 5 \), the \( x^2 \) and \( x \) terms have coefficient \( 0 \). Include them in synthetic division.

Quick Reference Summary

Scenario Value of \( p(a) \) Conclusion
\( p(a) = 0 \) Remainder is 0 \( (x - a) \) is a factor; \( a \) is a root
\( p(a) \neq 0 \) Remainder is nonzero \( (x - a) \) is not a factor
Divisor is \( (x + a) \) Evaluate at \( -a \) Rewrite as \( (x - (-a)) \) first
Synthetic division Last entry = remainder Should match direct substitution

Question 1 of 5

What is the remainder when \( p(x) = x^3 + 2x^2 - 5x - 6 \) is divided by \( (x - 2) \)?

Question 2 of 5

If \( (x + 4) \) is a factor of \( p(x) \), what must be true?

Question 3 of 5

Find the remainder when \( p(x) = 3x^3 - x^2 + 2x - 5 \) is divided by \( (x + 1) \).

Question 4 of 5

Which of the following is a factor of \( p(x) = x^3 - 6x^2 + 11x - 6 \)?

Question 5 of 5

A student divides \( p(x) \) by \( (x - 5) \) using synthetic division and gets a remainder of 7. What does this tell you?