The Rational Root Theorem Worksheet for 11th Grade | Lesson Planet - Free Printable

Let’s go through the first problem step by step.

We are given:
f(x) = x³ + 3x² + x + 3

We need to find all rational zeros using the Rational Root Theorem.

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Step 1: Understand the Rational Root Theorem

The Rational Root Theorem says that any possible rational zero (root) of a polynomial is of the form:

> p/q, where
> - p is a factor of the constant term (last number),
> - q is a factor of the leading coefficient (number in front of the highest power of x).

In this case:
- Constant term = 3 → factors: ±1, ±3
- Leading coefficient = 1 (since it’s x³) → factors: ±1

So possible rational roots = ±1/1, ±3/1 → ±1, ±3

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Step 2: Test each possible root

We plug each into f(x) and see if we get 0.

#### Try x = -1:
f(-1) = (-1)³ + 3(-1)² + (-1) + 3
= -1 + 3(1) -1 + 3
= -1 + 3 -1 + 3 = 4 → Not zero.

Wait — let me recalculate:

(-1)³ = -1
3*(-1)² = 3*1 = 3
+ (-1) = -1
+3 = +3

Total: -1 + 3 = 2; 2 -1 = 1; 1 + 3 = 4 → Not zero.

#### Try x = 1:
f(1) = 1 + 3 + 1 + 3 = 8 → Not zero.

#### Try x = -3:
f(-3) = (-3)³ + 3(-3)² + (-3) + 3
= -27 + 3*(9) -3 + 3
= -27 + 27 -3 + 3 = 0 → YES!

So x = -3 is a root.

#### Try x = 3:
f(3) = 27 + 27 + 3 + 3 = 60 → Not zero.

So only x = -3 works so far.

But wait — maybe there are more? Let’s factor the polynomial now that we know (x + 3) is a factor.

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Step 3: Factor using synthetic division or grouping

Let’s try factoring by grouping:

f(x) = x³ + 3x² + x + 3
Group: (x³ + 3x²) + (x + 3)
Factor each group:
x²(x + 3) + 1(x + 3)
Now factor out (x + 3):
→ (x + 3)(x² + 1)

Set equal to zero:
(x + 3)(x² + 1) = 0

So solutions:
- x + 3 = 0 → x = -3
- x² + 1 = 0 → x² = -1 → x = ±i → not real, definitely not rational.

So the only rational zero is x = -3

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Final Check

We tested all possible rational roots from the theorem: ±1, ±3
Only x = -3 gave us zero.
Factoring confirmed only one real root, and the other two are imaginary.

✔ So the answer is correct.

Final Answer:
-3