Welcome! Today we're tackling Evaluating Polynomials at Any Input — the foundational skill that powers everything else in this chapter.

Every technique in this chapter depends on one basic operation: plugging a number into a polynomial and computing the result.

Whether you are checking if something is a zero, verifying a relationship, or applying the remainder theorem — it all comes down to evaluation.

The catch is that most errors happen not because students don't understand the concept, but because they:

Make sign mistakes with negative inputs
Mess up fraction arithmetic

In this lesson, you will build an error-proof evaluation procedure that handles:

Input Type	Example
Integers	$2$ , $5$ , $- 3$
Negatives	$- 1$ , $- 4$
Fractions	$\frac{1}{2}$ , $- \frac{2}{3}$

1. Basic evaluation with positive inputs

We're building the evaluation skill from the ground up.

The first step is the basic procedure:

Substitute → Compute powers → Multiply by coefficients → Add

Even with positive inputs, there is a sequence that must be followed.

📋 Given Info

Here's how it works:

To evaluate $p (x) = 2 x^{2} - 3 x + 5$ at $x = 2$ , replace every $x$ with $2$ and compute:

$p (2) = 2 (2)^{2} - 3 (2) + 5 = 2 (4) - 6 + 5 = 8 - 6 + 5 = 7$

Step-by-step breakdown:

Substitute: Replace $x$ with $2$
Powers first: $2^{2} = 4$
Multiply coefficients: $2 (4) = 8$ and $3 (2) = 6$
Combine: $8 - 6 + 5 = 7$ answer

✍️ Question

Your turn! 📝

Evaluate $p(x) = x^3 + 2x^2 - x + 3$ at $x = 2$ input.

Show each step.

Here is the step-by-step procedure for evaluating a polynomial:

Step 1: Substitute the value, wrapping it in parenthesesProtect negative numbers and ensure correct order of operations.

So for $p (x) = x^{3} + 2 x^{2} - x + 3$ at $x = 2$ , we write: $p (2) = (2)^{3} + 2 (2)^{2} - (2) + 3$

Step 2: Simplify powers first.Exponents are handled before multiplying by coefficients

$(2)^{3} = 8$ and $(2)^{2} = 4$

So now we have: $p (2) = 8 + 2 (4) - 2 + 3$

✍️ FIB

Fill in the blank

In the expression

8 + 2(4) - 2 + 3

, what is

2(4)

?

✓

Step 3: Multiply each power by its coefficient.

$2(4) = 8$

So now: $p (2) = 8 + 8 - 2 + 3$

Step 4: Add all terms from left to right.

$8 + 8 - 2 + 3 = 16 - 2 + 3 = 14 + 3 = 17$

So $p(2) = 17$ AnswerThe four-step method to follow every time

✍️ MCQ

Choose one

What is the correct order of operations when evaluating a polynomial?

✓

For $p (x) = x^{3} + 2 x^{2} - x + 3$ at $x = 2$ :

Step 1: Substitute $x = 2$ into every $x$ $p (2) =$ $(2)^3 + 2(2)^2 - (2) + 3$ Always use parentheses to avoid sign errors

Step 2: Evaluate the powers firstOrder of operations — evaluate exponents first $= 8 + 2 (4) - 2 + 3$

Step 3: Multiply $= 8 +$ $8$ $- 2 + 3$

Step 4: Add and subtract left to rightSame sequence every time you evaluate

= 17

Answer!(Powers, multiplication, then add and subtract)

So, $p (2) = 17$

✍️ MCQ

Choose one

Evaluate

p(x) = x^2 + 3x - 4

at

x = 3

.

✓

Order of Operations for Polynomial Evaluation

The order matters: powers firstThis must come before anything else, then multiplicationSecond step after computing powers, then addition/subtractionFinal step in the sequence.

This is the golden ruleFollow this exact order every single time — mess up the order, and your answer will be wrongWrong order means wrong answer even if every calculation is correct!

⚠️ Common Mistake Alert

Skipping steps leads to arithmetic errors.

Always follow: Powers → Multiplication → Addition/Subtraction

Write out each lineYour safety net against arithmetic slips — it takes 10 extra seconds but saves you from silly mistakesPrevents losing marks in the exam!

✍️ MCQ

Choose one

When evaluating

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

at

x = 2

, which step should you do FIRST after substituting?

✓

2. Evaluation with negative inputs and sign handling

Let's test your skills with negative inputs 🎯

Positive inputs are straightforward. The real challenge comes with negative inputs, where:

Even powers produce positive results: $(- 2)^{2} = + 4$ , $(- 3)^{2} = + 9$
The double-negative in terms like $- 3 (- 1)$ trips many students

This is the single most common source of errors in polynomial evaluation.

📋 Given Info

Key rules to remember:

When substituting a negative number, wrap it in parentheses before applying the exponent
Even powers of any number are positive:
- $(- 1)^{2} = + 1$
- $(- 3)^{2} = + 9$
- ${(- \frac{1}{2})}^{2} = + \frac{1}{4}$
Negative times negative is positive: $- 3 (- 1) = + 3$

✍️ Question

Your turn! ✏️

Evaluate $p(x) = 3x^2 - 7x + 2$ at $x = -2$ input.

Show every step, especially how you handle the signs.

This is where most evaluation mistakes happen. Let's be very careful.

$p (x) = 3 x^{2} - 7 x + 2$ at $x = -2$ Signs trip people up with negative inputs.

Step 1: Substitute $x = - 2$ with parenthesesAlways wrap negative numbers in parentheses first

Write: $p (- 2) = 3 (- 2)^{2} - 7 (- 2) + 2$

Notice how we wrap $- 2$ in parentheses everywhere — this is crucial for handling the signs correctly!

✍️ MCQ

Choose one

What is

(-2)^2

?

✓

Now handle each term:

Term 1: $3 (- 2)^{2}$

First compute $(- 2)^{2}$ . This is $(- 2) \times (- 2) =$ $+4$ Two negatives multiply to positive.

Negative times negative is positiveThe key rule for squaring negatives.

Then $3 \times 4 =$ $12$ Term 1Always do exponents first, then multiply.

Term 2: $- 7 (- 2)$

This is negative times negative = positiveTwo negatives always give positive.

$- 7 \times - 2 =$ $+14$ Term 2Common exam mistake to watch for

✍️ MCQ

Choose one

In the expression

-7(-2)

, why does the result become

+14

instead of

-14

?

✓

Term 3: $+ 2$

Final stepCombine all simplified terms — Add all terms:

$12 + 14 + 2 =$ (Answer) $28$ The final result of our evaluation

So $p(-2) = 28$ The complete method for polynomials

✍️ MCQ

Choose one

Evaluate

q(x) = 2x^2 - 5x + 1

at

x = -3

.

✓

The two danger pointsWhere exam marks get lost:

$(-2)^2 = +4$ Even powers always give positive results, NOT $-4$ Common mistake to avoid. Even powers of any number are positive.The rule to remember

$-7(-2) = +14$ Negatives multiply to give positive, NOT $- 14$ . The two negatives cancel.The result flips to positive

✍️ MCQ

Choose one

What is

(-3)^2

?

✓

The safeguard: Always wrap the substituted value in parentheses.Parentheses protect against sign errors

Write $(- 2)^{2}$ , not $- 2^{2}$ .

Why this matters:

Without parentheses, $-2^2$ Negative stays outside the exponent means $- (2^{2}) =$ $- 4$

With parentheses, $(-2)^2$ The negative gets squared along with the number means $(- 2) \times (- 2) =$ $+ 4$

These are completely different values!An 8-mark difference on one question

✍️ MCQ

Choose one

Which of the following equals

+4

?

✓

3. Evaluation with fractional inputs

We've handled positive and negative integer inputs. The final category is fractions, which require:

Computing fractional powers first
Converting to a common denominator before adding

This completes the evaluation toolkit.

📋 Given Info

When evaluating at a fraction like $x = \frac{3}{2}$ :

Compute the power first: ${(\frac{3}{2})}^{2} = \frac{9}{4}$
Multiply by the coefficient: $6 \times \frac{9}{4} = \frac{54}{4} = \frac{27}{2}$
Convert all terms to a common denominator before adding

✍️ Question

Your Turn ✏️

Evaluate $q(x) = 6x^2 - 7x - 3$ at $x = \frac{3}{2}$ .

Show every step.

Fractional inputsOne small mistake with fractions can throw off your entire answer require careful step-by-step work. Let's evaluate $q (x) = 6 x^{2} - 7 x - 3$ at $x = \frac{3}{2}$ .

Step 1: Compute the power. ${(\frac{3}{2})}^{2} = \frac{9}{4}$

When squaring a fraction, we square both the numerator and denominator separatelySquare the top and bottom separately, don't do it all at once: $3^{2} = 9$ and $2^{2} = 4$ .

✍️ MCQ

Choose one

What is

6 \times \frac{9}{4}

?

✓

Step 2: Multiply by the coefficient. $6 \times \frac{9}{4} = \frac{54}{4} =$ $\frac{27}{2}$ simplifiedAlways simplify before moving to the next step

We write 6 as $\frac{6}{1}$ Writing whole numbers as fractions lets us multiply across easily, multiply across to get $\frac{54}{4}$ , then simplify by dividing both by 2.

Step 3: Compute the next term.

-7 \times \frac{3}{2}</pen> = <pen actions="highlight" highlight-color="green" narrationText="negative 21 over 2;;the second term is negative 21 over 2">-\frac{21}{2}

Step 4: Convert the constant to the same denominator.

(whole number)

-3</pen> = <pen actions="highlight" highlight-color="green" narrationText="negative 6 over 2" commentary="Same denominator lets you add numerators directly">-\frac{6}{2}

(Convert whole number to fraction form)

✍️ MCQ

Choose one

We have

\frac{27}{2} + \left(-\frac{21}{2}\right) + \left(-\frac{6}{2}\right)

. What is the final answer?

✓

Step 5: Add all terms with the common denominator 2: $\frac{27}{2} - \frac{21}{2} - \frac{6}{2} = \frac{27 - 21 - 6}{2} = \frac{0}{2} = 0$

27 - 21 = 6

6 - 6 = 0

0Answer!

The result is 0When the output equals zero — which means $x = \frac{3}{2}$ This specific input gives us zero is a zeroSpecial name for inputs that give zero output of this polynomial. That is a concept we explore next.

✍️ MCQ

Choose one

If

p(a) = 0

for some polynomial

p(x)

, what do we call

a

?

✓

The key habit with fractions: always convert to a common denominator before addingThe one habit that prevents most calculation errors.

Do not try to add $\frac{27}{2} - \frac{21}{2} - 3$ in your head.

Step-by-step approach:

Convert ALL terms to the same denominator first
Here, the common denominator is 2We identified 2 as the common denominator:
- $\frac{27}{2}$ ✓ (already has denominator 2)
- $\frac{21}{2}$ ✓ (already has denominator 2)
- $3 = \frac{6}{2}$ Convert the whole number before combining (convert the whole number)
Now combine: $\frac{27 - 21 - 6}{2} = \frac{0}{2} = 0$ Touch numerators only after denominators match

✍️ MCQ

Choose one

Convert

5

to a fraction with denominator

4

. What do you get?

✓