Polynomial Notation and Evaluation

Let $P(x) = x^2 - 4x + 1$. Work out $P(3)$, then $P(-3)$. Most people get one of them wrong on the first attempt. Before you check, write down which term you are least confident about and why.

The symbol $P(x)$ is a name for the whole polynomial, not $P$ multiplied by $x$. Replacing the $x$ inside the brackets with a number is an instruction: substitute that number everywhere $x$ appears, and evaluate.

When you write

$$P(x) = 2x^3 - 5x + 1$$

you are giving the expression on the right a name. The letter $P$ is the name and the $x$ in brackets announces which letter is the variable. $P(x)$ is not $P$ multiplied by $x$. There is no quantity called $P$ standing alone, so there is nothing to multiply.

Any capital letter will do, and questions often use several at once: $P(x)$, $Q(x)$, $A(x)$, $R(x)$. In Lesson 6 you will meet $P(x) = A(x)Q(x) + R(x)$, four different polynomials in one line, and that sentence is unreadable unless the notation is second nature.

The payoff is that you can now say things about the whole expression in a few symbols. "$P(3) = 0$" replaces "when you substitute $3$ into $2x^3 - 5x + 1$ you get zero". Every theorem in this focus area is written in that shorthand.

$P(a)$ means: replace every $x$ by $a$, then simplify. Take $P(x) = x^3 - 4x^2 + 2$.

$$P(3) = 3^3 - 4(3)^2 + 2 = 27 - 36 + 2 = -7$$

Negatives are where marks are lost, and one habit fixes it: write the number in brackets every time.

$$P(-2) = (-2)^3 - 4(-2)^2 + 2 = -8 - 4(4) + 2 = -8 - 16 + 2 = -22$$

Compare $(-2)^2 = 4$ with $-2^2 = -4$. Without brackets the power binds to the $2$ alone and the minus sign is applied afterwards, which is a different calculation. Substitution means the whole number goes in, sign included, so the brackets are not optional decoration.

Fractions work the same way:

$$P\left(\tfrac{1}{2}\right) = \left(\tfrac{1}{2}\right)^3 - 4\left(\tfrac{1}{2}\right)^2 + 2 = \tfrac{1}{8} - 1 + 2 = \tfrac{9}{8}$$

Take $P(x) = a_n x^n + \cdots + a_1 x + a_0$.

$P(0)$ is the constant term. Every term except the last contains a factor of $x$, so every one of them becomes $0$:

$$P(0) = a_n(0)^n + \cdots + a_1(0) + a_0 = a_0$$

$P(1)$ is the sum of the coefficients. Every power of $1$ is $1$, so each term collapses to its own coefficient:

$$P(1) = a_n + a_{n-1} + \cdots + a_1 + a_0$$

Both are worth using as checks. For $P(x) = 3x^3 - x^2 + 4x - 5$, you can read off $P(0) = -5$ instantly, and $P(1) = 3 - 1 + 4 - 5 = 1$ in one line of mental arithmetic. If a long substitution gives you a different $P(0)$, the error is in the substitution, not in the shortcut.

A third is nearly as quick: $P(-1)$ is the alternating sum, $a_0 - a_1 + a_2 - a_3 + \cdots$, because odd powers of $-1$ are $-1$ and even powers are $+1$.

Nothing in the definition of $P(a)$ requires $a$ to be a number. Substituting an expression works identically: replace every $x$ by the whole expression, in brackets.

With $P(x) = x^2 - 3x$:

$$P(2t) = (2t)^2 - 3(2t) = 4t^2 - 6t$$

$$P(x+1) = (x+1)^2 - 3(x+1) = x^2 + 2x + 1 - 3x - 3 = x^2 - x - 2$$

Two warnings follow from those lines. First, $P(x+1)$ is not $P(x) + 1$: the substitution goes inside the polynomial, not after it. Second, $P(2t) \neq 2P(t)$ in general, since $4t^2 - 6t$ is not $2(t^2 - 3t)$.

These are the same manipulations you will use in Lesson 12 to shift and stretch a polynomial's graph, so it is worth being comfortable with them now, while the polynomials are small.

Two polynomials are equal when they take the same value at every $x$. That turns out to be a very strong condition:

$$P(x) = Q(x) \text{ for all } x \quad \Longleftrightarrow \quad \text{every matching coefficient is equal}$$

So if $ax^2 + bx + c = 3x^2 - 7$ for all $x$, then $a = 3$, $b = 0$ and $c = -7$. Comparing coefficients like this is called equating coefficients, and it is the standard way to find unknown constants.

The alternative is to substitute clever values. From the identity $x^2 + px + q = (x+2)(x+r)$, putting $x = -2$ kills the right side and gives $4 - 2p + q = 0$ in one step. Choosing values that make brackets vanish is often faster than expanding.

Both methods rest on the same fact: an identity holds for every $x$, so you may substitute whatever value is most convenient. An ordinary equation such as $x^2 = 4$ holds only for particular values, and substituting into it proves nothing.

Five items on notation and evaluation. Work each one, then reveal the answer.

1. 1 $P(x) = x^2 - 6x + 5$. Find $P(-1)$.

   Show answer
   $(-1)^2 - 6(-1) + 5 = 1 + 6 + 5$.$12$

2. 2 $P(x) = 4x^3 - 2x + 9$. Find $P(0)$ without expanding anything.

   Show answer
   $P(0)$ is always the constant term.$9$

3. 3 $P(x) = 2x^4 - 3x^3 + x - 6$. Find $P(1)$.

   Show answer
   Add the coefficients: $2 - 3 + 0 + 1 - 6$.$-6$

4. 4 $P(x) = x^2 + x$. Find $P(x - 1)$ in expanded form.

   Show answer
   $(x-1)^2 + (x-1) = x^2 - 2x + 1 + x - 1$.$x^2 - x$

5. 5 If $ax^2 + 3x + b = 5x^2 + 3x - 2$ for all $x$, find $a$ and $b$.

   Show answer
   Equate matching coefficients.$a = 5$, $b = -2$