Polynomials — degree, types (monomial, binomial, trinomial): Exercises

Before we start, one reminder in plain words, because the whole page leans on it:

Level 1 — Recognition

Exercise 1.1

State the degree of $P (x) = 6 x^{4} - 9 x^{7} + 2 x - 5$ .

Recall Solution

WHAT we do: list the exponent of every term.

$6 x^{4} \to$ exponent $4$
$- 9 x^{7} \to$ exponent $7$
$2 x = 2 x^{1} \to$ exponent $1$
$- 5 = - 5 x^{0} \to$ exponent $0$

WHY: the degree is defined as the highest exponent whose coefficient is not zero. The terms being written out of order ( $x^{4}$ before $x^{7}$ ) does not matter — we only hunt for the biggest.

Biggest exponent $= 7$ , and its coefficient $- 9 \neq = 0$ . Degree $= 7$ .

Exercise 1.2

Classify each as monomial, binomial, or trinomial: (a) $- 3 x^{5}$ (b) $x^{2} + 9$ (c) $4 x^{3} - x + 8$

Recall Solution

WHAT / WHY: count the terms (chunks separated by $+$ or $-$ ). Each is already simplified — no like terms to combine.

(a) $- 3 x^{5}$ → one term → monomial.
(b) $x^{2} + 9$ → two terms ( $x^{2}$ and $9$ ) → binomial.
(c) $4 x^{3} - x + 8$ → three terms → trinomial.

Exercise 1.3

What is the degree of the constant polynomial $P (x) = 12$ ?

Recall Solution

WHAT: rewrite the constant using the invisible power. Since $x^{0} = 1$ , we have $12 = 12 x^{0}$ . WHY: every polynomial is secretly a sum of $x$ -powers; a lone number is the $x^{0}$ slot. Its exponent is $0$ . Degree $= 0$ . (Not "undefined" — only the zero polynomial $0$ is undefined-degree.)

Level 2 — Application

Exercise 2.1

Simplify, then classify by number of terms and state the degree: $S (x) = 5 x^{2} - 3 x^{2} + 4 x - 4 x + 7.$

Recall Solution

WHAT — combine like terms (see Combining Like Terms). Like terms share the same power of $x$ .

$x^{2}$ group: $5 x^{2} - 3 x^{2} = 2 x^{2}$
$x^{1}$ group: $4 x - 4 x = 0$ (they cancel completely)
$x^{0}$ group: $7$

WHY simplify first: type and degree are defined on the simplified form. The $x$ terms vanished, so they cannot count. $S (x) = 2 x^{2} + 7.$ Two terms → binomial. Degree $= 2$ .

Exercise 2.2

Multiply the monomials: $(4 x^{3}) (- 2 x^{5})$ .

Recall Solution

WHAT: multiply the numbers, then the powers. $4 \cdot (- 2) = - 8, x^{3} \cdot x^{5} = x^{3 + 5} = x^{8} .$ WHY the exponents add: $x^{3} \cdot x^{5}$ means $(x x x) (x x x x x)$ — eight $x$ 's multiplied, i.e. $x^{8}$ . That is the product rule of Exponent Rules. $(4 x^{3}) (- 2 x^{5}) = - 8 x^{8} .$

Exercise 2.3

Expand $(x + 4) (x - 3)$ using FOIL and classify the result.

Recall Solution

WHAT — FOIL = First, Outer, Inner, Last (see FOIL Method):

First: $x \cdot x = x^{2}$
Outer: $x \cdot (- 3) = - 3 x$
Inner: $4 \cdot x = 4 x$
Last: $4 \cdot (- 3) = - 12$

$x^{2} - 3 x + 4 x - 12.$ WHY combine now: $- 3 x$ and $4 x$ are like terms. $- 3 x + 4 x = 1 x = x$ . $= x^{2} + x - 12.$ Three terms → trinomial. Degree $= 2$ .

Level 3 — Analysis

Exercise 3.1

Factor the trinomial $x^{2} + 7 x + 12$ .

Recall Solution

WHAT — reverse FOIL. We want $(x + p) (x + q)$ such that it expands back to $x^{2} + 7 x + 12$ . Expanding $(x + p) (x + q) = x^{2} + (p + q) x + pq$ , so we need $p + q = 7 and p \cdot q = 12.$ WHY these two conditions: the middle coefficient is the sum of the roots-shifts and the constant is their product — that is just what FOIL produces. Test factor pairs of $12$ : $(1, 12), (2, 6), (3, 4)$ . Only $3 + 4 = 7$ . ✓ $x^{2} + 7 x + 12 = (x + 3) (x + 4) .$ Check by expanding: $x^{2} + 4 x + 3 x + 12 = x^{2} + 7 x + 12$ . ✓ (See Factoring.)

Exercise 3.2

A student writes: " $x^{50} + 1$ is degree 2 because it is a binomial." Diagnose and correct.

Recall Solution

WHAT is wrong: they tied degree to the number of terms. Those are the two independent questions from the top of this page.

Number of terms $= 2$ → yes, binomial.
Highest exponent $= 50$ → degree 50, not 2.

WHY the mix-up is impossible to sustain: "bi" counts terms, degree counts powers. $x^{50} + 1$ proves they can be wildly different (2 terms, degree 50).

Exercise 3.3

Recognise and factor $x^{2} - 10 x + 25$ as a perfect square.

Recall Solution

WHAT — pattern match against $(x - y)^{2} = x^{2} - 2 x y + y^{2}$ .

First term $x^{2}$ → $x$ is the first piece.
Last term $25 = 5^{2}$ → $y = 5$ .
Middle should be $- 2 x y = - 2 \cdot x \cdot 5 = - 10 x$ . ✓ It matches!

WHY the middle-term check matters: without it, $x^{2} - 26 x + 25$ would look like a perfect square (ends are squares) but isn't, since $- 2 \cdot 1 \cdot 25$ -type middle wouldn't line up. Here it does. $x^{2} - 10 x + 25 = (x - 5)^{2} .$

Level 4 — Synthesis

Exercise 4.1

Build a binomial of degree 6 whose two terms have coefficients that add to $0$ but the polynomial is not the zero polynomial. Then state its degree.

Recall Solution

WHAT we must satisfy at once: (i) exactly two terms, (ii) highest power $6$ , (iii) coefficients summing to $0$ , (iv) not the zero polynomial. A term of power $6$ cannot cancel a term of a different power — cancellation needs like terms. So pick two different powers, e.g. $x^{6}$ and $- x^{2}$ : their coefficients $1$ and $- 1$ add to $0$ , yet they never combine. $P (x) = x^{6} - x^{2} .$

Two terms → binomial. ✓
Coefficients $1 + (- 1) = 0$ . ✓
Highest power $6$ , coefficient $1 \neq = 0$ → degree 6. ✓ Not zero polynomial. ✓

WHY it works: "coefficients add to $0$ " does not force cancellation — cancellation only happens between matching powers.

Exercise 4.2

Multiply $(2 x^{2} + 3) (x^{2} - 5 x + 1)$ and state the degree and type of the answer.

Recall Solution

WHAT — distribute each term of the first factor across the second (extended FOIL). $2 x^{2} \cdot (x^{2} - 5 x + 1) = 2 x^{4} - 10 x^{3} + 2 x^{2}$ $3 \cdot (x^{2} - 5 x + 1) = 3 x^{2} - 15 x + 3$ WHY add them and combine like terms: distribution over a sum, then merge equal powers. $2 x^{4} - 10 x^{3} + 2 x^{2} + 3 x^{2} - 15 x + 3.$ Combine the $x^{2}$ terms: $2 x^{2} + 3 x^{2} = 5 x^{2}$ . $= 2 x^{4} - 10 x^{3} + 5 x^{2} - 15 x + 3.$ Five terms → a (general) polynomial (not mono/bi/trinomial). Degree $= 4$ — because degrees add when you multiply: $2 + 2 = 4$ .

Exercise 4.3

Find real numbers $a$ and $b$ so that $a x^{2} + b x + a$ equals $(x + 3)^{2}$ after expansion, or explain why it's impossible.

Recall Solution

WHAT — expand the target $(x + 3)^{2} = x^{2} + 6 x + 9$ . WHY match coefficients: two polynomials are equal only if every matching power has equal coefficients (an identity, true for all $x$ ).

$x^{2}$ : $a = 1$ .
$x^{0}$ (constant): $a = 9$ .

These demand $a = 1$ and $a = 9$ at the same time — a contradiction. Conclusion: impossible. The form $a x^{2} + b x + a$ forces the leading and constant coefficients to be equal, but $(x + 3)^{2}$ has them $1$ and $9$ , which differ.

Level 5 — Mastery

Exercise 5.1

For $P (x) = (k - 4) x^{3} + 5 x^{2} - 2 x + 1$ , discuss the degree and type for every possible value of the number $k$ .

Recall Solution

The only tricky term is $(k - 4) x^{3}$ ; its very existence depends on whether $k - 4$ is zero.

Case A — $k \neq = 4$ : then $k - 4 \neq = 0$ , so the $x^{3}$ term survives. Four terms remain → general polynomial; highest power $3$ → degree 3.

Case B — $k = 4$ : then $k - 4 = 0$ , and $(k - 4) x^{3} = 0 \cdot x^{3} = 0$ vanishes. $P (x) = 5 x^{2} - 2 x + 1.$ Three terms → trinomial; highest power $2$ → degree 2.

WHY this matters: the leading coefficient must be non-zero to count. A single value of $k$ silently deletes the top term and drops the degree. Covering every case means checking exactly where the leading coefficient can hit zero.

Exercise 5.2

Classify $\frac{6 x ^{5} - 4 x ^{3} + 2 x ^{2}}{2 x ^{2}}$ (assume $x \neq = 0$ ) and give its degree.

Recall Solution

WHAT — divide each term of the top by $2 x^{2}$ (allowed because the divisor is a single monomial): $\frac{6 x ^{5}}{2 x ^{2}} - \frac{4 x ^{3}}{2 x ^{2}} + \frac{2 x ^{2}}{2 x ^{2}} .$ WHY subtract exponents: $\frac{x ^{m}}{x ^{n}} = x^{m - n}$ (Exponent Rules) — cancelling shared copies of $x$ .

$\frac{6 x ^{5}}{2 x ^{2}} = 3 x^{3}$
$\frac{4 x ^{3}}{2 x ^{2}} = 2 x^{1} = 2 x$
$\frac{2 x ^{2}}{2 x ^{2}} = 1$

$= 3 x^{3} - 2 x + 1.$ The result has three terms → trinomial, degree 3. (Note: dividing by the variable $x^{2}$ is fine here only because it divides evenly; the original fraction itself isn't a polynomial until simplified.)

Exercise 5.3

The zero polynomial $Z (x) = 0$ . State how many terms it "has," its degree, and why it is the lone exception.

Recall Solution

WHAT: $Z (x) = 0$ has no term with a non-zero coefficient — you cannot point to any power of $x$ that is genuinely present.

Type: it fits none of monomial/binomial/trinomial, since those need $1$ , $2$ , or $3$ non-zero terms.
Degree: undefined (some texts write $- \infty$ ).

WHY undefined and not $0$ : degree means "highest exponent with a non-zero coefficient." The constant $5$ has one such term ( $5 x^{0}$ ) so its degree is $0$ . But $0$ has zero such terms — there is nothing to take the maximum of. An empty search has no answer, hence undefined.

Recall Quick self-check

A polynomial has 3 non-zero terms and highest power 8. Name and degree? ::: Trinomial, degree 8. Why does $x^{6} - x^{2}$ stay a binomial despite coefficients $1$ and $- 1$ ? ::: They are different powers, so they never combine. Degree of the zero polynomial? ::: Undefined (or $- \infty$ ).