Worked examples — Polynomials — degree, types (monomial, binomial, trinomial)

This page is the practice gymnasium for the parent topic on polynomials. We are going to hunt down every kind of situation a polynomial problem can put you in — friendly ones, sneaky ones, and the ones that make people lose marks in exams.

Before we solve anything, we build a map of all the traps. Then every worked example gets a label saying which trap it defuses. By the end, there is no scenario you haven't seen.

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The scenario matrix

Each row below is a distinct case class. The last column says which worked example lands on it. If you can handle one example per row, you can handle the whole topic.

#	Case class	What makes it tricky	Example
A	Clean degree + type read-off	nothing — the warm-up	Ex 1
B	Hidden simplification — like terms cancel/combine	count/degree changes after tidying	Ex 2
C	Zero leading coefficient (degenerate)	the top term vanishes, degree drops	Ex 3
D	Constant & zero polynomial (degree edge cases)	degree $0$ vs undefined	Ex 4
E	Multiply monomials/binomials — degree grows	exponents add (a common sign/rule slip)	Ex 5
F	Divide monomials — degree shrinks, sign of exponent	subtracting exponents, can hit $0$ or negative	Ex 6
G	Type changes shape (binomial² → trinomial)	expanding relabels the polynomial	Ex 7
H	Word problem — build the polynomial from a story	translating words → algebra, then classify	Ex 8
I	Exam twist — unknown exponent / "for what value…"	reasoning about the rules themselves	Ex 9
J	Big-degree / limiting behaviour — which term dominates	intuition about "who wins" as $x$ grows	Ex 10

Prerequisites you may want open: Combining Like Terms, Exponent Rules, FOIL Method, Factoring.

The picture above shows the two independent "rulers" every polynomial is measured by: how many terms (left, counting blocks) and how tall the tallest block is (right, the degree). Keep them separate — mixing them up is the #1 mistake.

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Ex 1 — Case A: clean read-off (warm-up)

Forecast: Guess before reading on — how many terms, and what degree? Say it out loud.

1. Count the terms. They are $-4x^3$, $+2x$, and $-9$ — that's three terms.
   • Why this step? Type is decided purely by the number of terms after simplifying, and nothing here combines (different powers).
2. List each term's exponent. $-4x^3\to 3$, $2x\to 1$, $-9=-9x^0\to 0$.
   • Why this step? The exponent of each term is what "degree" measures; a bare constant secretly carries $x^0$.
3. Take the largest exponent. $\max (3,1,0)=3$.
   • Why this step? Degree = highest power with a non-zero coefficient. Here $-4\ne 0$, so the top term is real.

Answer: trinomial, degree $3$.

Verify: Three separate powers ($3,1,0$), all coefficients non-zero → 3 terms confirms trinomial. The tallest, $x^3$, gives degree $3$. ✓

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Ex 2 — Case B: hidden simplification

Forecast: It looks like a five-term, degree-2 polynomial. Trust that? (Don't.)

1. Group like terms (same variable-and-power): $(5x^2-5x^2)+3x+(7-1)$.
   • Why this step? Like terms must be merged before counting — the definition of a polynomial refers to its simplified form.
2. Combine. $5x^2-5x^2=0$, and $7-1=6$. We are left with $3x+6$.
   • Why this step? The $x^2$ terms cancel completely — the "degree-2 look" was an illusion.
3. Now count and rank. Two terms ($3x$, $6$); exponents $1$ and $0$; max $=1$.
   • Why this step? Only now is the expression in its true, simplified form.

Answer: binomial, degree $1$.

Verify: Plug $x=2$ into both forms. Original: $5(4)+3(2)-5(4)+7-1=20+6-20+7-1=12$. Simplified: $3(2)+6=12$. ✓ Same value → same polynomial.

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Ex 3 — Case C: zero leading coefficient (degenerate)

Forecast: Is the degree really $4$? Which term is "fake"?

1. Kill the phantom term. $0\cdot x^4=0$, so that whole term disappears.
   • Why this step? Degree needs a non-zero coefficient on the top power. A zero coefficient means the term does not exist.
2. Combine the real like terms. $3x^2-2x^2=x^2$, plus $5$. So $R(x)=x^2+5$.
   • Why this step? Same tidy-first rule as Case B.
3. Read the degree. Exponents $2$ and $0$, max $=2$.

Answer: binomial, degree $2$ — not $4$.

Verify: At $x=3$: written form $0\cdot 81+27-18+5=14$; simplified $9+5=14$. ✓ The $x^4$ contributed nothing, proving it was never really there.

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Ex 4 — Case D: constant and zero-polynomial edge cases

Forecast: Do constants have degree $0$, or "no degree"? Are $7$ and $0$ treated the same?

1. Rewrite a non-zero constant with its hidden power. $7=7x^0$ and $-\frac{1}{2}=-\frac{1}{2}{x}^0$, since $x^0=1$ for any $x\ne 0$.
   • Why this step? This shows the constant is a legal term with exponent $0$.
2. Assign degrees to (a) and (c). Highest (only) exponent is $0$, so both have degree $0$.
   • Why this step? Consistency — a constant fits the pattern once you expose the $x^0$.
3. Handle the zero polynomial separately. $B(x)=0$ has no term with a non-zero coefficient at all.
   • Why this step? There is literally nothing to measure, so its degree is undefined (some books write $-\infty$).

Answer: (a) degree $0$; (b) degree undefined; (c) degree $0$.

Verify: $7$ evaluates to $7$ everywhere (a flat, non-zero constant — a genuine $x^0$ term), while $0$ evaluates to $0$ everywhere (no term survives). Different objects → different degree status. ✓

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Ex 5 — Case E: multiplying (degrees add)

Forecast: For the first, do the exponents add or multiply? For the second, will it stay a binomial?

Part 1 — monomial × monomial.

1. Multiply coefficients, then variables. $3\times (-2)=-6$, and $x^2\cdot x^5=x^{2+5}=x^7$.
   • Why this step? Exponent rule $x^m\cdot x^n=x^{m+n}$ — we add exponents because multiplying stacks the factors: $x^2{x}^5$ is seven $x$'s multiplied.
2. Assemble. $-6x^7$.

Answer 1: monomial, degree $7$.

Part 2 — binomial × binomial (FOIL).

1. First·Outer·Inner·Last: $x\cdot x+x\cdot (-4)+4\cdot x+4\cdot (-4)=x^2-4x+4x-16$.
   • Why this step? Every term of one bracket must meet every term of the other.
2. Combine middle terms. $-4x+4x=0$, leaving $x^2-16$.
   • Why this step? This is the difference of squares — the middle cancels by design.

Answer 2: binomial, degree $2$.

Verify: $(3x^2)(-2x^5)$ at $x=2$: $(3\cdot 4)(-2\cdot 32)=12\cdot (-64)=-768$; and $-6x^7=-6\cdot 128=-768$. ✓ For part 2 at $x=5$: $(9)(1)=9$ and $25-16=9$. ✓

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Ex 6 — Case F: dividing (degrees subtract)

Forecast: What happens to the degree when you divide? What if the exponent hits $0$?

1. Divide coefficients, subtract exponents. $\frac{20}{5}=4$ and $\frac{x^6}{x^2}=x^{6-2}=x^4$.
   • Why this step? Exponent rule $\frac{x^m}{x^n}=x^{m-n}$ — dividing cancels matched factors, so we subtract. First result: $4x^4$.
2. Second quotient. $\frac{8}{8}=1$ and $\frac{x^3}{x^3}=x^{3-3}=x^0=1$.
   • Why this step? When exponents match, subtraction gives $x^0$, which equals $1$ — this is why $x^0=1$ is defined that way: it must make division consistent.
3. Assemble. $\frac{8x^3}{8x^3}=1$.

Answer: $4x^4$ (monomial, degree $4$); and $1$ (constant monomial, degree $0$).

Verify: At $x=2$: $\frac{20\cdot 64}{5\cdot 4}=\frac{1280}{20}=64$ and $4\cdot 16=64$. ✓ Second is trivially $1$ for any non-zero $x$. ✓

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Ex 7 — Case G: type changes shape (binomial² → trinomial)

Forecast: A binomial is squared — does it stay a binomial?

1. Write the square as a product. $(2x+3{)}^2=(2x+3)(2x+3)$.
   • Why this step? "Squared" means "times itself" — we can only FOIL a product.
2. FOIL. $(2x)(2x)+(2x)(3)+(3)(2x)+(3)(3)=4x^2+6x+6x+9$.
   • Why this step? Distribute every term against every term.
3. Combine like terms. $6x+6x=12x$, giving $4x^2+12x+9$.
   • Why this step? Merging the two $x$ terms is what produces a three-term result.

Answer: trinomial, degree $2$. A two-term thing squared became three terms — matching $(a+b{)}^2=a^2+2ab+b^2$ with $a=2x,b=3$.

Verify: At $x=1$: $(2+3{)}^2=25$ and $4+12+9=25$. ✓

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Ex 8 — Case H: word problem (build it, then classify)

Forecast: Guess the degree from the story: area of a square hints at $x^2$… but is it the top term?

1. Garden area. A square of side $x$ has area $x\cdot x=x^2$.
   • Why this step? Area of a square is side².
2. Two paths. Each path runs the full length $x$ with width $3$, so each has area $3x$; two of them give $3x+3x=6x$.
   • Why this step? Path is a rectangle, area = length × width; two identical paths add.
3. Patio. A fixed $4{\text{m}}^2$ square adds the constant $4$.
   • Why this step? It doesn't depend on $x$, so it's a constant ($4x^0$) term.
4. Total and classify. $A(x)=x^2+6x+4$. Three terms, top power $2$.

Answer: trinomial, degree $2$ (a quadratic in disguise).

Verify: Units: $x^2$ is m², $6x$ is m·(m) = m², $4$ is m² — all areas, consistent. ✓ Numeric: for $x=5$: garden $25$, paths $30$, patio $4$, total $59$; formula $25+30+4=59$. ✓

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Ex 9 — Case I: exam twist (reason about the rules)

Forecast: What single value of $k$ makes the top term vanish?

1. When is the $x^3$ term real? The degree is $3$ only if its coefficient $(k-2)\ne 0$.
   • Why this step? Degree needs a non-zero leading coefficient (Case C logic).
2. Solve $(k-2)=0$. That gives $k=2$. At $k=2$ the $x^3$ term dies, so it is not cubic.
   • Why this step? Setting the coefficient to zero removes the top term. Solve like a tiny linear equation.
3. What remains at $k=2$? $D(x)=5x-1$ — two terms.
   • Why this step? Once the cubic term is gone, count what's left: a binomial, degree $1$.

Answer: cubic for all $k\ne 2$; at $k=2$ it collapses to the binomial $5x-1$ (degree $1$).

Verify: Put $k=2$, $x=3$: $(0)(27)+15-1=14$; and $5(3)-1=14$. ✓ Put $k=5$, $x=1$: $(3)(1)+5-1=7$, a genuine cubic contribution present. ✓

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Ex 10 — Case J: limiting behaviour (who dominates?)

Forecast: For small $x$ the $-100x^2$ looks scary. Does it stay the boss as $x$ grows?

1. Evaluate each term at $x=1000$. $x^3=10^9$, $-100x^2=-100\cdot 10^6=-10^8$, and $+1$.
   • Why this step? Comparing actual sizes shows which term dominates.
2. Compare magnitudes. $10^9$ is ten times larger than $10^8$; the constant $1$ is negligible.
   • Why this step? The highest power grows fastest, so for large $x$ the degree term wins — this is exactly why degree describes a polynomial's long-run behaviour.
3. Conclude. $F(1000)=10^9-10^8+1=900,000,001$, dominated by $x^3$.

Answer: the $x^3$ (degree-$3$) term dominates for large $x$; the leading term always eventually wins, which is why the degree tells you the shape of the tails.

Verify: $10^9-10^8+1=900000001$, and the $x^3$ term alone is $10^9$ — within about $10\%$ of the total, confirming dominance. ✓

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Recall

Recall Which ruler decides "type", which decides "degree"?

Type is decided by ::: the number of terms after full simplification. Degree is decided by ::: the highest exponent with a non-zero coefficient.

Recall After simplifying

$5x^2-5x^2+3x+6$, what type is it? Binomial ($3x+6$) — the $x^2$ terms cancel. :::

Recall What is the degree of the zero polynomial vs the constant

$7$? Zero polynomial: undefined. Constant $7$: degree $0$. :::