Polynomials and Their Graphs: Practice

Degrees add when polynomials multiply, because the two leading terms produce the highest power and nothing cancels it.

$\deg = 2 + 3 = 5$

The leading term is $(x^2)(x^3) = x^5$, so the product is a degree-$5$ polynomial.

The leading term of a product is the product of the leading terms; no other pairing of terms reaches as high a power.

$(2x)(3x^2) = 6x^3$

Leading coefficients multiply ($2 \cdot 3 = 6$) and degrees add ($1 + 2 = 3$).

Parenthesize the input and take each power carefully: $(-2)^4 = 16$ and $(-2)^2 = 4$.

$f(-2) = 16 - 3(4) + 2(-2) - 6 = 16 - 12 - 4 - 6 = -6$

The even powers erase the minus sign; the answer $-38$ comes from wrongly treating $(-2)^4$ as $-16$.

Every power of $1$ equals $1$, so evaluating at $1$ adds up exactly the coefficients.

$f(1) = 7 - 4 + 2 - 9 + 3 = -1$

One substitution replaces five separate readings, signs included.

Only the leading term speaks far from the origin, and here it is $-4x^6$.

$n = 6 \text{ (even)}, \qquad a_n = -4 < 0$

An even degree makes the ends agree and the negative coefficient points them down, so both ends fall.

The large coefficient on $x^2$ shapes the middle of the graph but has no say far out, where the higher power always wins.

$n = 3 \text{ (odd)}, \qquad a_n = 5 > 0$

Odd degree with a positive leading coefficient falls on the left and rises on the right, no matter how big the lower coefficients are.

Combine like terms before reading the degree: the two cubic terms cancel.

$x^3 + 2x - x^3 + 7 = 2x + 7$

What remains is linear, so the degree is $1$, not $3$.

Each crossing is a real zero, and a polynomial of degree $n$ has at most $n$ real zeros.

$5 \text{ zeros} \;\Rightarrow\; n \ge 5$

Degree $5$ suffices, for example $x(x-1)(x-2)(x-3)(x-4)$, so the least possible degree is $5$.

A degree-$n$ graph has at most $n - 1$ turning points, so the turning points force the degree up.

$n - 1 \ge 4 \;\Rightarrow\; n \ge 5$

A degree-$5$ polynomial can indeed have $4$ turning points, so the least possible degree is $5$.

For even degree, $x^n$ is positive on both far ends, so both ends carry the sign of the leading coefficient.

$n = 4 \text{ even} \;\Rightarrow\; \text{ends agree}$

Whether both rise or both fall depends on that sign, which the degree alone does not reveal.

Count the terms and read the exponent separately.

$5x^4 \;\Rightarrow\; 1 \text{ term}, \quad \text{degree } 4$

One term makes it a monomial, and the exponent $4$ (not the coefficient $5$) is its degree.

Add the polynomials and watch the leading terms cancel.

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

The sum is linear, so its degree is $1$. A sum's degree is at most the larger degree, and cancellation can drop it lower.