Adding and Subtracting Polynomials: Free Response

The four terms have degrees $4$, $3$, $1$, and $0$ (reading $9x$ as $9x^1$ and $-11$ as $-11x^0$). Sorting from the highest degree down to the lowest, with each term keeping its own sign:

$p(x) = -5x^4 + 2x^3 + 9x - 11.$

Only the ORDER of the terms changed; the value of $p(x)$ at any $x$ is the same as it was before.

Standard form puts the highest-degree term first, so each feature can be read straight off it:

$p(x) = \underbrace{-5x^4}_{\text{degree }4} + \underbrace{2x^3}_{\text{degree }3} + \underbrace{9x}_{\text{degree }1} + \underbrace{-11}_{\text{degree }0}.$

The degree of $p$ is the largest of these four, $4$. The leading term is that degree-$4$ term, $-5x^4$, so the leading coefficient is $-5$. The constant term, the $x^0$ term, is $-11$.

The leading coefficient is defined by DEGREE, not by writing order. In the original expression, the first term written down happens to be $9x$, whose coefficient is $9$, but $9x$ only has degree $1$; it is nowhere near the highest power present.

Once $p$ is sorted into standard form,

$p(x) = -5x^4 + 2x^3 + 9x - 11,$

the term of highest degree is visibly $-5x^4$, so the leading coefficient is its coefficient, $-5$. Reading '$9$' off as the leading coefficient mistakes the first-written term for the highest-degree term, exactly the trap that writing a polynomial in standard form before naming anything is meant to avoid.

A leading plus sign changes nothing inside the parentheses, so drop them and group by power:

$4x^3 - 6x + 9 + 2x^2 + 5x - 3.$

Only the $x$-terms and the constants have a partner to combine:

$4x^3 + 2x^2 + (-6x+5x) + (9-3) = 4x^3+2x^2-x+6.$

The $x^3$ term and the $x^2$ term each came from only one of the two polynomials, so they carry straight down unchanged.

Compare the two polynomials term by term. The first, $4x^3-6x+9$, has degrees $3$, $1$, $0$. The second, $2x^2+5x-3$, has degrees $2$, $1$, $0$.

Degree $3$ shows up only in the first polynomial, and degree $2$ shows up only in the second, so each of those two terms carries straight down into the sum with no partner:

$4x^3 \;\text{(no partner)}, \qquad 2x^2 \;\text{(no partner)}.$

Degree $1$ and degree $0$ each show up in BOTH polynomials, which is exactly why the $x$-terms and the constants were the ones that combined in part A.

A polynomial's definition places two requirements on every term: its exponent must be a nonnegative whole number, and its coefficient must be a real number. Check what addition does to each.

Exponents: combining like terms never changes the shared power, it only merges terms that already have it,

$(-6x)+(5x) = (-6+5)x = -x;$

a term with no partner, like $4x^3$ or $2x^2$ above, simply carries down with its own exponent untouched. Every exponent in the sum is therefore one that was already present, still a nonnegative whole number.

Coefficients: adding two real numbers, whether they are matching coefficients such as $-6$ and $5$, or any other pair, always produces another real number.

Since neither requirement is ever broken, the sum of $4x^3-6x+9$ and $2x^2+5x-3$, or of any two polynomials at all, satisfies the definition of a polynomial. That is what it means to say the polynomials are closed under addition.

Test each line against the one before it.

Line 1 distributes the minus sign to every term of the second polynomial, correctly: sound.

Line 2 combines like terms from Line 1. The two $x^3$ terms cancel ($5x^3-5x^3=0$), the $x^2$ terms give $-2x^2-3x^2=-5x^2$, and the constants give $0+4=4$:

$-5x^2+x+4.$

That matches Line 2 exactly: sound.

Line 3 is where the chain breaks. It asserts the difference has degree $3$ by appealing to 'a difference always keeps the higher of the two degrees,' but Line 2's own expression, $-5x^2+x+4$, has no $x^3$ term at all: its degree is $2$. Line 3 contradicts the work directly above it. The rule it quotes is not universal; it is guarded, and this pair of polynomials is exactly the exception, since both have degree $3$ with the same leading term $5x^3$, which cancels in the subtraction.

Line 2 already gives the fully simplified difference; nothing further needs combining. It is in standard form,

$-5x^2+x+4,$

with degree $2$. Its leading term is $-5x^2$, so its leading coefficient is $-5$, not the $5$ that Line 4 reports.

The rule that survives every case is guarded, not absolute: the degree of a difference is at most the larger of the two degrees, and drops below it only when the two polynomials share that same degree AND their leading terms cancel.

Check the condition here. Both $(5x^3-2x^2+x)$ and $(5x^3+3x^2-4)$ have degree $3$, and the leading term of each is $5x^3$. Subtracting lines them up with opposite signs,

$5x^3 - 5x^3 = 0,$

so the leading terms cancel exactly. Both halves of the guarded condition hold, which is why the degree drops to $2$ instead of staying at $3$.

Substitute the two given polynomials and distribute the minus sign to EVERY term of $C(x)$, not only its first:

$P(x) = \left(5x^2+40x\right) - \left(2x^2+15x+300\right) = 5x^2+40x-2x^2-15x-300.$

Combine like terms:

$(5x^2-2x^2)+(40x-15x)-300 = 3x^2+25x-300.$

The $3x^2$, the $25x$, and the $-300$ all keep their own sign; only the $2x^2$, $15x$, and $300$ of $C(x)$ had their signs flipped by the subtraction.

Substitute $x=10$ into the standard-form polynomial from part A, not into $R$ and $C$ separately:

$P(10) = 3(10)^2+25(10)-300 = 300+250-300.$

$P(10) = 250.$

The result is positive, so at $10$ units sold the business has a profit that month.

$R$ and $C$ each have degree $2$: their highest powers are $5x^2$ and $2x^2$. The guarded rule for a difference says the result has degree at most the larger of the two, here $2$, and drops below that only if the two degree-$2$ leading terms are equal and cancel. Check whether they do:

$5x^2 - 2x^2 = 3x^2 \neq 0.$

Since the leading terms do not cancel, nothing forces the degree down, and $P$ keeps degree $2$ exactly, which matches the $3x^2$ term found in part A without needing to redo the whole subtraction.

Drop the parentheses, since the connecting sign is a plus, and group by power:

$\left(2x^4-3x^2+x-5\right)+\left(-2x^4+5x^2-4\right) = (2x^4-2x^4)+(-3x^2+5x^2)+x+(-5-4).$

The $x^4$ terms cancel completely, since $2+(-2)=0$:

$f(x)+g(x) = 2x^2+x-9.$

Subtract $h(x)=3x^3-x+2$ from the sum found in part A, flipping the sign of every one of its three terms:

$\left(2x^2+x-9\right) - \left(3x^3-x+2\right) = 2x^2+x-9-3x^3+x-2.$

Combine like terms, in standard form:

$k(x) = -3x^3+2x^2+(x+x)+(-9-2) = -3x^3+2x^2+2x-11.$

Read off each polynomial's degree from its own leading term: $f$ and $g$ both have degree $4$ (leading terms $2x^4$ and $-2x^4$), and $h$ has degree $3$ (leading term $3x^3$). A first guess might say $k$ inherits the largest of these, degree $4$.

But the guarded rule for a sum applies at the FIRST step: $f(x)+g(x)$ has the same degree, $4$, in both polynomials, and their leading terms $2x^4$ and $-2x^4$ are opposite, so

$2x^4 + (-2x^4) = 0,$

and the sum drops to degree $2$, as part A found. That degree-$4$ pair is gone before $h$ is ever involved, so the second step is really a degree-$2$ polynomial minus a degree-$3$ one: the guarded rule this time gives degree $3$, the larger of $2$ and $3$, with no further cancellation, matching part B's $-3x^3$ term. The degree of the whole expression is decided step by step, not by looking at the three original degrees $4$, $4$, $3$ and taking the largest.