Lesson 3 ~40 min Polynomials · Path +85 XP

Adding, Subtracting and Multiplying

Adding, subtracting and multiplying polynomials uses no technique you have not already met. What is new is the bookkeeping at length, and the habit of predicting the degree, the leading coefficient and the constant term of the answer before writing a single line.

Today's hook: Multiply a polynomial of degree $3$ by one of degree $5$ and the answer has degree $8$, always, with no exceptions. Add them instead and the answer has degree $5$, unless it does not. One of those operations is completely predictable and the other is not, and the difference is worth understanding before you rely on either.

0/5QUESTS

Think First

warm-up

Let $A(x) = x^3 + 2x - 1$ and $B(x) = x^3 - 5x^2 + 2x$. Without expanding anything, write down what you expect the degree of $A(x) + B(x)$ to be, and then the degree of $A(x) - B(x)$. Now check. If your two predictions were the same, one of them is wrong.

Record your answer in your workbook.

The Big Idea

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Adding and subtracting means collecting like terms: same power with same power. Multiplying means every term of one polynomial times every term of the other, then collecting. The only genuinely new skill is staying organised while doing it.

$$\deg(PQ) = \deg P + \deg Q$$

Multiplication is predictable: the degree of the product is the sum of the degrees, the leading coefficient is the product of the leading coefficients, and the constant term is the product of the constant terms. Addition is not: the leading terms can cancel, so $\deg(P + Q)$ is at most the larger degree, and sometimes less.

$\deg(PQ) = \deg P + \deg Q$

Bracket before subtracting

Write the whole second polynomial in brackets, then distribute the minus to every term.

Use a grid

One cell per pair of terms. A missing product becomes a visibly empty cell.

Check the ends

Leading coefficients multiply, constant terms multiply. Both are one-second checks.

What You'll Master

objectives

Know

How to add, subtract and multiply polynomials and present the answer in descending powers
That the degree of a product is the sum of the degrees
That the degree of a sum is at most the larger degree, and can be less

Understand

Why the leading terms of a product can never cancel but those of a sum can
Why every term of one polynomial must meet every term of the other

Can Do

Expand a product of two polynomials accurately using a grid or systematic distribution
Predict the degree, leading coefficient and constant term of a result before expanding
Use those predictions as checks on a completed expansion

Words You Need

vocabulary

Like termsTerms with the same power of the variable, so they can be combined.

DistributeMultiply a bracket's contents term by term.

Partial productOne of the individual products formed when expanding, before collecting.

Descending powersWritten from the highest power down to the constant. The standard presentation.

ClosureThe property that adding, subtracting or multiplying polynomials always gives another polynomial.

Adding and Subtracting

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Adding is collecting like terms. Line the two polynomials up by power, including any missing ones, and add coefficient to coefficient.

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

Subtracting is the same, once the minus sign has been dealt with properly. Bracket the whole second polynomial first, then distribute:

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

$$= 2x^3 - 5x^2 + x + 4$$

Notice that the $+5x^2$ became $-5x^2$ and the $-2x$ became $+2x$. Every term changes sign, not just the first. Applying the minus to the leading term alone is the single most common error in this lesson, and it is invisible on the page unless you look for it.

Why a column layout helps

Writing the two polynomials one above the other, with a column for every power from the highest down to the constant, makes a missing term appear as a gap rather than as nothing at all. It is the same reason you learned to line up place value in primary arithmetic.

The Degree of a Sum Can Collapse

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If $\deg P = m$ and $\deg Q = n$, what is the degree of $P + Q$?

When $m \neq n$, the answer is clean: the larger of the two, since only one polynomial contributes a term of that power and there is nothing to cancel it.

When $m = n$, the leading terms might cancel. Take

$$P(x) = x^4 + 3x, \qquad Q(x) = -x^4 + 2$$

Both have degree $4$, but $P(x) + Q(x) = 3x + 2$, of degree $1$. Push it further and $P(x) = x^4 + 3x$ with $Q(x) = -x^4 - 3x$ gives the zero polynomial, which has no degree at all.

So the safe statement is

$$\deg(P + Q) \leq \max(\deg P, \deg Q)$$

with equality guaranteed only when the degrees differ, or when the leading coefficients do not cancel. Never assert the degree of a sum without checking the leading terms.

Multiplying: Every Term Meets Every Term

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To multiply two polynomials, multiply each term of the first by each term of the second, then collect. A polynomial with $3$ terms times one with $2$ terms produces $6$ partial products, and all six must appear.

$$(x^2 + 3x - 2)(x - 4)$$

Working through the grid in the diagram above:

$$x^3 - 4x^2 + 3x^2 - 12x - 2x + 8$$

$$= x^3 - x^2 - 14x + 8$$

The grid matters because it makes omission visible. Distributing in your head across a cubic times a quadratic means tracking $12$ products with nothing to check against; a grid with $12$ cells cannot silently lose one.

Sign discipline is the other half. In the cell for $3x$ times $(-4)$, the answer is $-12x$; in the cell for $(-2)$ times $(-4)$, it is $+8$. Carry each term's sign into its cell and the collecting step becomes ordinary addition.

The Degree of a Product Never Collapses

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Multiplication is far better behaved than addition. If $P$ has leading term $ax^m$ and $Q$ has leading term $bx^n$, then the highest power the product can contain is

$$(ax^m)(bx^n) = ab\,x^{m+n}$$

and nothing else in the expansion produces $x^{m+n}$, because every other partial product uses a lower power from at least one of the two. So there is nothing available to cancel it. The only way $ab\,x^{m+n}$ could vanish is if $ab = 0$, and a product of two non-zero real numbers is never zero. Therefore

$$\deg(PQ) = \deg P + \deg Q \quad \text{always}$$

Three free checks follow from the same reasoning, and each takes about a second:

The degree of the answer is the sum of the degrees. The leading coefficient of the answer is the product of the leading coefficients. The constant term of the answer is the product of the constant terms, since the only way to reach $x^0$ is to take the constant from both.

For $(x^2 + 3x - 2)(x - 4)$: degree $2 + 1 = 3$, leading coefficient $1 \times 1 = 1$, constant term $(-2)(-4) = 8$. Compare with $x^3 - x^2 - 14x + 8$ and all three agree. Two of the four coefficients are confirmed before any checking of the middle.

Common Pitfalls

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Subtracting a polynomial by changing the sign of its first term only.

Fix: bracket the whole polynomial being subtracted, then distribute the minus into every term. $-(x^3 + 5x^2 - 2x) = -x^3 - 5x^2 + 2x$.

Adding the exponents when adding terms, so that $3x^2 + 4x^2$ becomes $7x^4$.

Fix: adding like terms adds the coefficients and keeps the power: $7x^2$. Exponents add only when terms are multiplied.

Claiming the degree of a sum equals the larger degree, without checking whether the leading terms cancel.

Fix: it is at most the larger degree. Look at the two leading coefficients: only if they do not sum to zero is the degree what you expect.

Missing a partial product in a long expansion and never noticing.

Fix: count. An $m$-term polynomial times an $n$-term one gives exactly $mn$ partial products before collecting. If you wrote fewer, one is missing.

Watch Me Solve It · 3 examples

Watch Me Solve It · Subtracting carefully

+15 XP per step

PROBLEM

Given $A(x) = 4x^3 - 2x^2 + 7$ and $B(x) = 4x^3 + x^2 - 5x + 7$, find $A(x) - B(x)$ and state its degree.

1

Bracket the second polynomial

$A(x) - B(x) = (4x^3 - 2x^2 + 7) - (4x^3 + x^2 - 5x + 7)$

The bracket is what forces every term to receive the minus sign.
2

Distribute the minus into all four terms

$= 4x^3 - 2x^2 + 7 - 4x^3 - x^2 + 5x - 7$

Each sign flips: $+x^2 \to -x^2$, $-5x \to +5x$, $+7 \to -7$.
3

Collect like terms

$= (4 - 4)x^3 + (-2 - 1)x^2 + 5x + (7 - 7)$

$= -3x^2 + 5x$
4

State the degree, and note what happened

The degree is $2$, not $3$: both polynomials had degree $3$ and their leading terms cancelled exactly. This is why the degree of a difference must always be read off the answer, never predicted from the inputs.

Answer$A(x) - B(x) = -3x^2 + 5x$, of degree $2$

Watch Me Solve It · Expanding with a grid

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PROBLEM

Expand $(2x^2 - x + 3)(x^2 + 4x - 1)$, giving the answer in descending powers.

1

Predict the answer's shape first

$\deg = 2 + 2 = 4, \quad \text{leading coefficient} = 2 \times 1 = 2, \quad \text{constant} = 3 \times (-1) = -3$

Now you know the answer starts $2x^4$ and ends $-3$ before doing any work.
2

Form all nine partial products

$2x^4 + 8x^3 - 2x^2$

$-x^3 - 4x^2 + x$

$3x^2 + 12x - 3$

Three rows of three, one row per term of the first polynomial. Three times three is nine, and nine products are written.
3

Collect like powers

$x^3: \ 8 - 1 = 7, \qquad x^2: \ -2 - 4 + 3 = -3, \qquad x: \ 1 + 12 = 13$
4

Write the answer and check the ends

$2x^4 + 7x^3 - 3x^2 + 13x - 3$

Leading term $2x^4$ and constant $-3$ both match the prediction, so the two hardest-to-spot errors are already ruled out.

Answer$2x^4 + 7x^3 - 3x^2 + 13x - 3$

Watch Me Solve It · Working with the structure, not the terms

+15 XP per step

PROBLEM

$P(x)$ has degree $3$ with leading coefficient $-2$ and constant term $5$. $Q(x)$ has degree $4$ with leading coefficient $3$ and constant term $-1$. Without knowing anything else about them, state the degree, leading coefficient and constant term of (a) $P(x)Q(x)$, and (b) $P(x) + Q(x)$.

1

(a) Use the product rules

$\deg(PQ) = 3 + 4 = 7$

Degrees add, with no exceptions.
2

(a) Multiply the ends

$\text{leading coefficient} = (-2)(3) = -6, \qquad \text{constant} = (5)(-1) = -5$

The only source of $x^7$ is leading times leading, and the only source of $x^0$ is constant times constant.
3

(b) Compare the degrees for the sum

$\deg(P + Q) = 4$

The degrees differ, so nothing can cancel the $3x^4$: $P$ contributes no $x^4$ term at all. The leading coefficient is $3$.
4

(b) Add the constants

$5 + (-1) = 4$

Constant terms add directly. Note that had both degrees been $4$, part (b) would have been unanswerable without more information.

Answer(a) degree $7$, leading coefficient $-6$, constant term $-5$. (b) degree $4$, leading coefficient $3$, constant term $4$

Brain Trainer · 5 problems

Brain Trainer · Expand and collect

5 problems

Five items on the three operations. Work each one, then reveal the answer.

1 Simplify $(x^2 + 5x) + (3x^2 - 5x + 2)$.

Collect like terms; the $x$ terms cancel.$4x^2 + 2$
2 Simplify $(2x^3 - x) - (2x^3 + 4x - 6)$.

Distribute the minus into all three terms.$-5x + 6$
3 Expand $(x + 3)(x^2 - 2x + 5)$.

Six partial products, then collect.$x^3 + x^2 - x + 15$
4 $P$ has degree $5$ and $Q$ has degree $2$. State the degree of $P(x)Q(x)$.

Degrees add for a product, always.$7$
5 Without expanding, find the constant term of $(3x^2 - 7)(x^3 + 2x - 4)$.

Constant times constant: $(-7)(-4)$.$28$

Complete in your workbook.

Multiple Choice · 5 questions

MC1

Subtracting

+10 XP

$(x^2 + 2x - 1) - (x^2 - 3x + 4)$ simplifies to:

MC2

Degree of a product

+10 XP

If $\deg P = 6$ and $\deg Q = 3$, then $\deg(PQ)$ is:

MC3

Degree of a sum

+10 XP

$P$ and $Q$ both have degree $4$. The degree of $P + Q$ is:

MC4

Counting partial products

+10 XP

Expanding $(x^3 + 2x^2 - x + 4)(x^2 - 5)$ before collecting produces how many partial products?

MC5

Reading the ends

+10 XP

The constant term of $(4x^3 - x + 6)(2x^2 + 5x - 3)$ is:

Short Answer · 3 questions

The three operations

+15 XP

SHORT ANSWER

Let $A(x) = 3x^3 - x^2 + 4$ and $B(x) = x^3 + 2x^2 - x - 4$.
(a) Find $A(x) + B(x)$.
(b) Find $A(x) - B(x)$.
(c) Find the degree, leading coefficient and constant term of $A(x)B(x)$, without expanding.
(d) State the number of partial products that a full expansion of $A(x)B(x)$ would produce before collecting.

Write your working in your book.

Expand and verify

+15 XP

SHORT ANSWER

(a) Expand $(x^2 - 3x + 2)(2x^2 + x - 5)$, giving the answer in descending powers.
(b) Verify your leading coefficient and constant term using the shortcut rules.
(c) Verify the whole expansion by evaluating both the original product and your answer at $x = 1$.
(d) Explain why the check in (c) would not, on its own, prove the expansion is correct.

Write your working in your book.

Reasoning about degree

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SHORT ANSWER

(a) Give two polynomials of degree $5$ whose sum has degree $2$.
(b) Explain why no two polynomials of degree $5$ can have a product of degree $2$.
(c) $P$ has degree $m$ and $Q$ has degree $n$, with $m > n$. Prove that $\deg(P + Q) = m$ exactly, with no exceptions.
(d) A student writes: "The degree of $P - Q$ is $\deg P - \deg Q$." Give a counterexample and explain the confusion.

Write your working in your book.

Stretch Challenge · What the operations preserve

+25 XP

CHALLENGE

(a) Explain why adding, subtracting or multiplying two polynomials always produces another polynomial, referring to the definition from Lesson 1.
(b) Show that dividing one polynomial by another does not always produce a polynomial, and give the simplest example you can.
(c) Let $P$ and $Q$ be polynomials of degrees $m$ and $n$ with $m > n$, both with positive leading coefficients. Show that $P(x) + Q(x)$ and $P(x)$ have the same leading coefficient and the same degree, and explain what this means about their graphs far out to the right.

Quick Review

recap

Subtracting

Bracket first, then flip every sign

Multiplying

$m$ terms times $n$ terms gives $mn$ products

Product degree

Degrees add, always, with no cancellation

Sum degree

At most the larger, and sometimes less

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