Multiplying Polynomials

(a) $-6a^9$    (b) $7x^5$    (c) $4x^5$

This section gives the rule for multiplying two monomials: multiply the coefficients and add the exponents,

The exponents are added rather than multiplied because of the Product Rule for exponents from the earlier section: $x^n\cdot x^m=x^{n+m}$.

(a) $\left(3a^4\right)(-2a^5)$.

Coefficients: $3\times (-2)=-6$. A positive times a negative is negative.

Exponents: $4+5=9$.

(b) $(-x^2)(-7x^3)$.

First recover the hidden coefficient: $-x^2$ means $-1\cdot x^2$, so the coefficient is $-1$.

Coefficients: $(-1)\times (-7)=7$. A negative times a negative is positive.

Exponents: $2+3=5$.

The invisible $-1$ is the thing most often mishandled here. It is easy to see the two minus signs and forget that one of them carries a factor of $1$, not nothing.

(c) $\left(2\sqrt{2}x\right)\left(\sqrt{2}{x}^4\right)$.

Coefficients: $2\sqrt{2}\times \sqrt{2}$. Group the numbers so that the two roots meet:

since $\sqrt{2}\cdot \sqrt{2}=2$ by the definition of a square root from the earlier section on exponents.

Exponents: $1+4=5$, remembering that a bare $x$ means $x^1$.

Note how a pair of irrational coefficients has produced a perfectly ordinary whole-number coefficient. Always look for such simplifications before assuming the answer must stay messy.

$12x^5-8x^3+24x^2$

This section says to use the distributive property, multiplying the monomial by each term of the polynomial. There are three terms inside, so there will be three products.

Product with the first term.

Product with the second term. Keep the minus sign attached to the term it belongs to:

Product with the third term. The constant $6$ is $6x^0$, so the exponents add as $2+0=2$:

Adding the three products.

There are no like terms to collect, since the three powers $5$, $3$ and $2$ are all different, so this is the final answer.

Check with a number. Substitute $x=1$ into the original:

and into the answer:

The two commonest errors. The first is multiplying only the leading term, writing $12x^5-2x+6$; the distributive property reaches every term inside the bracket. The second is multiplying the exponents instead of adding them, writing $4x^2\cdot 3x^3=12x^6$. Exponents are multiplied only when a power is raised to a power, which is not what is happening here.

(a) $x^2+3x-10$    (b) $6y^2-5y-4$

FOIL, as this section explains, is a way of remembering the four products that arise when two binomials are multiplied: First, Outer, Inner, Last. Each term of the first bracket must meet each term of the second, and $2\times 2=4$ products result.

(a) $(x+5)(x-2)$. Read the second bracket as $x+(-2)$, so that the "last" term is $-2$ and its sign travels with it.

• First: $x\times x=x^2$
• Outer: $x\times (-2)=-2x$
• Inner: $5\times x=5x$
• Last: $5\times (-2)=-10$

Add the four results:

The middle two are like terms, so collect them: $-2x+5x=3x$. Hence

Check at $x=1$: the original gives $(6)(-1)=-6$, and the answer gives $1+3-10=-6$. ✓

(b) $(3y-4)(2y+1)$.

• First: $3y\times 2y=6y^2$
• Outer: $3y\times 1=3y$
• Inner: $(-4)\times 2y=-8y$
• Last: $(-4)\times 1=-4$

Add them:

Collect the middle terms: $3y-8y=-5y$. Hence

Check at $y=1$: the original gives $(-1)(3)=-3$, and the answer gives $6-5-4=-3$. ✓

A warning about FOIL. As this section stresses, FOIL is only a mnemonic for the distributive property, and it applies only when both factors are binomials. With three terms in a bracket there are six products, not four, and the acronym gives no help.

$2x^4-x^3+11x^2-4x+12$

Follow the method of the worked example in this section: distribute each term of the first polynomial across the whole of the second.

Step 1: split the first polynomial into its three terms.

Note that the middle term carries its minus sign with it.

Step 2: expand each piece.

Step 3: write all six products together.

Step 4: collect like terms, power by power.

• $x^4$: only $2x^4$.
• $x^3$: only $-x^3$.
• $x^2$: $8x^2+3x^2=11x^2$.
• $x$: only $-4x$.
• constant: only $12$.

Two checks.

Degree. This section states $\mathrm{deg}(A\cdot B)=\mathrm{deg}(A)+\mathrm{deg}(B)$. Here $2+2=4$, and our answer has degree $4$. ✓

Value. At $x=1$ the original gives $(2-1+3)(1+4)=4\cdot 5=20$, and the answer gives $2-1+11-4+12=20$. ✓

How many products should there be? Three terms times two terms gives $3\times 2=6$ products, and six is exactly what Step 3 shows. Counting them before you start is a good way to be sure none has been dropped.

(a) $9$    (b) $11$    (c) $6$

This section records the rule

so all we need is the degree of each factor, which is its largest exponent.

(a) The first factor has largest exponent $5$, so its degree is $5$. The second has largest exponent $4$, so its degree is $4$. The product therefore has degree

(b) The first factor $7x^2$ is a monomial of degree $2$. The second has largest exponent $9$, so degree $9$. The product has degree

(c) Both factors have degree $3$, so the product has degree

Why the rule works. The highest power in the product can only come from multiplying the highest power of one factor by the highest power of the other, and by the Product Rule for exponents those exponents add. In (a), the term of highest degree is $x^5\cdot 3x^4=3x^9$, and no other pairing can reach as high.

Why no cancellation can spoil it. In the earlier section on adding polynomials we saw that degrees can drop when leading terms cancel. That cannot happen here, because there is only one way to produce the top power — a single product of the two leading terms — so it has nothing to cancel against. Its coefficient is the product of the two leading coefficients, and a product of two non-zero numbers is never zero, by the Zero-Product Property.

So for multiplication the degree rule is exact, with no "at most" needed, unlike the rule for addition.

$11$

This section gives the rule: the coefficient of $x^i$ in a product is the sum of all products $a_j{b}_k$ for which $j+k=i$, where $a_j$ is the coefficient of $x^j$ in the first polynomial and $b_k$ that of $x^k$ in the second.

Step 1: list the coefficients.

From the first polynomial $x^3-2x^2+4x-1$:

From the second polynomial $3x^2+x-5$:

Take the signs from the polynomial exactly as written, and remember that $x^3$ has coefficient $1$, not none.

Step 2: find all pairs with $j+k=2$. The available values are $j=0,1,2,3$ and $k=0,1,2$. The pairs adding to $2$ are

The pair $(3,-1)$ is impossible, since there is no $k=-1$.

Step 3: work out each product.

Step 4: add them.

So the coefficient of $x^2$ is $11$.

Why the rule is true. A term $x^2$ can only arise from multiplying some $a_j{x}^j$ by some $b_k{x}^k$ where the exponents add to $2$, since the Product Rule for exponents adds them. Every such pairing contributes, and they are like terms, so their coefficients are added.

Why this is worth doing. The full product here has degree $3+2=5$ and six terms. Expanding it all to read off one coefficient means twelve multiplications; the method above needs three. On the section's own example with degrees $75$ and $60$, full expansion is out of the question and this is the only practical route.

A caution. Be systematic when listing the pairs. Missing one, such as $(0,2)$, changes the answer, and nothing in the arithmetic afterwards would reveal the omission.

$a=4$

Step 1: expand the left side, keeping $a$ as a letter. Using FOIL:

• First: $x\cdot x=x^2$
• Outer: $x\cdot 3=3x$
• Inner: $a\cdot x=ax$
• Last: $a\cdot 3=3a$

Adding,

The two middle terms are like terms — both are multiples of $x$ — so they combine. By the distributive property,

Hence

Step 2: compare with the required answer. We want this to be the same expression as

for every $x$, that is, an identity in the sense of the earlier section. Two polynomials agree for all $x$ exactly when their matching coefficients agree, so we compare power by power:

• coefficient of $x^2$: $1=1$, which is automatic;
• coefficient of $x$: $3+a=7$;
• constant term: $3a=12$.

Step 3: solve. From the constant terms, $3a=12$, so dividing both sides by $3$ gives

Step 4: confirm with the other equation. The middle coefficients require $3+a=7$, and with $a=4$ this reads $3+4=7$. ✓ Both conditions are met by the same value, which is a good sign that no error has crept in. Had the two disagreed, no value of $a$ would have worked.

Full check. Substituting $a=4$ into the original left side and expanding:

The technique. Matching coefficients of like powers is a powerful method that will be used repeatedly. It converts a single statement about polynomials into several simple statements about numbers.

(a) $2x^2+9x+4$    (b) $49$ square metres    (c) Degree $2$

(a) The area of a rectangle is length times width:

Both factors are binomials, so FOIL applies:

• First: $2x\cdot x=2x^2$
• Outer: $2x\cdot 4=8x$
• Inner: $1\cdot x=x$
• Last: $1\cdot 4=4$

Adding,

and collecting the like terms $8x+x=9x$:

(b) Substitute $x=3$, applying the exponent before the multiplication:

The area is $49$ square metres.

Check against the dimensions directly: the sides are $2\cdot 3+1=7$ metres and $3+4=7$ metres, and $7\times 7=49$. ✓ The garden happens to be square at this value of $x$.

(c) Each side is a polynomial of degree $1$, and this section states that degrees add under multiplication:

That matches the answer, whose largest exponent is $2$.

This is exactly what one should expect on physical grounds too. Length is a one-dimensional quantity and area is two-dimensional, so multiplying two lengths must produce something that grows like the square of the scale. If the answer had come out with degree $3$, it would have been describing a volume, and something would have gone wrong.

A remark on the constant term. The $4$ says that when $x=0$ the area is $4$ square metres — the garden would then measure $1$ by $4$ metres. Note that $x=0$ is geometrically acceptable here, but $x$ could not be very negative, since a side length cannot be negative.

(a) $x^2-16$

(a) Expanding by FOIL.

• First: $x\cdot x=x^2$
• Outer: $x\cdot 4=4x$
• Inner: $(-4)\cdot x=-4x$
• Last: $(-4)\cdot 4=-16$

Adding the four products:

The two middle terms are like terms, and they are exact opposites:

They cancel completely, leaving

(b) Comparison with the formula. The formula from the first section of this chapter,

with $a=x$ and $b=4$ gives

the same answer. So the general formula and the FOIL expansion agree, as they must.

What happens to the middle terms. FOIL produces four terms, and in the general case the Outer and Inner products are

These are opposites, so they add to zero and vanish. That is precisely why the answer has only two terms instead of the usual three — and it is why the formula is worth remembering as a shortcut.

When does this happen? Only when the two binomials are identical apart from the sign between the terms, as in $(x-4)$ and $(x+4)$. Change one number and the cancellation fails: for instance $(x-4)(x+5)$ gives Outer $5x$ and Inner $-4x$, which combine to $x$ rather than disappearing, leaving $x^2+x-20$.

Using it in reverse. Since the formula is an equality, it can also be read backwards to turn a difference of two squares into a product. That will be useful whenever a quantity such as $x^2-16$ needs to be written as a product.

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

The long multiplication method takes each term of one polynomial in turn, multiplies it across the whole of the other, and writes the partial products in rows with like powers lined up in columns. With three terms in each factor this is far less error-prone than working in a single line.

Row 1: multiply by $x^2$.

Row 2: multiply by $2x$.

Row 3: multiply by $-1$. Every sign inside flips:

Adding the columns. Line the three rows up by power and add each column separately.

• $x^4$: only Row 1 contributes, giving $2x^4$.
• $x^3$: Rows 1 and 2 give $-1$ and $4$, so $(-1+4)x^3=3x^3$.
• $x^2$: Rows 1, 2 and 3 give $3$, $-2$ and $-2$, so $(3-2-2)x^2=-x^2$.
• $x$: Rows 2 and 3 give $6$ and $1$, so $(6+1)x=7x$.
• constant: only Row 3 contributes, giving $-3$.

The result.

Two checks.

Degree. Both factors have degree $2$, so the product should have degree $2+2=4$. It does. ✓

Value. At $x=1$ the original gives $(1+2-1)(2-1+3)=2\cdot 4=8$, and the answer gives $2+3-1+7-3=8$. ✓

Why the layout helps. Three terms times three terms produces $3\times 3=9$ products, and nine separate terms strung along one line are easy to miscount or misalign. Arranging them in three rows of three, with columns by power, turns the final stage into ordinary column addition — exactly as with long multiplication of numbers, which is where the method gets its name.

The correct expansion is $x^3+5x^2+7x+2$.

What went wrong. FOIL names exactly four products — First, Outer, Inner, Last — because it was designed for two binomials, where $2\times 2=4$ products arise. Here the second factor is a trinomial, so the correct count is

and the student produced only four. Two products were never formed at all: the ones involving the final term $1$ of the trinomial.

This section warns about exactly this, noting that FOIL is only a way of visualising the distributive property, and that once there are more terms the acronym no longer applies. The principle that always applies is: multiply every term of the first polynomial by every term of the second.

The correct expansion. Distribute each term of $(x+2)$ across the whole trinomial.

That is six products in all. Now add them:

Collect like terms:

• $x^3$: only one, giving $x^3$.
• $x^2$: $3x^2+2x^2=5x^2$.
• $x$: $x+6x=7x$.
• constant: $2$.

Check at $x=1$. The original gives $(3)(1+3+1)=3\cdot 5=15$, and the answer gives $1+5+7+2=15$. ✓ Note that the student's expression gives $1+3+2+6=12$, which fails the check — so the error would have been caught at once.

A reliable habit. Before starting, multiply the number of terms in each factor to find how many products to expect, then count them when finished. Two by three means six; three by three means nine. Any shortfall signals a missing product.

The correct expansion is $x^2+6x+9$; the lost term is $6x$.

The correct expansion. The notation $(x+3{)}^2$ means $(x+3)$ multiplied by itself:

That is a product of two binomials, so FOIL applies:

• First: $x\cdot x=x^2$
• Outer: $x\cdot 3=3x$
• Inner: $3\cdot x=3x$
• Last: $3\cdot 3=9$

Adding,

and collecting the like terms $3x+3x=6x$:

The lost term. The student's answer keeps only the First and Last products and throws away the Outer and Inner ones. Those two are what produce the middle term $6x$, and it is exactly this term that has been lost.

A numerical demonstration. Substitute $x=1$:

Since $16\ne 10$, the student's rule is false. Our expansion gives $1+6+9=16$ ✓, matching the correct value.

The underlying mistake. The student has treated the exponent as though it could be applied to each part of the sum separately. As the earlier section on exponents established, an exponent may never be distributed over a sum: the rule $(bc{)}^n=b^n{c}^n$ concerns a product, and there is no corresponding rule for $(b+c{)}^n$.

A useful mental picture. Think of a square of side $x+3$. It splits into a square of side $x$ (area $x^2$), a square of side $3$ (area $9$), and two rectangles measuring $x$ by $3$ (area $3x$ each). The student's answer accounts for the two squares but forgets the two rectangles, and those rectangles are the missing $6x$.

Sometimes true.

The answer is sometimes.

A case where it holds. Take $(x+1)(x+2)$. By FOIL,

which has three terms and so is a trinomial. Here the statement is true.

A case where it fails. Take $(x+2)(x-2)$. By FOIL,

The Outer and Inner products are exact opposites, so they cancel, leaving

which has only two terms and so is a binomial, not a trinomial. Here the statement is false.

Since both outcomes occur, the statement is sometimes true.

What decides it. FOIL always produces four products, and the Outer and Inner ones are like terms, so they combine into a single middle term. The count of terms in the answer therefore depends on whether that middle term survives:

• if the Outer and Inner products do not cancel, three terms remain and the result is a trinomial;
• if they cancel exactly, only two remain and the result is a binomial.

Cancellation happens precisely in the case of the difference of two squares, $(a-b)(a+b)=a^2-b^2$, met in the first section of this chapter.

Can the count fall to one? Not for binomials of the kind met here. The First product and the Last product are each a product of two non-zero coefficients, so by the Zero-Product Property neither can be zero, and they involve different powers of $x$, so they cannot combine with each other. At least two terms therefore always survive.

The lesson. Never predict the shape of an answer before doing the algebra. Carry out the expansion, collect the like terms, and count what is actually left.

$a_8{b}_2+a_7{b}_3+a_6{b}_4+a_5{b}_5+a_4{b}_6$

By the rule stated in this section, the coefficient of $x^{10}$ is the sum of all products $a_j{b}_k$ with

The work lies in finding which pairs are actually available, since $j$ and $k$ cannot take any values we please.

The permitted ranges. In the first polynomial the powers run from $0$ up to $8$, so

In the second they run from $0$ up to $6$, so

Finding the pairs. Since $k=10-j$, the condition $0\le k\le 6$ becomes

The right-hand inequality $10-j\le 6$ gives $j\ge 4$, and the left-hand one $10-j\ge 0$ gives $j\le 10$. Combined with $j\le 8$ from the first polynomial, we get

So the permitted pairs are

five in all.

The coefficient.

Why the obvious pairs are excluded. One might start with $j=10,k=0$ or $j=9,k=1$, but the first polynomial has no $x^{10}$ or $x^9$ term, so $a_{10}$ and $a_9$ do not exist. At the other end, $j=3,k=7$ fails because the second polynomial stops at $x^6$. Checking both ranges is the whole difficulty of this type of problem.

A check on the count. As $j$ runs from $4$ to $8$ there are $8-4+1=5$ values, matching the five terms listed.

A related question. The product has degree $8+6=14$, and the coefficient of the top power $x^{14}$ comes from the single pair $(8,6)$, so it is simply $a_8{b}_6$. There is only one way to reach the highest power, which is why no cancellation can reduce the degree.

(b) $x^3+8$

Part (a). Distribute each term of $(x+1)$ across the trinomial. Two terms times three terms gives $2\times 3=6$ products.

Adding all six terms:

Now collect like terms:

• $x^3$: only $x^3$.
• $x^2$: $-x^2+x^2=0$. These cancel.
• $x$: $x-x=0$. These cancel too.
• constant: $1$.

Everything in the middle disappears, leaving

Check at $x=2$: the original gives $(3)(4-2+1)=3\cdot 3=9$, and the answer gives $8+1=9$. ✓

Part (b). Now distribute $(x+2)$ across $x^2-2x+4$.

Adding and collecting:

• $x^3$: only $x^3$.
• $x^2$: $-2x^2+2x^2=0$.
• $x$: $4x-4x=0$.
• constant: $8$.

Again the middle vanishes:

Check at $x=1$: the original gives $(3)(1-2+4)=3\cdot 3=9$, and the answer gives $1+8=9$. ✓

What to notice. Both products collapse to just two terms, and in each case the answer is $x^3$ plus a cube: $1=1^3$ in part (a) and $8=2^3$ in part (b). The pattern is

with $b=1$ and $b=2$ in the two parts.

Why the middle terms die. Expanding the general form, the $x^2$ terms are $-b{x}^2$ and $+b{x}^2$, and the $x$ terms are $b^{2}x$ and $-b^{2}x$. Each pair consists of exact opposites, so each pair sums to zero. The trinomial is built precisely so that this happens — which is why its middle sign is negative and its last term is $b^2$.

A test of the pattern. Predict $(x+3)(x^2-3x+9)$. The rule says it should be $x^3+27$. Verifying at $x=1$: the product is $(4)(1-3+9)=4\cdot 7=28$, and $1+27=28$. ✓