Multiplication of algebraic expressions — monomial × polynomial, polynomial × polynomial: Question bank

Every reveal line works like this: read the left side (the question), think, then read the right side (the reasoning) after the :::.

Quick toolkit (read this first)

Before the traps, here are the four ideas they attack — each stated plainly so the references below actually carry weight.

Definition The four tools you will need

Distributive Property: to multiply a single term by a sum, multiply it by each addend and add: $a (b + c) = ab + a c$ . Picture a rectangle of width $a$ and length $b + c$ split into two smaller rectangles.
Combining Like Terms: you may only add terms with the same variable and exponent. $3 x$ and $- 8 x$ combine (both are " $x$ -blocks"); $x$ and $x^{2}$ do not, like apples vs. oranges.
Exponent Rules: multiplying like bases adds exponents ( $x^{2} \cdot x^{3} = x^{5}$ ); a power of a power multiplies them ( $(x^{2})^{3} = x^{6}$ ). Figure 3 below makes this concrete: on the left it draws $2$ copies of $x$ next to $3$ copies of $x$ and you simply count the boxes — $2 + 3 = 5$ copies, so the exponent adds; on the right it draws $3$ separate groups each holding $2$ copies, so you multiply $3 \times 2 = 6$ groups' worth. Adding lines them up, multiplying stacks groups — two different pictures.

Figure 1 — the FOIL grid for $(a + b) (c + d)$ :

Figure — Multiplication of algebraic expressions — monomial × polynomial, polynomial × polynomial

The $2 \times 2$ grid in Figure 1 is the whole idea: each cell is one product, so $(a + b) (c + d)$ has exactly four products $a c, a d, b c, b d$ before you combine anything.

Figure 2 — the four cells of $(a + b)^{2}$ , diagonal vs. cross:

Figure 3 — adding exponents vs. multiplying exponents:

Figure 4 — an identity can still take negative values, $(x + 2) (x - 2)$ :

Grids also generalise beyond binomials. A trinomial times a trinomial has $3$ rows and $3$ columns, so $3 \times 3 = 9$ cells — Figure 5 shows this larger grid so the "nine products" claim later on is something you can see.

Figure 5 — a $3 \times 3$ grid: trinomial × trinomial has nine cells:

True or false — justify

(a + b)^{2} = a^{2} + b^{2}

False. Squaring means

(a + b) (a + b)

, so every term meets every term — the off-diagonal cells in Figure 2 give the two cross terms

ab + ba = 2 ab

. The correct expansion is

(a + b)^{2} = a^{2} + 2 ab + b^{2}

(the

b^{2}

is right, but the missing

2 ab

is the whole point).

3 (x + y + z) = 3 x + 3 y + 3 z

True. The Distributive Property makes the

3

"touch" every addend inside the package, not just the first one.

(a + b) (c + d)

has exactly

4

product terms before combining.

True. As the Figure 1 grid shows, each of the

2

left terms pairs with each of the

2

right terms:

2 \times 2 = 4

cells, giving

a c, a d, b c, b d

Multiplying two binomials always gives a trinomial.

False. It gives four terms first; you only reach three if two of them combine. In

(x + 1) (x - 1) = x^{2} - 1

the Outer and Inner cells cancel, leaving only two terms.

x^{2} \cdot x^{3} = x^{6}

False. When multiplying like bases you add exponents:

x^{2 + 3} = x^{5}

(count the copies of

x

in Figure 3, left panel). Multiplying exponents is the separate power-of-a-power rule.

The order of multiplication changes the answer:

(x + 2) (x + 5) \neq = (x + 5) (x + 2)

False. Multiplication is commutative, so both give

x^{2} + 7 x + 10

. FOIL just visits the four cells in a different order.

(- 2 a) (- 3 a) = - 6 a^{2}

False. Negative times negative is positive, so the coefficient is

+ 6

: the answer is

6 a^{2}

5 x \cdot 0 \cdot (x + 7) = 5 x (x + 7)

False. Any product containing a factor of

0

0

. The whole expression collapses to

0

, no distribution survives.

(a + b) (a - b) = a^{2} - b^{2}

only works when

a > b

False. It is an identity — true for all values. The Inner (

- ab

) and Outer (

+ ab

) cells cancel regardless of size or sign.

Distributing

- 1

over

(x - 4)

gives

- x - 4

False. The

- 1

multiplies both terms:

- 1 \cdot x + (- 1) (- 4) = - x + 4

. The sign of the constant flips too.

Spot the error

3 x (2 x + 5) = 6 x^{2} + 5

. Where is the slip?

The

3 x

was distributed only to

2 x

, not to the

5

. The second product is

3 x \cdot 5 = 15 x

, so the answer is

6 x^{2} + 15 x

(x - 3) (x - 2) = x^{2} - 6

. Where is the slip?

Only the Last cell (the two constants,

- 3 \cdot - 2 = 6

) was kept, skipping the Outer and Inner cells. All four cells give

x^{2} - 2 x - 3 x + 6 = x^{2} - 5 x + 6

(x + 3)^{2} = x^{2} + 9

. Where is the slip?

Only the diagonal cells were kept; both cross terms were dropped.

(x + 3) (x + 3)

needs the middle

2 \cdot x \cdot 3 = 6 x

, giving

x^{2} + 6 x + 9

(x^{2})^{3} = x^{5}

. Where is the slip?

This is a power of a power, so exponents multiply:

x^{2 \cdot 3} = x^{6}

. Adding exponents is only for products like

x^{2} \cdot x^{3}

(compare the two panels of Figure 3).

- 2 a^{2} (3 a^{2} - 4 a) = - 6 a^{4} - 8 a^{3}

. Where is the slip?

The second product has two negatives:

(- 2 a^{2}) (- 4 a) = + 8 a^{3}

. The correct answer is

- 6 a^{4} + 8 a^{3}

(2 x + 3) (x - 4) = 2 x^{2} - 12

(I "cancelled the middle"). Where is the slip?

The Outer cell (

- 8 x

) and Inner cell (

3 x

) are not equal and opposite, so they do not cancel — they combine to

- 5 x

. Answer:

2 x^{2} - 5 x - 12

4 (x + 2) + 3 = 4 x + 20

. Where is the slip?

The student illegally pulled the outside

+ 3

into the parentheses, computing

4 (x + 2 + 3) = 4 (x + 5) = 4 x + 20

. But the

+ 3

is added after the multiplication, not multiplied by

4

. Correct:

4 (x + 2) + 3 = 4 x + 8 + 3 = 4 x + 11

Why questions

Why must the monomial touch every term inside the parentheses?

Because addition inside the brackets keeps the terms separate — there is no shortcut "through" a sum, so the Distributive Property reaches each addend independently (each split rectangle in the toolkit picture).

Why does

(a + b)^{2}

produce a middle term but

ab

alone does not?

Squaring fills all four cells of the Figure 2 table, producing two cross cells

ab

and

ba

; commutativity fuses them into

2 ab

(Combining Like Terms).

Why can we only combine

- 8 x

and

3 x

but not

6 x^{2}

and

15 x

Like terms must share the same variable and exponent.

x

and

x^{2}

are different "objects", like adding apples to oranges.

Why does polynomial × polynomial reduce to repeated use of one property?

Picture the grid growing: distributing the first bracket lays down the rows, and distributing inside each piece fills the columns of that row. So double-distribution is just "sweep every row across every column" — exactly what filling all cells of the Figure 5 grid looks like, no cell skipped.

Why does difference-of-squares lose its middle term?

(a + b) (a - b)

the Outer cell

- ab

and Inner cell

+ ab

are exact opposites, so they sum to zero, leaving

a^{2} - b^{2}

Why is "every term with every term" the safest rule for large polynomials?

It guarantees no cell of the grid is missed — a trinomial times a trinomial has the

3 \times 3 = 9

cells drawn in Figure 5, and FOIL (a

2 \times 2

trick only) would leave most of them empty.

Edge cases

What is

m (0)

when

m = 7 a^{3}

It is

0

. Multiplying any expression by zero annihilates it, regardless of how complicated

m

is.

What does

(x + 0) (x + 5)

equal, and what does the

0

do?

It equals

x^{2} + 5 x

. The

0

fills two cells with zero (

x \cdot 0 = 0

and

0 \cdot 5 = 0

), so it silently vanishes.

1 \cdot (a + 4 b)

a "real" distribution?

Yes — the monomial happens to be

1

, so each term is unchanged:

a + 4 b

. This is exactly the step that catches people in a trinomial like

(2 a - 3 b + 1) (a + 4 b)

What is

(x + 3)^{1}

, and why is there no cross term?

It is just

x + 3

. A first power is a single copy of the bracket, so there is no second bracket to form cells with — cross terms only appear from power

2

upward.

If a binomial is multiplied by a monomial of degree

0

(a plain number), does the degree change?

No. A constant multiplier scales coefficients but adds

0

to every exponent, so

5 (x^{2} + x) = 5 x^{2} + 5 x

keeps degree

2

What happens to

(a - b)^{2}

versus

(b - a)^{2}

They are equal.

(b - a)^{2} = (- (a - b))^{2}

, and squaring removes the sign, so both give

a^{2} - 2 ab + b^{2}

Can

(x + 2) (x - 2)

ever be negative?

Yes — it equals

x^{2} - 4

, negative whenever

- 2 < x < 2

(the shaded dip in Figure 4, e.g.

x = 0

gives

- 4

). An identity in form can still take negative values.