Mathematics • Year 9 • Unit 1 • Lesson 12
Simplifying with the Product Rule
Build fluency multiplying algebraic terms — split every problem into THREE jobs: sign, coefficient and each variable. Move from a fully worked example through a guided one to eight independent practice problems.
1. I do — fully worked example
Read every line. Notice the THREE columns: sign, coefficient, and each variable handled separately.
Problem. Simplify $2x^2 y^3 \times 3x^4 y$.
Step 1 — Decide the sign.
Both terms are positive, so the answer is positive.
Reason: $(+) \times (+) = (+)$.
Step 2 — Multiply the coefficients.
$2 \times 3 = 6$
Reason: coefficients use normal multiplication, NOT index laws.
Step 3 — Add the $x$ indices.
$x^2 \times x^4 = x^{2+4} = x^6$
Reason: product rule — same base, add the indices.
Step 4 — Add the $y$ indices (remember $y = y^1$).
$y^3 \times y = y^{3+1} = y^4$
Reason: a lone $y$ counts as $y^1$ — the index of $1$ is implicit.
Step 5 — Combine.
$2x^2 y^3 \times 3x^4 y = 6 x^6 y^4$
Reason: coefficient out the front, then each variable with its added index.
Answer: $\mathbf{6 x^6 y^4}$.
2. We do — fill in the missing steps
Same structure as Section 1, but with the working faded. Fill in each blank. 4 marks
Problem. Simplify $3a^2 b \times 5 a^3 b^2$.
Step 1 — Sign: both terms positive, so the answer is __________________ .
Step 2 — Multiply the coefficients:
$3 \times 5 = \_\_\_\_\_$
Step 3 — Add the $a$ indices:
$a^2 \times a^3 = a^{\,\_\_\,+\,\_\_\,} = a^{\_\_\_}$
Step 4 — Add the $b$ indices (remember $b = b^{\_\_\_}$):
$b \times b^2 = b^{\,\_\_\,+\,\_\_\,} = b^{\_\_\_}$
Step 5 — Combine:
$3a^2 b \times 5 a^3 b^2 = \_\_\_\_\_\_\_\_$
3. You do — independent practice
Show your working under each problem. The first four are foundation (one variable, positive coefficients). The middle two are standard (signs or two variables). The last two are extension (combine ideas with a bracket expansion).
Foundation
3.1 Simplify $5 x^3 \times 4 x^2$. 1 mark
3.2 Simplify $4 a^3 b \times 2 a b^4$. 1 mark
3.3 Simplify $7 m^2 \times 3 m^5$. 1 mark
3.4 Simplify $2p \times 6 p^3 q^2$. 1 mark
Standard — with signs and two variables
3.5 Simplify $-3 p^2 \times 4 p^5$. 2 marks
3.6 Simplify $-5 m \times -6 m^4$. 2 marks
Extension — expand and simplify
3.7 Expand and simplify $3y(2y^2 + 5)$. 2 marks
3.8 Expand and simplify $-2a(3a^3 - 4)$. 3 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (faded $3a^2 b \times 5 a^3 b^2$)
Step 1: positive ($+\times+ = +$).
Step 2: $3 \times 5 = \mathbf{15}$.
Step 3: $a^{2+3} = \mathbf{a^5}$.
Step 4: $b = b^1$, so $b^{1+2} = \mathbf{b^3}$.
Step 5: $3a^2 b \times 5 a^3 b^2 = \mathbf{15 a^5 b^3}$.
3.1 — $5 x^3 \times 4 x^2$
$(5 \times 4) \times x^{3+2} = \mathbf{20 x^5}$.
3.2 — $4 a^3 b \times 2 a b^4$
$(4 \times 2) \times a^{3+1} \times b^{1+4} = \mathbf{8 a^4 b^5}$.
3.3 — $7 m^2 \times 3 m^5$
$(7 \times 3) \times m^{2+5} = \mathbf{21 m^7}$.
3.4 — $2p \times 6 p^3 q^2$
$(2 \times 6) \times p^{1+3} \times q^2 = \mathbf{12 p^4 q^2}$ ($q$ has no partner — its index of $2$ carries through unchanged).
3.5 — $-3 p^2 \times 4 p^5$
Sign: $(-)(+) = (-)$. Coefficients: $3 \times 4 = 12$. Indices: $p^{2+5} = p^7$. Answer: $\mathbf{-12 p^7}$.
3.6 — $-5 m \times -6 m^4$
Sign: $(-)(-) = (+)$. Coefficients: $5 \times 6 = 30$. Indices: $m^{1+4} = m^5$. Answer: $\mathbf{30 m^5}$. (Common slip: leaving it negative — two negatives multiplied make a positive.)
3.7 — $3y(2y^2 + 5)$
Distribute: $3y \cdot 2y^2 + 3y \cdot 5 = 6 y^{1+2} + 15 y = \mathbf{6 y^3 + 15 y}$. The two terms aren't like terms ($y^3$ vs $y^1$), so leave them.
3.8 — $-2a(3a^3 - 4)$
Distribute the $-2a$ to both terms.
$-2a \cdot 3a^3 = -6 a^{1+3} = -6 a^4$.
$-2a \cdot (-4) = +8a$ (negative times negative is positive).
Final: $\mathbf{-6 a^4 + 8 a}$.