Single brackets (OCR GCSE Maths): Revision Notes

Key Rule: You must multiply EVERYTHING inside the bracket by the term on the outside.

This rule is the foundation of expanding single brackets and is essential for simplifying algebraic expressions.

Example 1: Expanding with Positive Numbers

Problem: Expand $3(2a+6)$.

Solution:

• Step 1: Remember, the $3$ outside the bracket is multiplying both the $2a$ and the $6$.

  

• Step 2: Multiply$3$ by $2a:$

• Step 3: Multiply $3$ by $6$:

• Final Answer:

Explanation:

• The term outside the bracket ($3$) is positive, and both terms inside the bracket are positive, so the final expression is simply $6a+12$.

Example 2: Expanding with a Negative Number Inside the Bracket

Problem: Expand $5(7d−4)$.

Solution:

• Step 1: The $5$ outside the bracket multiplies both the $7d$ and the $−4$.

  

• Step 2: Multiply $5$ by $7d$:

• Step 3: Multiply $5$ by $−4$:

• Final Answer:

Explanation:

• The key here is to pay close attention to the negative sign before the $4$. When multiplying, a positive times a negative gives a negative, hence $−20$.

Example 3: Expanding with a Negative Outside the Bracket

Problem: Expand $−4(t+2)$.

Solution:

• Step 1: The $−4$ outside the bracket multiplies both the $t$ and the $2$.

  

• Step 2: Multiply $−4$ by $t$:

• Step 3: Multiply $−4$ by $2$:

• Final Answer:

Explanation:

• A negative number multiplied by a positive number results in a negative product. Here, $−4$ times $t$ is $−4t$, and $−4$ times $2$ is $−8$.

Example 4: Expanding with Two Negative Numbers

Problem: Expand $−10(2c−4)$.

Solution:

• Step 1: The $−10$ outside the bracket multiplies both the $2c$ and the $−4$.

  

• Step 2: Multiply $−10$ by $2c$

• Step 3: Multiply $−10$ by $−4$:

• Final Answer:

Explanation:

• Be careful with the signs: A negative times a positive gives a negative, but a negative times a negative gives a positive. Here, $−10$ times $2c$ is $−20c$, and $−10$ times $−4$ is $40$.

Example 5: Expanding with a Negative Term Inside the Bracket

Problem: Expand $5a(2b−c)$.

Solution:

• Step 1: The $5a$ outside the bracket multiplies both the $2b$ and the $−c$.

  

• Step 2: Multiply $5a$by $2b$:

• Step 3: Multiply $5a$ by $−c$:

• Final Answer:

Explanation:

• The term $5a$is distributed to both $2b$ and $−c$, resulting in $10ab$ and $−5ac$. The negative sign is critical to include in the final expression.

Example 6: Expanding with Three Terms Inside the Bracket

Problem: Expand 7ar$$(10st+2b−5).

Solution:

• Step 1: The $7ar$ outside the bracket multiplies all three terms inside the bracket: $10st$, $2b$, and $−5$.

  

• Step 2: Multiply $7ar$ by $10st$:

• Step 3: Multiply $7ar$ by $2b$:

• Step 4: Multiply $7ar$ by $−5$:

• Final Answer:

Explanation:

• The term $7ar$ is distributed to each term inside the bracket. Notice that the order of multiplication doesn't matter, but the signs and the arrangement of variables should be consistent.

Example 7: Expanding with Variables and Squares

Problem: Expand $4r(2r−9t)$.

Solution:

• Step 1: The $4r$ outside the bracket multiplies both the $2r$ and the $−9t$.

  

• Step 2: Multiply $4r$ by $2r$:

• Step 3: Multiply $4r$by $−9t$:

• Final Answer:

Explanation:

• The term $4r$is distributed to each term inside the bracket. When $4r$is multiplied by $2r$, the result is $8r²$ because $r×r=r²$. The second multiplication involves a negative sign, leading to a negative result.

Example 8: Expanding with Multiple Variables and Squares

Problem: Expand $2ab(4a−3ab²)$.

Solution:

• Step 1: The $2ab$ outside the bracket multiplies both the $4a$ and the $−3ab²$.

  

• Step 2: Multiply $2ab$by $4a$:

• Step 3: Multiply $2ab$by $−3ab²$:

• Final Answer:

Explanation:

• Here, the multiplication of $2ab$by $4a$results in $8a²b$ because $a×a=a²$. When multiplying $2ab$by $−3ab²$, we get $−6a²b³$ because the powers of both $a$ and $b$ are added together.