What Are the Parentheses Rules?

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On this page, we can see all the rules for parentheses in one spot.

Rule

\begin{array}{l} (a + b) (c + d) & = a \cdot c + a \cdot d \\ + b \cdot c + b \cdot d \\ a + (c + d) & = a + c + d \\ a - (c + d) & = a - c - d \\ a \cdot (b + c) & = a \cdot b + a \cdot c \end{array}

\begin{array}{l} (a + b) (c + d) & = a \cdot c + a \cdot d + b \cdot c + b \cdot d \\ a + (c + d) & = a + c + d \\ a - (c + d) & = a - c - d \\ a \cdot (b + c) & = a \cdot b + a \cdot c \end{array}

Note! Remember to write the terms in descending order! Then it will look like this:

2 x^{3} - 4 x^{2} + 5 x - 3

2 x^{3} - 4 x^{2} + 5 x - 3

Example 1

Evaluate $x + (- 2 x + 5)$

\begin{array}{l} x + (- 2 x + 5) & = x - 2 x + 5 \\ = - x + 5 \end{array}

Example 2

Evaluate $4 - (12 - 3 x^{2})$

\begin{array}{l} 4 - (12 - 3 x^{2}) & = 4 - 12 + 3 x^{2} \\ = 3 x^{2} - 8 \end{array}

Example 3

Evaluate $- 4 x^{2} (2 - x)$

\begin{array}{l} - 4 x^{2} (2 - x) & = - 8 x^{2} + 4 x^{3} \\ = 4 x^{3} - 8 x^{2} \end{array}

Example 4

Evaluate $(x + 1) (x - 2)$

\begin{array}{l} (x + 1) (x - 2) \\ = x^{2} - 2 x + x - 2 \\ = x^{2} - x - 2 \end{array}

\begin{array}{l} (x + 1) (x - 2) = x^{2} - 2 x + x - 2 = x^{2} - x - 2 \end{array}

Example 5

Evaluate $(3 x^{2} + y) (2 - x)$

\begin{array}{l} (3 x^{2} + y) (2 - x) \\ = 6 x^{2} - 3 x^{3} + 2 y - x y \\ = - 3 x^{3} + 6 x^{2} - x y + 2 y \end{array}

\begin{array}{l} (3 x^{2} + y) (2 - x) & = 6 x^{2} - 3 x^{3} + 2 y - x y \\ = - 3 x^{3} + 6 x^{2} - x y + 2 y \end{array}

Example 6

Evaluate $(2 x + 3) (4 x - 5)$

\begin{array}{l} (2 x + 3) (4 x - 5) \\ = 2 x \cdot 4 x + 2 x \cdot (- 5) \\ + 3 \cdot 4 x + 3 \cdot (- 5) \\ = 8 x^{2} + (- 10 x) + 12 x - 15 \\ = 8 x^{2} + 2 x - 15 \end{array}

\begin{array}{l} (2 x + 3) (4 x - 5) & = 2 x \cdot 4 x + 2 x \cdot (- 5) + 3 \cdot 4 x + 3 \cdot (- 5) \\ = 8 x^{2} + (- 10 x) + 12 x - 15 \\ = 8 x^{2} + 2 x - 15 \end{array}

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