Expanding Brackets
p.1. Rules of Expanding Parentheses
Expanding parentheses in an expression is an identity transformation that allows removing a pair of parentheses.
For addition:
b + (-a) = b - a
b - (-a) = b + a
(-a) + b = -a + b
For multiplication:
(-a)b = -ab
a(-b) = -ab
(-a)(-b) = ab
Remove the parentheses without changing the signs of the terms inside, if:
1. there is a «+» sign in front of the left parenthesis: a + (b - c + d) = a + b - c + d
2. the expression begins with a parenthesis without any signs:
(a+b-c)+d=a+b-c+d
Remove the parentheses changing the signs of the terms inside to the opposites, if:
1. there is a «-» sign in front of the left parenthesis: a - (b - c + d) = a - b + c - d
2. the expression begins with a «-» sign in front of a parenthesis:
-(a + b - c) + d = -a - b + c + d
Using these rules, you can expand parentheses in any expression. For example:
a +b(c + d - f + e) = a + bc + bd - bf + be
a - b(c + d - f + e) = a + bc + bd - bf + be
-a(b + c - d) + f = -ab - ac + ad + f
(a + b)(c - d) = a(c - d) + b(c - d) = ac - ad + bc - bd
(-a + b)(c + d) = -a(c + d) + b(c + d)= -ac - ad + bc + bd
$(a + b)^2 = (a + b)(a + b) = a(a + b) + b(a + b) = a^2 + ab + ab + b^2=$
$=a^2 + 2ab + b^2$
p.2. Order of Expanding Parentheses
The order of expanding parentheses is consistent with the order of operations:
- first, raise the polynomials in the parentheses to natural powers;
- then, carry out multiplication and division from left to right;
- finally, when only the terms remain inside, expand the parentheses and combine like terms.
Example 1. Expand the parentheses and simplify the expression:
-(2a + 5b) + (3a - 2b + 1) - (2a + 4) = -2a - 5b + 3a - 2b + 1 - 2a - 4 = (-2a + 3a - 2a) + (-5b - 2b) + (1 - 4) = -a - 7b - 3
Example 2. Prove that the numerical value of the expression 3(2a - 7) - (a - (5a + 4)) is negative for any a.
Proof:
● 3(2a - 7) - (a - 5(a + 1)) = 6a - 21 - a + 5(a + 1) = 6a - 21- a + 5a + 5 = (6a - a + 5a) + (-21 + 5) = 0 - 16 = -16
Thus, the numerical value of the expression does not depend on the a value, and always negative.
Q.E.D. ○
Example 3. Find the cube of the difference $(2a-1)^3$
Solution:
$(2a - 1)^3 = (2a - 1)(2a - 1)(2a - 1)=(2a - 1)(2a(2a - 1) - (2a - 1))=$
$=(2a - 1)(4a^2 - 2a - 2a + 1)=(2a - 1)(4a^2 - 4a + 1)=$
$=2a(4a^2 - 4a + 1) - (4a^2 - 4a + 1)=8a^3 - 8a^2 + 2a - 4a^2 + 4a - 1=$
$=8a^3 - 12a^2 + 6a - 1$
Answer: $8a^3 - 12a^2 + 6a - 1$
p.3. Expanding Absolute Value Brackets
For a variable:
$ |x| = \left[ \begin{array}{cc} {x,}&{x \ge 0}\\ {-x,}&{x<0} \end{array} \right. $
For an expression:
$ |f(x)| = \left[ \begin{array}{cc} {f(x),}&{f(x) \ge 0}\\ {-f(x),}&{f(x)<0} \end{array} \right. $
Example 4. Find the solutions of the equation 2(x + 1) - 3|x - 1| = -1
Solution:
First, we find the x value that returns 0 from the absolute value brackets:
$x - 1 = 0 \Rightarrow x = 1$
Therefore, the brackets are expanded as follows:
$ |x-1| = \left[ \begin{array}{cc} {x-1,}&{f(x) \ge 1}\\ {-(x-1),}&{f(x)<1} \end{array} \right. $
Now, we solve two equations with constraints:
$$ |f(x)| = \left[ \begin{array}{cc} {\left\{ \begin{array}{c} 2(x+1)-3(x-1)=-1 \\ x \ge 1 \end{array} \right.} \\ {\left\{ \begin{array}{c} 2(x+1)+3(x-1)=-1 \\ x < 1 \end{array} \right.} \end{array} \right. \Rightarrow \left[ \begin{array}{cc} {\left\{ \begin{array}{c} 2x+2-3x+3=-1 \\ x \ge 1 \end{array} \right.} \\ {\left\{ \begin{array}{c} 2x+2+3x-3=-1 \\ x < 1 \end{array} \right.} \end{array} \right. \Rightarrow $$
$$ \Rightarrow \left[ \begin{array}{cc} {\left\{ \begin{array}{c} -x=-6 \\ x \ge 1 \end{array} \right.} \\ {\left\{ \begin{array}{c} 5x=0 \\ x < 1 \end{array} \right.} \end{array} \right. \Rightarrow \left[ \begin{array}{cc} {\left\{ \begin{array}{c} x=6 \\ x \ge 1 \end{array} \right.} \\ {\left\{ \begin{array}{c} x=0 \\ x < 1 \end{array} \right.} \end{array} \right. \Rightarrow \left[ \begin{array}{cc} {x_1=0}\\ {x_2=6} \end{array} \right. $$
Answer: $x_1=0,x_2=6$
Example 5. Is the equality $|a - b| \cdot |b - a| = (a - b)^2$ an identity?
Solution:
Expand the absolute value brackets:
$$|a - b| = \left[ \begin{array}{cc} {a-b,}&{a \ge b}\\ {b-a,}&{a < b} \end{array} \right. , |b - a| = \left[ \begin{array}{cc} {b-a,}&{a \ge b}\\ {a-b,}&{a < b} \end{array} \right. $$
Find the product:
$$|a - b| \cdot |b - a|= \left[ \begin{array}{cc} {(b-a)(b-a),}&{a < b}\\ {(a-b)(a-b),}&{a \ge b} \end{array} \right. $$
$$(b - a)(b -a)=-(a - b) \cdot (-(a - b)) = (a - b)(a - b)=(a - b)^2$$
Therefore:
$$|a - b| \cdot |b - a| = \left[ \begin{array}{cc} {(a-b)^2,}&{a < b}\\ {(a-b)^2,}&{a \ge b} \end{array} \right. = (a - b)^2, ∀a \in \mathbb R and b \in \mathbb R $$
Thus, the equality is an identity.
Answer: Yes
