show that (x+1)^2-(x-1)^2=4x

Let's verify whether

{(x + 1)}^{2} - {(x - 1)}^{2} = 4 x

⇥Let's evaluate:

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

⇥ⓘ Raising a sum to a positive integer power means multiplying it by itself as many times as the exponent indicates:

{(x + 1)}^{2}

(x + 1) \cdot (x + 1)

{(x - 1)}^{2}

(x - 1) \cdot (x - 1)

⇥

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

⇥ⓘ Multiply each term of the sum by the other factor, using the distributive law:

(x + 1) \cdot (x + 1)

x \cdot x

+ x \cdot 1

+ 1 \cdot x

+ 1 \cdot 1

- (x - 1) \cdot (x - 1)

- (x \cdot x

+ x \cdot (-1)

- 1 \cdot x

- 1 \cdot (-1))

⇥

x \cdot x

+ x \cdot 1

+ 1 \cdot x

+ 1 \cdot 1

- (x \cdot x

+ x \cdot (-1)

- 1 \cdot x

- 1 \cdot (-1))

⇥ⓘ Multiplying powers with the same base adds the exponents:

x \cdot x

x^{2}

■Perform multiplications before additions and subtractions:

1 \cdot 1

1

■With a minus sign before parentheses, change the sign of every term inside:

- (x \cdot x

+ x \cdot (-1)

- 1 \cdot x

- 1 \cdot (-1))

- x \cdot x

- x \cdot (-1)

+ 1 \cdot x

+ 1 \cdot (-1)

⇥

x^{2}

+ x

+ x

+ 1

- x \cdot x

- x \cdot (-1)

+ 1 \cdot x

+ 1 \cdot (-1)

⇥ⓘ Group like terms by adding their coefficients:

x + x - x \cdot x

2 x - x^{2}

x^{2} - x^{2}

0

■Multiplying by -1 gives the opposite:

- x \cdot (-1)

x

⇥

0

+ 2 x

+ 1

+ x

+ x

- 1

⇥ⓘ Adding 0 changes nothing:

0 + 2 x

2 x

■Start from

1

and move back

1

on the number line, down to zero:

1 - 1

0

■Group like terms by adding their coefficients:

2 x + x + x

4 x

⇥

4 x + 0

⇥ⓘ Adding 0 changes nothing:

4 x + 0

4 x

⇥

4 x

▶Both sides reduce to $4 x$ : ${(x + 1)}^{2} - {(x - 1)}^{2} = 4 x$ is indeed verified

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