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Mathematics, 04.05.2021 02:30 mackdoolittle1

Ophelia was asked whether the following equation is an identity: 2(x-3)(4x-1)=(4x-6)^2-4(6+x)2(xβˆ’3)( 4xβˆ’1)=(4xβˆ’6)
2
βˆ’4(6+x)2, left parenthesis, x, minus, 3, right parenthesis, left parenthesis, 4, x, minus, 1, right parenthesis, equals, left parenthesis, 4, x, minus, 6, right parenthesis, squared, minus, 4, left parenthesis, 6, plus, x, right parenthesis
She performed the following steps:
\begin{aligned} &\phantom{=}2(x-3)(4x-1) \xhookrightarrow{\text{Step }1}\quad&=(2x-6)(8x-2) \xhookrightarrow{\text{Step }2}\quad&=16x^2-4x-48x+12 \xhookrightarrow{\text{Step }3}\quad&=16x^2-48x+36-24-4x \xhookrightarrow{\text{Step }4}\quad&=(4x-6)^2-4(6+x) \end{aligned}
Step 1
​

Step 2
​

Step 3
​

Step 4
​

​

=2(xβˆ’3)(4xβˆ’1)
=(2xβˆ’6)(8xβˆ’2)
=16x
2
βˆ’4xβˆ’48x+12
=16x
2
βˆ’48x+36βˆ’24βˆ’4x
=(4xβˆ’6)
2
βˆ’4(6+x)
​

For this reason, Ophelia stated that the equation is a true identity.
Is Ophelia correct? If not, in which step did she make a mistake?
Choose 1
Choose 1

(Choice A)
A
Ophelia is correct.

(Choice B)
B
Ophelia is incorrect. She made a mistake in step 111.

(Choice C)
C
Ophelia is incorrect. She made a mistake in step 333.

(Choice D)
D
Ophelia is incorrect. She made a mistake in step 444.

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Answers: 2

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Ophelia was asked whether the following equation is an identity: 2(x-3)(4x-1)=(4x-6)^2-4(6+x)2(xβˆ’...
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