expand (3x+1)^2

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`x`²	`x`^□	log_□	√☐	^□√☐	≤	≥	□□	·	÷	`x`^◦	π
(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

☐²

x^☐

√☐

^□√☐

□□

log_□

θ

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ddx

≥	≤	·	÷	`x`^◦	(☐)	\|☐\|	(`f` ◦ `g`)	`f`(`x`)	ln	`e`^☐
(☐)^′	∂∂`x`	∫_□^□	lim	∑	sin	cos	tan	cot	csc	sec

simplify solve for inverse tangent line

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expand (3x+1)²

Solution

9x²+6x+1

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(3x+1)²

Apply Perfect Square Formula: (a+b)²=a²+2ab+b²

(3x+1)²=(3x)²+2· 3x· 1+1²

=(3x)²+2· 3x· 1+1²

Simplify (3x)²+2· 3x· 1+1²: 9x²+6x+1

=9x²+6x+1

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
▭ \|▭	)	.	0	=	+	`y`

expand (3x+1)^2

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