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i^{22}

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

☐²

x^☐

√☐

^□√☐

□□

log_□

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∞

∫

ddx

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

simplify solve for partial fractions long division line

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2x⁵+x⁴−2x−1=0

Solution

x=−12 , x=i, x=−i, x=−1, x=1

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2x⁵+x⁴−2x−1=0

Factor 2x⁵+x⁴−2x−1: (2x+1)(x²+1)(x+1)(x−1)

(2x+1)(x²+1)(x+1)(x−1)=0

Using the Zero Factor Principle: If ab=0 then a=0 or b=0

2x+1=0 or x²+1=0 or x+1=0 or x−1=0

Solve 2x+1=0: x=−12

Solve x²+1=0: x=i, x=−i

Solve x+1=0: x=−1

Solve x−1=0: x=1

The solutions are

x=−12 , x=i, x=−i, x=−1, x=1

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

i^{22}

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