テイラー (2x-5)e^x

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

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

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synthetic division 4x³−7x²−11x+54x+5

Solução

x²−3x+1

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4x³−7x²−11x+54x+5

Factor out coefficient of the denominator

4x+5=4· (x+54 )

4x³−7x²−11x+54(x+54 )

Coefficients of the numerator polynom

4 −7 −11 5

Find the zeros of the denominator: x=−54

Write the problem in synthetic division format

−54 ¦ 4 −7 −11 5

Carry down the leading coefficient, unchanged, to below the division symbol

−54 ¦ 4 −7 −11 5

Multiply the carry-down value by the zero of the denominator, and carry the result up into the next column:

4(−54 )=−5

−54 ¦ 4 −7 −11 5

¦ −5

Add down the column:

−7−5=−12

−54 ¦ 4 −7 −11 5

¦ −5

4 −12

Multiply the carry-down value by the zero of the denominator, and carry the result up into the next column:

(−12)(−54 )=15

−54 ¦ 4 −7 −11 5

¦ −5 15

4 −12

Add down the column:

−11+15=4

−54 ¦ 4 −7 −11 5

¦ −5 15

4 −12 4

Multiply the carry-down value by the zero of the denominator, and carry the result up into the next column:

4(−54 )=−5

−54 ¦ 4 −7 −11 5

¦ −5 15 −5

4 −12 4

Add down the column:

5−5=0

−54 ¦ 4 −7 −11 5

¦ −5 15 −5

4 −12 4 0

The last carry-down value is the remainder

Write the result in polynom format

4x²−12x+4

Divide the result by the denominator factor

4x²−12x+44

Simplify 4x²−12x+44 : x²−3x+1

x²−3x+1

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Oi, faça uma pergunta sobre synthetic division 4x³−7x²−11x+54x+5

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

テイラー (2x-5)e^x

Solução

Passos da solução

Reta numérica