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boolean algebra (A∨ C)∧(A∧ D∨ C∨ A∧¬ D)

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

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log_□

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ddx

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

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synthetic division 7x³+4x+8x+2

Solution

7x²−14x+32−56x+2

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Solution steps

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7x³+4x+8x+2

Coefficients of the numerator polynom

7 4 8

Find the zeros of the denominator: x=−2

Write the problem in synthetic division format

−2¦ 7 0 4 8

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

−2¦ 7 0 4 8

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

7(−2)=−14

−2¦ 7 0 4 8

¦ −14

Add down the column:

0−14=−14

−2¦ 7 0 4 8

¦ −14

7 −14

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

(−14)(−2)=28

−2¦ 7 0 4 8

¦ −14 28

7 −14

Add down the column:

4+28=32

−2¦ 7 0 4 8

¦ −14 28

7 −14 32

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

32(−2)=−64

−2¦ 7 0 4 8

¦ −14 28 −64

7 −14 32

Add down the column:

8−64=−56

−2¦ 7 0 4 8

¦ −14 28 −64

7 −14 32 −56

The last carry-down value is the remainder

−56

Write the result in polynom format

7x²−14x+32−56x+2

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Notebook

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

boolean algebra (A∨ C)∧(A∧ D∨ C∨ A∧¬ D)

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