Respuesta :
Steps:
Use the divisor and find x :
x - 1 = 0 **add 1
x= 1
Now we will use the 1 in dividing:
take the coefficients from in front of all terms
** make sure you include 0's for x^2 and x since you have to have all terms
set it up with a 1 in a box:
1| 1 0 0 1 **bring the first number down
____________
1 **multiply the boxed number by the first number and add it to the second number
1| 1 0 0 1
____+1_____ **repeat with the rest of the terms
1 1
1| 1 0 0 1
___+1_+1_+1
1 1 1 2
**when you're done, use the new numbers to write an equation starting with a term with a degree one less than the previous equation.
**since there is a remainder, rewrite it divided by the original divisor
final answer:
x^2 + x + 1 + (2/ x -1)
Use the divisor and find x :
x - 1 = 0 **add 1
x= 1
Now we will use the 1 in dividing:
take the coefficients from in front of all terms
** make sure you include 0's for x^2 and x since you have to have all terms
set it up with a 1 in a box:
1| 1 0 0 1 **bring the first number down
____________
1 **multiply the boxed number by the first number and add it to the second number
1| 1 0 0 1
____+1_____ **repeat with the rest of the terms
1 1
1| 1 0 0 1
___+1_+1_+1
1 1 1 2
**when you're done, use the new numbers to write an equation starting with a term with a degree one less than the previous equation.
**since there is a remainder, rewrite it divided by the original divisor
final answer:
x^2 + x + 1 + (2/ x -1)
The quotient is of the division of [tex]{x^3} + 1[/tex] and [tex]x - 1 \:is \:\boxed{{x^2} + x + 1}[/tex] and the remainder is [tex]\boxed2[/tex].
Further explanation:
The numerator of the division is [tex]{x^3} + 1[/tex] and the denominator isx - 1.
Solve the given polynomial [tex]P\left( x \right) = {x^3} + 1[/tex] to obtain the quotient by the use of synthetic division.
Now obtain the value of [tex]x[/tex] from the denominator.
[tex]\begin{aligned} x - 1&= 0 \\x &= 1\\\end{aligned}[/tex]
Divide the coefficients of the polynomial by , to check whether is a zero of the polynomial.
[tex]\begin{aligned}1\left| \!{\nderline {\,{1\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,1\,} \,}} \right.\hfill \\\,\,\,\,\underline {\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,1}\hfill \\\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,2 \hfill\\\end{aligned}[/tex]
The last entry of the synthetic division told about remainder and the last entry of the synthetic division is 2. Therefore, the remainder of the synthetic division is .
The entry in the last line of synthetic division gives us the coefficients of the quotient.
The new numbers are the coefficients of the quotient.
The quotient is of the division of [tex]{x^3} + 1 \:and \:x - 1 \:is\: \boxed{{x^2} + x + 1}[/tex] and the remainder is [tex]\boxed2[/tex].
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Answer details:
Grade: High School
Subject: Mathematics
Chapter: Synthetic Division
Keywords: division, synthetic division, long division method, coefficients, quotients, remainders, numerator, denominator, polynomial, zeros, degree.
