(x3-3x2+3x-2)/(x2-x+1)
This deals with finding the root (zeroes) of polynomials.
You are watching: What is the quotient (x3 – 3x2 + 3x – 2) ÷ (x2 – x + 1)?
Step by action Solution
Reformatting the input :
Changes made to her input have to not impact the solution: (1): "x2" was replaced by "x^2". 2 more similar replacement(s).Step 1 :
Equation in ~ the finish of action 1 :
Step 2 :
x3 - 3x2 + 3x - 2 simplify ————————————————— x2 - x + 1 Checking because that a perfect cube :2.1x3 - 3x2 + 3x - 2 is no a perfect cube Trying to variable by pulling the end :2.2 Factoring: x3 - 3x2 + 3x - 2 Thoughtfully split the expression at your disposal into groups, each team having 2 terms:Group 1: 3x - 2Group 2: -3x2 + x3Pull the end from each group separately :Group 1: (3x - 2) • (1)Group 2: (x - 3) • (x2)Bad news no Factoring through pulling out fails : The teams have no usual factor and can not be included up to form a multiplication.
Polynomial roots Calculator :
2.3 uncover roots (zeroes) the : F(x) = x3 - 3x2 + 3x - 2Polynomial root Calculator is a collection of techniques aimed at finding values ofxfor i beg your pardon F(x)=0 Rational root Test is one of the over mentioned tools. It would only find Rational Roots that is numbers x which deserve to be expressed as the quotient of 2 integersThe Rational root Theorem says that if a polynomial zeroes because that a reasonable numberP/Q then p is a variable of the Trailing consistent and Q is a variable of the leading CoefficientIn this case, the leading Coefficient is 1 and the Trailing consistent is -2. The factor(s) are: of the leading Coefficient : 1of the Trailing constant : 1 ,2 Let united state test ....
| -1 | 1 | -1.00 | -9.00 | ||||||
| -2 | 1 | -2.00 | -28.00 | ||||||
| 1 | 1 | 1.00 | -1.00 | ||||||
| 2 | 1 | 2.00 | 0.00 | x - 2 |
The element Theorem states that if P/Q is root of a polynomial then this polynomial have the right to be separated by q*x-p keep in mind that q and also p originate from P/Q diminished to the lowest terms In our situation this means that x3 - 3x2 + 3x - 2can be separated with x - 2
Polynomial Long department :
2.4 Polynomial Long division dividing : x3 - 3x2 + 3x - 2("Dividend") By:x - 2("Divisor")
| dividend | x3 | - | 3x2 | + | 3x | - | 2 | ||
| -divisor | * x2 | x3 | - | 2x2 | |||||
| remainder | - | x2 | + | 3x | - | 2 | |||
| -divisor | * -x1 | - | x2 | + | 2x | ||||
| remainder | x | - | 2 | ||||||
| -divisor | * x0 | x | - | 2 | |||||
| remainder | 0 |
Quotient : x2-x+1 Remainder: 0
Trying to aspect by splitting the center term2.5Factoring x2-x+1 The first term is, x2 that coefficient is 1.The center term is, -x the coefficient is -1.The critical term, "the constant", is +1Step-1 : multiply the coefficient of the first term by the constant 1•1=1Step-2 : find two factors of 1 whose sum equates to the coefficient the the middle term, i m sorry is -1.
| -1 | + | -1 | = | -2 | ||
| 1 | + | 1 | = | 2 |
Observation : No two such factors can be found !! Conclusion : Trinomial have the right to not it is in factored
Trying to factor by separating the middle term2.6Factoring x2-x+1 The first term is, x2 that coefficient is 1.The middle term is, -x the coefficient is -1.The last term, "the constant", is +1Step-1 : multiply the coefficient of the an initial term by the consistent 1•1=1Step-2 : discover two factors of 1 who sum equals the coefficient the the center term, which is -1.
See more: Which Group On The Periodic Table Has Two Elements That Exist As Gases At Stp ?
| -1 | + | -1 | = | -2 | ||
| 1 | + | 1 | = | 2 |
Observation : No two such determinants can be discovered !! Conclusion : Trinomial have the right to not it is in factored
Canceling out :2.7 Cancel the end (x2-x+1) which shows up on both political parties of the fraction line.