### (2x^3-9x^2+11x-6)/(x-3)

This faces finding the root (zeroes) the polynomials.

You are watching: What is the result when 2x^3-9x^2+11x-6 is divided by x-3

## Step by step Solution

## Step 1 :

Equation in ~ the finish of action 1 :## step 2 :

Equation at the end of step 2 :## Step 3 :

2x3 - 9x2 + 11x - 6 leveling ——————————————————— x - 3 Checking because that a perfect cube :3.12x3 - 9x2 + 11x - 6 is not a perfect cube Trying to element by pulling the end :3.2 Factoring: 2x3 - 9x2 + 11x - 6 Thoughtfully break-up the expression at your disposal into groups, each team having 2 terms:Group 1: 11x - 6Group 2: 2x3 - 9x2Pull out from each group separately :Group 1: (11x - 6) • (1)Group 2: (2x - 9) • (x2)Bad news !! Factoring through pulling out falls short : The teams have no usual factor and can not be added up to kind a multiplication.

### Polynomial root Calculator :

3.3 find roots (zeroes) of : F(x) = 2x3 - 9x2 + 11x - 6Polynomial root Calculator is a set of approaches aimed in ~ finding worths ofxfor which F(x)=0 Rational root Test is one of the above mentioned tools. It would only find Rational Roots that is numbers x which have the right to be expressed as the quotient of 2 integersThe Rational source Theorem claims that if a polynomial zeroes because that a reasonable numberP/Q then p is a aspect of the Trailing consistent and Q is a variable of the top CoefficientIn this case, the top Coefficient is 2 and the Trailing continuous is -6. The factor(s) are: of the top Coefficient : 1,2 the the Trailing consistent : 1 ,2 ,3 ,6 Let us test ....

PQP/QF(P/Q)Divisor-1 | 1 | -1.00 | -28.00 | ||||||

-1 | 2 | -0.50 | -14.00 | ||||||

-2 | 1 | -2.00 | -80.00 | ||||||

-3 | 1 | -3.00 | -174.00 | ||||||

-3 | 2 | -1.50 | -49.50 | ||||||

-6 | 1 | -6.00 | -828.00 | ||||||

1 | 1 | 1.00 | -2.00 | ||||||

1 | 2 | 0.50 | -2.50 | ||||||

2 | 1 | 2.00 | -4.00 | ||||||

3 | 1 | 3.00 | 0.00 | x - 3 | |||||

3 | 2 | 1.50 | -3.00 | ||||||

6 | 1 | 6.00 | 168.00 |

The variable Theorem says that if P/Q is root of a polynomial climate this polynomial can be separated by q*x-p note that q and p originate indigenous P/Q lessened to the lowest terms In our case this way that 2x3 - 9x2 + 11x - 6can be divided with x - 3

### Polynomial Long department :

3.4 Polynomial Long division dividing : 2x3 - 9x2 + 11x - 6("Dividend") By:x - 3("Divisor")

dividend | 2x3 | - | 9x2 | + | 11x | - | 6 | ||

-divisor | * 2x2 | 2x3 | - | 6x2 | |||||

remainder | - | 3x2 | + | 11x | - | 6 | |||

-divisor | * -3x1 | - | 3x2 | + | 9x | ||||

remainder | 2x | - | 6 | ||||||

-divisor | * 2x0 | 2x | - | 6 | |||||

remainder | 0 |

Quotient : 2x2-3x+2 Remainder: 0

Trying to aspect by separating the center term3.5Factoring 2x2-3x+2 The very first term is, 2x2 that coefficient is 2.The center term is, -3x its coefficient is -3.The critical term, "the constant", is +2Step-1 : main point the coefficient of the an initial term by the consistent 2•2=4Step-2 : find two components of 4 who sum equals the coefficient the the center term, which is -3.

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-4 | + | -1 | = | -5 | ||

-2 | + | -2 | = | -4 | ||

-1 | + | -4 | = | -5 | ||

1 | + | 4 | = | 5 | ||

2 | + | 2 | = | 4 | ||

4 | + | 1 | = | 5 |

Observation : No 2 such components can be found !! Conclusion : Trinomial have the right to not be factored

Canceling out :3.6 Cancel the end (x-3) which shows up on both political parties of the fraction line.