## Step 1 :

Equation at the end of step 1 : (((2 • (x3)) - 3x2) - 6x) - 9## Step 2 :

Equation at the end of step 2 : ((2x3 - 3x2) - 6x) - 9## Step 3 :

Checking for a perfect cube :3.12x3-3x2-6x-9 is not a perfect cube Trying to factor by pulling out :3.2 Factoring: 2x3-3x2-6x-9 Thoughtfully split the expression at hand into groups, each group having two terms:Group 1: -6x-9Group 2: 2x3-3x2Pull out from each group separately :Group 1: (2x+3) • (-3)Group 2: (2x-3) • (x2)Bad news !! Factoring by pulling out fails : The groups have no common factor and can not be added up to form a multiplication.

### Polynomial Roots Calculator :

3.3 Find roots (zeroes) of : F(x) = 2x3-3x2-6x-9Polynomial Roots Calculator is a set of methods aimed at finding values ofxfor which F(x)=0 Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers x which can be expressed as the quotient of two integersThe Rational Root Theorem states that if a polynomial zeroes for a rational numberP/Q then P is a factor of the Trailing Constant and Q is a factor of the Leading CoefficientIn this case, the Leading Coefficient is 2 and the Trailing Constant is -9. The factor(s) are: of the Leading Coefficient : 1,2 of the Trailing Constant : 1 ,3 ,9 Let us test ....

PQP/QF(P/Q)Divisor-1 | 1 | -1.00 | -8.00 | ||||||

-1 | 2 | -0.50 | -7.00 | ||||||

-3 | 1 | -3.00 | -72.00 | ||||||

-3 | 2 | -1.50 | -13.50 | ||||||

-9 | 1 | -9.00 | -1656.00 | ||||||

-9 | 2 | -4.50 | -225.00 | ||||||

1 | 1 | 1.00 | -16.00 | ||||||

1 | 2 | 0.50 | -12.50 | ||||||

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

3 | 2 | 1.50 | -18.00 | ||||||

9 | 1 | 9.00 | 1152.00 | ||||||

9 | 2 | 4.50 | 85.50 |

The Factor Theorem states that if P/Q is root of a polynomial then this polynomial can be divided by q*x-p Note that q and p originate from P/Q reduced to its lowest terms In our case this means that 2x3-3x2-6x-9can be divided with x-3

### Polynomial Long Division :

3.4 Polynomial Long Division Dividing : 2x3-3x2-6x-9("Dividend") By:x-3("Divisor")

dividend | 2x3 | - | 3x2 | - | 6x | - | 9 | ||

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

remainder | 3x2 | - | 6x | - | 9 | ||||

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

remainder | 3x | - | 9 | ||||||

-divisor | * 3x0 | 3x | - | 9 | |||||

remainder | 0 |

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

Trying to factor by splitting the middle term3.5Factoring 2x2+3x+3 The first term is, 2x2 its coefficient is 2.The middle term is, +3x its coefficient is 3.The last term, "the constant", is +3Step-1 : Multiply the coefficient of the first term by the consistent 2•3=6Step-2 : Find two factors of 6 whose sum equals the coefficient of the middle term, which is 3.

-6 | + | -1 | = | -7 | ||

-3 | + | -2 | = | -5 | ||

-2 | + | -3 | = | -5 | ||

-1 | + | -6 | = | -7 | ||

1 | + | 6 | = | 7 | ||

2 | + | 3 | = | 5 | ||

3 | + | 2 | = | 5 | ||

6 | + | 1 | = | 7 |

Observation : No two such factors can be found !! Conclusion : Trinomial can not be factored