Polynomial Long Division - Tiger Algebra
Step 1 :
Equation at the end of step 1 :
(((((2•(x5))-(2•(x4)))+(3•(x3)))-17x2)+23x)-9Step 2 :
Equation at the end of step 2 :
(((((2•(x5))-(2•(x4)))+3x3)-17x2)+23x)-9Step 3 :
Equation at the end of step 3 :
(((((2•(x5))-2x4)+3x3)-17x2)+23x)-9Step 4 :
Equation at the end of step 4 :
((((2x5 - 2x4) + 3x3) - 17x2) + 23x) - 9Step 5 :
Trying to factor by pulling out :
5.1 Factoring: 2x5-2x4+3x3-17x2+23x-9 Thoughtfully split the expression at hand into groups, each group having two terms :Group 1: 3x3-17x2 Group 2: 2x5-2x4 Group 3: 23x-9 Pull out from each group separately :Group 1: (3x-17) • (x2)Group 2: (x-1) • (2x4)Group 3: (23x-9) • (1) Looking for common sub-expressions : Group 1: (3x-17) • (x2) Group 3: (23x-9) • (1) Group 2: (x-1) • (2x4)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 :
5.2 Find roots (zeroes) of : F(x) = 2x5-2x4+3x3-17x2+23x-9Polynomial Roots Calculator is a set of methods aimed at finding values of x for 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 number P/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 ....
| P | Q | P/Q | F(P/Q) | Divisor |
|---|---|---|---|---|
| -1 | 1 | -1.00 | -56.00 | |
| -1 | 2 | -0.50 | -25.31 | |
| -3 | 1 | -3.00 | -960.00 | |
| -3 | 2 | -1.50 | -117.19 | |
| -9 | 1 | -9.00 | -135000.00 | |
| -9 | 2 | -4.50 | -5240.81 | |
| 1 | 1 | 1.00 | 0.00 | x-1 |
| 1 | 2 | 0.50 | -1.44 | |
| 3 | 1 | 3.00 | 312.00 | |
| 3 | 2 | 1.50 | 2.44 | |
| 9 | 1 | 9.00 | 105984.00 | |
| 9 | 2 | 4.50 | 2894.06 |
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 2x5-2x4+3x3-17x2+23x-9 can be divided with x-1
Polynomial Long Division :
5.3 Polynomial Long Division Dividing : 2x5-2x4+3x3-17x2+23x-9 ("Dividend") By : x-1 ("Divisor")
| dividend | 2x5 | - | 2x4 | + | 3x3 | - | 17x2 | + | 23x | - | 9 |
| - divisor | * 2x4 | 2x5 | - | 2x4 | |||||||
| remainder | 3x3 | - | 17x2 | + | 23x | - | 9 | ||||
| - divisor | * 0x3 | ||||||||||
| remainder | 3x3 | - | 17x2 | + | 23x | - | 9 | ||||
| - divisor | * 3x2 | 3x3 | - | 3x2 | |||||||
| remainder | - | 14x2 | + | 23x | - | 9 | |||||
| - divisor | * -14x1 | - | 14x2 | + | 14x | ||||||
| remainder | 9x | - | 9 | ||||||||
| - divisor | * 9x0 | 9x | - | 9 | |||||||
| remainder | 0 |
Quotient : 2x4+3x2-14x+9 Remainder: 0
Polynomial Roots Calculator :
5.4 Find roots (zeroes) of : F(x) = 2x4+3x2-14x+9 See theory in step 5.2 In 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 ....
| P | Q | P/Q | F(P/Q) | Divisor |
|---|---|---|---|---|
| -1 | 1 | -1.00 | 28.00 | |
| -1 | 2 | -0.50 | 16.88 | |
| -3 | 1 | -3.00 | 240.00 | |
| -3 | 2 | -1.50 | 46.88 | |
| -9 | 1 | -9.00 | 13500.00 | |
| -9 | 2 | -4.50 | 952.88 | |
| 1 | 1 | 1.00 | 0.00 | x-1 |
| 1 | 2 | 0.50 | 2.88 | |
| 3 | 1 | 3.00 | 156.00 | |
| 3 | 2 | 1.50 | 4.88 | |
| 9 | 1 | 9.00 | 13248.00 | |
| 9 | 2 | 4.50 | 826.88 |
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 2x4+3x2-14x+9 can be divided with x-1
Polynomial Long Division :
5.5 Polynomial Long Division Dividing : 2x4+3x2-14x+9 ("Dividend") By : x-1 ("Divisor")
| dividend | 2x4 | + | 3x2 | - | 14x | + | 9 |
| - divisor | * 2x3 | 2x4 | - | 2x3 | |||
| remainder | 2x3 | + | 3x2 | - | 14x | + | 9 |
| - divisor | * 2x2 | 2x3 | - | 2x2 | |||
| remainder | 5x2 | - | 14x | + | 9 | ||
| - divisor | * 5x1 | 5x2 | - | 5x | |||
| remainder | - | 9x | + | 9 | |||
| - divisor | * -9x0 | - | 9x | + | 9 | ||
| remainder | 0 |
Quotient : 2x3+2x2+5x-9 Remainder: 0
Polynomial Roots Calculator :
5.6 Find roots (zeroes) of : F(x) = 2x3+2x2+5x-9 See theory in step 5.2 In 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 ....
| P | Q | P/Q | F(P/Q) | Divisor |
|---|---|---|---|---|
| -1 | 1 | -1.00 | -14.00 | |
| -1 | 2 | -0.50 | -11.25 | |
| -3 | 1 | -3.00 | -60.00 | |
| -3 | 2 | -1.50 | -18.75 | |
| -9 | 1 | -9.00 | -1350.00 | |
| -9 | 2 | -4.50 | -173.25 | |
| 1 | 1 | 1.00 | 0.00 | x-1 |
| 1 | 2 | 0.50 | -5.75 | |
| 3 | 1 | 3.00 | 78.00 | |
| 3 | 2 | 1.50 | 9.75 | |
| 9 | 1 | 9.00 | 1656.00 | |
| 9 | 2 | 4.50 | 236.25 |
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+2x2+5x-9 can be divided with x-1
Polynomial Long Division :
5.7 Polynomial Long Division Dividing : 2x3+2x2+5x-9 ("Dividend") By : x-1 ("Divisor")
| dividend | 2x3 | + | 2x2 | + | 5x | - | 9 |
| - divisor | * 2x2 | 2x3 | - | 2x2 | |||
| remainder | 4x2 | + | 5x | - | 9 | ||
| - divisor | * 4x1 | 4x2 | - | 4x | |||
| remainder | 9x | - | 9 | ||||
| - divisor | * 9x0 | 9x | - | 9 | |||
| remainder | 0 |
Quotient : 2x2+4x+9 Remainder: 0
Trying to factor by splitting the middle term
5.8 Factoring 2x2+4x+9 The first term is, 2x2 its coefficient is 2 .The middle term is, +4x its coefficient is 4 .The last term, "the constant", is +9 Step-1 : Multiply the coefficient of the first term by the constant 2 • 9 = 18 Step-2 : Find two factors of 18 whose sum equals the coefficient of the middle term, which is 4 .
| -18 | + | -1 | = | -19 |
| -9 | + | -2 | = | -11 |
| -6 | + | -3 | = | -9 |
| -3 | + | -6 | = | -9 |
| -2 | + | -9 | = | -11 |
| -1 | + | -18 | = | -19 |
| 1 | + | 18 | = | 19 |
| 2 | + | 9 | = | 11 |
| 3 | + | 6 | = | 9 |
| 6 | + | 3 | = | 9 |
| 9 | + | 2 | = | 11 |
| 18 | + | 1 | = | 19 |
Observation : No two such factors can be found !! Conclusion : Trinomial can not be factored
Multiplying Exponential Expressions :
5.9 Multiply (x-1) by (x-1) The rule says : To multiply exponential expressions which have the same base, add up their exponents.In our case, the common base is (x-1) and the exponents are : 1 , as (x-1) is the same number as (x-1)1 and 1 , as (x-1) is the same number as (x-1)1 The product is therefore, (x-1)(1+1) = (x-1)2
Multiplying Exponential Expressions :
5.10 Multiply (x-1)2 by (x-1) The rule says : To multiply exponential expressions which have the same base, add up their exponents.In our case, the common base is (x-1) and the exponents are : 2 and 1 , as (x-1) is the same number as (x-1)1 The product is therefore, (x-1)(2+1) = (x-1)3
Final result :
(2x2 + 4x + 9) • (x - 1)3Từ khóa » G(x)=5x^9+17x^5
-
What Is Another Name For This Polynomial Based On The ...
-
What Is Another Name For This Polynomial Based On The Number Of ...
-
G(x)=5x-9 - Solution
-
3x+17x-5=4(5x+9) - Solution
-
Solve Approximation 17x^3+7x^2-5x-9=0 Tiger Algebra Solver
-
What Is Another Name For This Polynomial, Based On The ... - Wyzant
-
How Do You Simplify 2+17x -5x+9? - Math Homework Answers
-
Algebra Examples - Mathway
-
[PDF] ASSET_2_Solutions.pdf
-
Systems Of Linear Equations
-
[PDF] 3.2 The Factor Theorem And The Remainder Theorem