Solution - Finding the roots of polynomials
Other Ways to Solve
Finding the roots of polynomialsStep by Step Solution
Step 1 :
Equation at the end of step 1 :
(((4 • (x3)) - (2•3x2)) + 4x) - 1Step 2 :
Equation at the end of step 2 :
((22x3 - (2•3x2)) + 4x) - 1
Step 3 :
Checking for a perfect cube :
3.1 4x3-6x2+4x-1 is not a perfect cube
Trying to factor by pulling out :
3.2 Factoring: 4x3-6x2+4x-1
Thoughtfully split the expression at hand into groups, each group having two terms :
Group 1: 4x-1
Group 2: 4x3-6x2
Pull out from each group separately :
Group 1: (4x-1) • (1)
Group 2: (2x-3) • (2x2)
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) = 4x3-6x2+4x-1
Polynomial 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 integers
The 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 Coefficient
In this case, the Leading Coefficient is 4 and the Trailing Constant is -1.
The factor(s) are:
of the Leading Coefficient : 1,2 ,4
of the Trailing Constant : 1
Let us test ....
| P | Q | P/Q | F(P/Q) | Divisor | |||||
|---|---|---|---|---|---|---|---|---|---|
| -1 | 1 | -1.00 | -15.00 | ||||||
| -1 | 2 | -0.50 | -5.00 | ||||||
| -1 | 4 | -0.25 | -2.44 | ||||||
| 1 | 1 | 1.00 | 1.00 | ||||||
| 1 | 2 | 0.50 | 0.00 | 2x-1 | |||||
| 1 | 4 | 0.25 | -0.31 |
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
4x3-6x2+4x-1
can be divided with 2x-1
Polynomial Long Division :
3.4 Polynomial Long Division
Dividing : 4x3-6x2+4x-1
("Dividend")
By : 2x-1 ("Divisor")
| dividend | 4x3 | - | 6x2 | + | 4x | - | 1 | ||
| - divisor | * 2x2 | 4x3 | - | 2x2 | |||||
| remainder | - | 4x2 | + | 4x | - | 1 | |||
| - divisor | * -2x1 | - | 4x2 | + | 2x | ||||
| remainder | 2x | - | 1 | ||||||
| - divisor | * x0 | 2x | - | 1 | |||||
| remainder | 0 |
Quotient : 2x2-2x+1 Remainder: 0
Trying to factor by splitting the middle term
3.5 Factoring 2x2-2x+1
The first term is, 2x2 its coefficient is 2 .
The middle term is, -2x its coefficient is -2 .
The last term, "the constant", is +1
Step-1 : Multiply the coefficient of the first term by the constant 2 • 1 = 2
Step-2 : Find two factors of 2 whose sum equals the coefficient of the middle term, which is -2 .
| -2 | + | -1 | = | -3 | ||
| -1 | + | -2 | = | -3 | ||
| 1 | + | 2 | = | 3 | ||
| 2 | + | 1 | = | 3 |
Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored
Final result :
(2x2 - 2x + 1) • (2x - 1)
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