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 :
(((6 • (x3)) - 23x2) + 16x) - 3Step 2 :
Equation at the end of step 2 :
(((2•3x3) - 23x2) + 16x) - 3
Step 3 :
Checking for a perfect cube :
3.1 6x3-23x2+16x-3 is not a perfect cube
Trying to factor by pulling out :
3.2 Factoring: 6x3-23x2+16x-3
Thoughtfully split the expression at hand into groups, each group having two terms :
Group 1: 16x-3
Group 2: 6x3-23x2
Pull out from each group separately :
Group 1: (16x-3) • (1)
Group 2: (6x-23) • (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) = 6x3-23x2+16x-3
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 6 and the Trailing Constant is -3.
The factor(s) are:
of the Leading Coefficient : 1,2 ,3 ,6
of the Trailing Constant : 1 ,3
Let us test ....
| P | Q | P/Q | F(P/Q) | Divisor | |||||
|---|---|---|---|---|---|---|---|---|---|
| -1 | 1 | -1.00 | -48.00 | ||||||
| -1 | 2 | -0.50 | -17.50 | ||||||
| -1 | 3 | -0.33 | -11.11 | ||||||
| -1 | 6 | -0.17 | -6.33 | ||||||
| -3 | 1 | -3.00 | -420.00 | ||||||
| -3 | 2 | -1.50 | -99.00 | ||||||
| 1 | 1 | 1.00 | -4.00 | ||||||
| 1 | 2 | 0.50 | 0.00 | 2x-1 | |||||
| 1 | 3 | 0.33 | 0.00 | 3x-1 | |||||
| 1 | 6 | 0.17 | -0.94 | ||||||
| 3 | 1 | 3.00 | 0.00 | x-3 | |||||
| 3 | 2 | 1.50 | -10.50 |
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
6x3-23x2+16x-3
can be divided by 3 different polynomials,including by x-3
Polynomial Long Division :
3.4 Polynomial Long Division
Dividing : 6x3-23x2+16x-3
("Dividend")
By : x-3 ("Divisor")
| dividend | 6x3 | - | 23x2 | + | 16x | - | 3 | ||
| - divisor | * 6x2 | 6x3 | - | 18x2 | |||||
| remainder | - | 5x2 | + | 16x | - | 3 | |||
| - divisor | * -5x1 | - | 5x2 | + | 15x | ||||
| remainder | x | - | 3 | ||||||
| - divisor | * x0 | x | - | 3 | |||||
| remainder | 0 |
Quotient : 6x2-5x+1 Remainder: 0
Trying to factor by splitting the middle term
3.5 Factoring 6x2-5x+1
The first term is, 6x2 its coefficient is 6 .
The middle term is, -5x its coefficient is -5 .
The last term, "the constant", is +1
Step-1 : Multiply the coefficient of the first term by the constant 6 • 1 = 6
Step-2 : Find two factors of 6 whose sum equals the coefficient of the middle term, which is -5 .
| -6 | + | -1 | = | -7 | ||
| -3 | + | -2 | = | -5 | That's it |
Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -3 and -2
6x2 - 3x - 2x - 1
Step-4 : Add up the first 2 terms, pulling out like factors :
3x • (2x-1)
Add up the last 2 terms, pulling out common factors :
1 • (2x-1)
Step-5 : Add up the four terms of step 4 :
(3x-1) • (2x-1)
Which is the desired factorization
Final result :
(2x - 1) • (3x - 1) • (x - 3)
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