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 :
(((5 • (x3)) - (22•3x2)) - 11x) + 6Step 2 :
Equation at the end of step 2 :
((5x3 - (22•3x2)) - 11x) + 6
Step 3 :
Checking for a perfect cube :
3.1 5x3-12x2-11x+6 is not a perfect cube
Trying to factor by pulling out :
3.2 Factoring: 5x3-12x2-11x+6
Thoughtfully split the expression at hand into groups, each group having two terms :
Group 1: -11x+6
Group 2: -12x2+5x3
Pull out from each group separately :
Group 1: (-11x+6) • (1) = (11x-6) • (-1)
Group 2: (5x-12) • (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) = 5x3-12x2-11x+6
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 5 and the Trailing Constant is 6.
The factor(s) are:
of the Leading Coefficient : 1,5
of the Trailing Constant : 1 ,2 ,3 ,6
Let us test ....
| P | Q | P/Q | F(P/Q) | Divisor | |||||
|---|---|---|---|---|---|---|---|---|---|
| -1 | 1 | -1.00 | 0.00 | x+1 | |||||
| -1 | 5 | -0.20 | 7.68 | ||||||
| -2 | 1 | -2.00 | -60.00 | ||||||
| -2 | 5 | -0.40 | 8.16 | ||||||
| -3 | 1 | -3.00 | -204.00 | ||||||
| -3 | 5 | -0.60 | 7.20 | ||||||
| -6 | 1 | -6.00 | -1440.00 | ||||||
| -6 | 5 | -1.20 | -6.72 | ||||||
| 1 | 1 | 1.00 | -12.00 | ||||||
| 1 | 5 | 0.20 | 3.36 | ||||||
| 2 | 1 | 2.00 | -24.00 | ||||||
| 2 | 5 | 0.40 | 0.00 | 5x-2 | |||||
| 3 | 1 | 3.00 | 0.00 | x-3 | |||||
| 3 | 5 | 0.60 | -3.84 | ||||||
| 6 | 1 | 6.00 | 588.00 | ||||||
| 6 | 5 | 1.20 | -15.84 |
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
5x3-12x2-11x+6
can be divided by 3 different polynomials,including by x-3
Polynomial Long Division :
3.4 Polynomial Long Division
Dividing : 5x3-12x2-11x+6
("Dividend")
By : x-3 ("Divisor")
| dividend | 5x3 | - | 12x2 | - | 11x | + | 6 | ||
| - divisor | * 5x2 | 5x3 | - | 15x2 | |||||
| remainder | 3x2 | - | 11x | + | 6 | ||||
| - divisor | * 3x1 | 3x2 | - | 9x | |||||
| remainder | - | 2x | + | 6 | |||||
| - divisor | * -2x0 | - | 2x | + | 6 | ||||
| remainder | 0 |
Quotient : 5x2+3x-2 Remainder: 0
Trying to factor by splitting the middle term
3.5 Factoring 5x2+3x-2
The first term is, 5x2 its coefficient is 5 .
The middle term is, +3x its coefficient is 3 .
The last term, "the constant", is -2
Step-1 : Multiply the coefficient of the first term by the constant 5 • -2 = -10
Step-2 : Find two factors of -10 whose sum equals the coefficient of the middle term, which is 3 .
| -10 | + | 1 | = | -9 | ||
| -5 | + | 2 | = | -3 | ||
| -2 | + | 5 | = | 3 | That's it |
Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -2 and 5
5x2 - 2x + 5x - 2
Step-4 : Add up the first 2 terms, pulling out like factors :
x • (5x-2)
Add up the last 2 terms, pulling out common factors :
1 • (5x-2)
Step-5 : Add up the four terms of step 4 :
(x+1) • (5x-2)
Which is the desired factorization
Final result :
(5x - 2) • (x + 1) • (x - 3)
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