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 :
(((3 • (x3)) + (22•3x2)) + 3x) - 18Step 2 :
Equation at the end of step 2 :
((3x3 + (22•3x2)) + 3x) - 18
Step 3 :
Step 4 :
Pulling out like terms :
4.1 Pull out like factors :
3x3 + 12x2 + 3x - 18 =
3 • (x3 + 4x2 + x - 6)
Checking for a perfect cube :
4.2 x3 + 4x2 + x - 6 is not a perfect cube
Trying to factor by pulling out :
4.3 Factoring: x3 + 4x2 + x - 6
Thoughtfully split the expression at hand into groups, each group having two terms :
Group 1: x - 6
Group 2: x3 + 4x2
Pull out from each group separately :
Group 1: (x - 6) • (1)
Group 2: (x + 4) • (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 :
4.4 Find roots (zeroes) of : F(x) = x3 + 4x2 + x - 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 1 and the Trailing Constant is -6.
The factor(s) are:
of the Leading Coefficient : 1
of the Trailing Constant : 1 ,2 ,3 ,6
Let us test ....
| P | Q | P/Q | F(P/Q) | Divisor | |||||
|---|---|---|---|---|---|---|---|---|---|
| -1 | 1 | -1.00 | -4.00 | ||||||
| -2 | 1 | -2.00 | 0.00 | x + 2 | |||||
| -3 | 1 | -3.00 | 0.00 | x + 3 | |||||
| -6 | 1 | -6.00 | -84.00 | ||||||
| 1 | 1 | 1.00 | 0.00 | x - 1 | |||||
| 2 | 1 | 2.00 | 20.00 | ||||||
| 3 | 1 | 3.00 | 60.00 | ||||||
| 6 | 1 | 6.00 | 360.00 |
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
x3 + 4x2 + x - 6
can be divided by 3 different polynomials,including by x - 1
Polynomial Long Division :
4.5 Polynomial Long Division
Dividing : x3 + 4x2 + x - 6
("Dividend")
By : x - 1 ("Divisor")
| dividend | x3 | + | 4x2 | + | x | - | 6 | ||
| - divisor | * x2 | x3 | - | x2 | |||||
| remainder | 5x2 | + | x | - | 6 | ||||
| - divisor | * 5x1 | 5x2 | - | 5x | |||||
| remainder | 6x | - | 6 | ||||||
| - divisor | * 6x0 | 6x | - | 6 | |||||
| remainder | 0 |
Quotient : x2+5x+6 Remainder: 0
Trying to factor by splitting the middle term
4.6 Factoring x2+5x+6
The first term is, x2 its coefficient is 1 .
The middle term is, +5x its coefficient is 5 .
The last term, "the constant", is +6
Step-1 : Multiply the coefficient of the first term by the constant 1 • 6 = 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 | ||
| -2 | + | -3 | = | -5 | ||
| -1 | + | -6 | = | -7 | ||
| 1 | + | 6 | = | 7 | ||
| 2 | + | 3 | = | 5 | That's it |
Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, 2 and 3
x2 + 2x + 3x + 6
Step-4 : Add up the first 2 terms, pulling out like factors :
x • (x+2)
Add up the last 2 terms, pulling out common factors :
3 • (x+2)
Step-5 : Add up the four terms of step 4 :
(x+3) • (x+2)
Which is the desired factorization
Final result :
3 • (x + 3) • (x + 2) • (x - 1)
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