Solution - Finding the roots of polynomials
Other Ways to Solve
Finding the roots of polynomialsStep by Step Solution
Reformatting the input :
Changes made to your input should not affect the solution:
(1): "x2" was replaced by "x^2". 1 more similar replacement(s).
Step 1 :
Equation at the end of step 1 :
Step 2 :
Equation at the end of step 2 :
Step 3 :
-3x3 - 2x2 - x - 2
Simplify ——————————————————
x - 2
Step 4 :
Pulling out like terms :
4.1 Pull out like factors :
-3x3 - 2x2 - x - 2 =
-1 • (3x3 + 2x2 + x + 2)
Checking for a perfect cube :
4.2 3x3 + 2x2 + x + 2 is not a perfect cube
Trying to factor by pulling out :
4.3 Factoring: 3x3 + 2x2 + x + 2
Thoughtfully split the expression at hand into groups, each group having two terms :
Group 1: x + 2
Group 2: 3x3 + 2x2
Pull out from each group separately :
Group 1: (x + 2) • (1)
Group 2: (3x + 2) • (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) = 3x3 + 2x2 + x + 2
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 3 and the Trailing Constant is 2.
The factor(s) are:
of the Leading Coefficient : 1,3
of the Trailing Constant : 1 ,2
Let us test ....
| P | Q | P/Q | F(P/Q) | Divisor | |||||
|---|---|---|---|---|---|---|---|---|---|
| -1 | 1 | -1.00 | 0.00 | x + 1 | |||||
| -1 | 3 | -0.33 | 1.78 | ||||||
| -2 | 1 | -2.00 | -16.00 | ||||||
| -2 | 3 | -0.67 | 1.33 | ||||||
| 1 | 1 | 1.00 | 8.00 | ||||||
| 1 | 3 | 0.33 | 2.67 | ||||||
| 2 | 1 | 2.00 | 36.00 | ||||||
| 2 | 3 | 0.67 | 4.44 |
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
3x3 + 2x2 + x + 2
can be divided with x + 1
Polynomial Long Division :
4.5 Polynomial Long Division
Dividing : 3x3 + 2x2 + x + 2
("Dividend")
By : x + 1 ("Divisor")
| dividend | 3x3 | + | 2x2 | + | x | + | 2 | ||
| - divisor | * 3x2 | 3x3 | + | 3x2 | |||||
| remainder | - | x2 | + | x | + | 2 | |||
| - divisor | * -x1 | - | x2 | - | x | ||||
| remainder | 2x | + | 2 | ||||||
| - divisor | * 2x0 | 2x | + | 2 | |||||
| remainder | 0 |
Quotient : 3x2-x+2 Remainder: 0
Trying to factor by splitting the middle term
4.6 Factoring 3x2-x+2
The first term is, 3x2 its coefficient is 3 .
The middle term is, -x its coefficient is -1 .
The last term, "the constant", is +2
Step-1 : Multiply the coefficient of the first term by the constant 3 • 2 = 6
Step-2 : Find two factors of 6 whose sum equals the coefficient of the middle term, which is -1 .
| -6 | + | -1 | = | -7 | ||
| -3 | + | -2 | = | -5 | ||
| -2 | + | -3 | = | -5 | ||
| -1 | + | -6 | = | -7 | ||
| 1 | + | 6 | = | 7 | ||
| 2 | + | 3 | = | 5 | ||
| 3 | + | 2 | = | 5 | ||
| 6 | + | 1 | = | 7 |
Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored
Step 5 :
Pulling out like terms :
5.1 Pull out like factors :
-3x2 + x - 2 = -1 • (3x2 - x + 2)
Final result :
(-3x2 + x - 2) • (x + 1)
————————————————————————
x - 2
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