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 :
(((x3) - 3x2) - 4x) - 30
Step 2 :
Checking for a perfect cube :
2.1 x3-3x2-4x-30 is not a perfect cube
Trying to factor by pulling out :
2.2 Factoring: x3-3x2-4x-30
Thoughtfully split the expression at hand into groups, each group having two terms :
Group 1: -4x-30
Group 2: x3-3x2
Pull out from each group separately :
Group 1: (2x+15) • (-2)
Group 2: (x-3) • (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 :
2.3 Find roots (zeroes) of : F(x) = x3-3x2-4x-30
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 -30.
The factor(s) are:
of the Leading Coefficient : 1
of the Trailing Constant : 1 ,2 ,3 ,5 ,6 ,10 ,15 ,30
Let us test ....
| P | Q | P/Q | F(P/Q) | Divisor | |||||
|---|---|---|---|---|---|---|---|---|---|
| -1 | 1 | -1.00 | -30.00 | ||||||
| -2 | 1 | -2.00 | -42.00 | ||||||
| -3 | 1 | -3.00 | -72.00 | ||||||
| -5 | 1 | -5.00 | -210.00 | ||||||
| -6 | 1 | -6.00 | -330.00 | ||||||
| -10 | 1 | -10.00 | -1290.00 | ||||||
| -15 | 1 | -15.00 | -4020.00 | ||||||
| -30 | 1 | -30.00 | -29610.00 | ||||||
| 1 | 1 | 1.00 | -36.00 | ||||||
| 2 | 1 | 2.00 | -42.00 | ||||||
| 3 | 1 | 3.00 | -42.00 | ||||||
| 5 | 1 | 5.00 | 0.00 | x-5 | |||||
| 6 | 1 | 6.00 | 54.00 | ||||||
| 10 | 1 | 10.00 | 630.00 | ||||||
| 15 | 1 | 15.00 | 2610.00 | ||||||
| 30 | 1 | 30.00 | 24150.00 |
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
x3-3x2-4x-30
can be divided with x-5
Polynomial Long Division :
2.4 Polynomial Long Division
Dividing : x3-3x2-4x-30
("Dividend")
By : x-5 ("Divisor")
| dividend | x3 | - | 3x2 | - | 4x | - | 30 | ||
| - divisor | * x2 | x3 | - | 5x2 | |||||
| remainder | 2x2 | - | 4x | - | 30 | ||||
| - divisor | * 2x1 | 2x2 | - | 10x | |||||
| remainder | 6x | - | 30 | ||||||
| - divisor | * 6x0 | 6x | - | 30 | |||||
| remainder | 0 |
Quotient : x2+2x+6 Remainder: 0
Trying to factor by splitting the middle term
2.5 Factoring x2+2x+6
The first term is, x2 its coefficient is 1 .
The middle term is, +2x its coefficient is 2 .
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 2 .
| -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
Final result :
(x2 + 2x + 6) • (x - 5)
How did we do?
Please leave us feedback.