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) + (2•3x2)) - 13x) - 42
Step 2 :
Checking for a perfect cube :
2.1 x3+6x2-13x-42 is not a perfect cube
Trying to factor by pulling out :
2.2 Factoring: x3+6x2-13x-42
Thoughtfully split the expression at hand into groups, each group having two terms :
Group 1: -13x-42
Group 2: 6x2+x3
Pull out from each group separately :
Group 1: (13x+42) • (-1)
Group 2: (x+6) • (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+6x2-13x-42
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 -42.
The factor(s) are:
of the Leading Coefficient : 1
of the Trailing Constant : 1 ,2 ,3 ,6 ,7 ,14 ,21 ,42
Let us test ....
| P | Q | P/Q | F(P/Q) | Divisor | |||||
|---|---|---|---|---|---|---|---|---|---|
| -1 | 1 | -1.00 | -24.00 | ||||||
| -2 | 1 | -2.00 | 0.00 | x+2 | |||||
| -3 | 1 | -3.00 | 24.00 | ||||||
| -6 | 1 | -6.00 | 36.00 | ||||||
| -7 | 1 | -7.00 | 0.00 | x+7 | |||||
| -14 | 1 | -14.00 | -1428.00 | ||||||
| -21 | 1 | -21.00 | -6384.00 | ||||||
| -42 | 1 | -42.00 | -63000.00 | ||||||
| 1 | 1 | 1.00 | -48.00 | ||||||
| 2 | 1 | 2.00 | -36.00 | ||||||
| 3 | 1 | 3.00 | 0.00 | x-3 | |||||
| 6 | 1 | 6.00 | 312.00 | ||||||
| 7 | 1 | 7.00 | 504.00 | ||||||
| 14 | 1 | 14.00 | 3696.00 | ||||||
| 21 | 1 | 21.00 | 11592.00 | ||||||
| 42 | 1 | 42.00 | 84084.00 |
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+6x2-13x-42
can be divided by 3 different polynomials,including by x-3
Polynomial Long Division :
2.4 Polynomial Long Division
Dividing : x3+6x2-13x-42
("Dividend")
By : x-3 ("Divisor")
| dividend | x3 | + | 6x2 | - | 13x | - | 42 | ||
| - divisor | * x2 | x3 | - | 3x2 | |||||
| remainder | 9x2 | - | 13x | - | 42 | ||||
| - divisor | * 9x1 | 9x2 | - | 27x | |||||
| remainder | 14x | - | 42 | ||||||
| - divisor | * 14x0 | 14x | - | 42 | |||||
| remainder | 0 |
Quotient : x2+9x+14 Remainder: 0
Trying to factor by splitting the middle term
2.5 Factoring x2+9x+14
The first term is, x2 its coefficient is 1 .
The middle term is, +9x its coefficient is 9 .
The last term, "the constant", is +14
Step-1 : Multiply the coefficient of the first term by the constant 1 • 14 = 14
Step-2 : Find two factors of 14 whose sum equals the coefficient of the middle term, which is 9 .
| -14 | + | -1 | = | -15 | ||
| -7 | + | -2 | = | -9 | ||
| -2 | + | -7 | = | -9 | ||
| -1 | + | -14 | = | -15 | ||
| 1 | + | 14 | = | 15 | ||
| 2 | + | 7 | = | 9 | That's it |
Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, 2 and 7
x2 + 2x + 7x + 14
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 :
7 • (x+2)
Step-5 : Add up the four terms of step 4 :
(x+7) • (x+2)
Which is the desired factorization
Final result :
(x + 7) • (x + 2) • (x - 3)
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