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 :
((x - 2)3) - 23x3
Step 2 :
2.1 Evaluate : (x-2)3 = x3-6x2+12x-8
Step 3 :
Pulling out like terms :
3.1 Pull out like factors :
-7x3 - 6x2 + 12x - 8 =
-1 • (7x3 + 6x2 - 12x + 8)
Checking for a perfect cube :
3.2 7x3 + 6x2 - 12x + 8 is not a perfect cube
Trying to factor by pulling out :
3.3 Factoring: 7x3 + 6x2 - 12x + 8
Thoughtfully split the expression at hand into groups, each group having two terms :
Group 1: 7x3 + 8
Group 2: 6x2 - 12x
Pull out from each group separately :
Group 1: (7x3 + 8) • (1)
Group 2: (x - 2) • (6x)
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 :
3.4 Find roots (zeroes) of : F(x) = 7x3 + 6x2 - 12x + 8
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 7 and the Trailing Constant is 8.
The factor(s) are:
of the Leading Coefficient : 1,7
of the Trailing Constant : 1 ,2 ,4 ,8
Let us test ....
| P | Q | P/Q | F(P/Q) | Divisor | |||||
|---|---|---|---|---|---|---|---|---|---|
| -1 | 1 | -1.00 | 19.00 | ||||||
| -1 | 7 | -0.14 | 9.82 | ||||||
| -2 | 1 | -2.00 | 0.00 | x + 2 | |||||
| -2 | 7 | -0.29 | 11.76 | ||||||
| -4 | 1 | -4.00 | -296.00 | ||||||
| -4 | 7 | -0.57 | 15.51 | ||||||
| -8 | 1 | -8.00 | -3096.00 | ||||||
| -8 | 7 | -1.14 | 19.10 | ||||||
| 1 | 1 | 1.00 | 9.00 | ||||||
| 1 | 7 | 0.14 | 6.43 | ||||||
| 2 | 1 | 2.00 | 64.00 | ||||||
| 2 | 7 | 0.29 | 5.22 | ||||||
| 4 | 1 | 4.00 | 504.00 | ||||||
| 4 | 7 | 0.57 | 4.41 | ||||||
| 8 | 1 | 8.00 | 3880.00 | ||||||
| 8 | 7 | 1.14 | 12.57 |
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
7x3 + 6x2 - 12x + 8
can be divided with x + 2
Polynomial Long Division :
3.5 Polynomial Long Division
Dividing : 7x3 + 6x2 - 12x + 8
("Dividend")
By : x + 2 ("Divisor")
| dividend | 7x3 | + | 6x2 | - | 12x | + | 8 | ||
| - divisor | * 7x2 | 7x3 | + | 14x2 | |||||
| remainder | - | 8x2 | - | 12x | + | 8 | |||
| - divisor | * -8x1 | - | 8x2 | - | 16x | ||||
| remainder | 4x | + | 8 | ||||||
| - divisor | * 4x0 | 4x | + | 8 | |||||
| remainder | 0 |
Quotient : 7x2-8x+4 Remainder: 0
Trying to factor by splitting the middle term
3.6 Factoring 7x2-8x+4
The first term is, 7x2 its coefficient is 7 .
The middle term is, -8x its coefficient is -8 .
The last term, "the constant", is +4
Step-1 : Multiply the coefficient of the first term by the constant 7 • 4 = 28
Step-2 : Find two factors of 28 whose sum equals the coefficient of the middle term, which is -8 .
| -28 | + | -1 | = | -29 | ||
| -14 | + | -2 | = | -16 | ||
| -7 | + | -4 | = | -11 | ||
| -4 | + | -7 | = | -11 | ||
| -2 | + | -14 | = | -16 | ||
| -1 | + | -28 | = | -29 | ||
| 1 | + | 28 | = | 29 | ||
| 2 | + | 14 | = | 16 | ||
| 4 | + | 7 | = | 11 | ||
| 7 | + | 4 | = | 11 | ||
| 14 | + | 2 | = | 16 | ||
| 28 | + | 1 | = | 29 |
Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored
Final result :
(-7x2 + 8x - 4) • (x + 2)
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