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 :
((((x4)-(11•(x3)))+25x2)-4x)-48Step 2 :
Equation at the end of step 2 :
((((x4) - 11x3) + 25x2) - 4x) - 48
Step 3 :
Polynomial Roots Calculator :
3.1 Find roots (zeroes) of : F(x) = x4-11x3+32x2-4x-48
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 -48.
The factor(s) are:
of the Leading Coefficient : 1
of the Trailing Constant : 1 ,2 ,3 ,4 ,6 ,8 ,12 ,16 ,24 ,48
Let us test ....
| P | Q | P/Q | F(P/Q) | Divisor | |||||
|---|---|---|---|---|---|---|---|---|---|
| -1 | 1 | -1.00 | 0.00 | x+1 | |||||
| -2 | 1 | -2.00 | 192.00 | ||||||
| -3 | 1 | -3.00 | 630.00 | ||||||
| -4 | 1 | -4.00 | 1440.00 | ||||||
| -6 | 1 | -6.00 | 4800.00 | ||||||
| -8 | 1 | -8.00 | 11760.00 | ||||||
| -12 | 1 | -12.00 | 44352.00 | ||||||
| -16 | 1 | -16.00 | 118800.00 | ||||||
| -24 | 1 | -24.00 | 502320.00 | ||||||
| -48 | 1 | -48.00 | 6598800.00 | ||||||
| 1 | 1 | 1.00 | -30.00 | ||||||
| 2 | 1 | 2.00 | 0.00 | x-2 | |||||
| 3 | 1 | 3.00 | 12.00 | ||||||
| 4 | 1 | 4.00 | 0.00 | x-4 | |||||
| 6 | 1 | 6.00 | 0.00 | x-6 | |||||
| 8 | 1 | 8.00 | 432.00 | ||||||
| 12 | 1 | 12.00 | 6240.00 | ||||||
| 16 | 1 | 16.00 | 28560.00 | ||||||
| 24 | 1 | 24.00 | 198000.00 | ||||||
| 48 | 1 | 48.00 | 4165392.00 |
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
x4-11x3+32x2-4x-48
can be divided by 4 different polynomials,including by x-6
Polynomial Long Division :
3.2 Polynomial Long Division
Dividing : x4-11x3+32x2-4x-48
("Dividend")
By : x-6 ("Divisor")
| dividend | x4 | - | 11x3 | + | 32x2 | - | 4x | - | 48 | ||
| - divisor | * x3 | x4 | - | 6x3 | |||||||
| remainder | - | 5x3 | + | 32x2 | - | 4x | - | 48 | |||
| - divisor | * -5x2 | - | 5x3 | + | 30x2 | ||||||
| remainder | 2x2 | - | 4x | - | 48 | ||||||
| - divisor | * 2x1 | 2x2 | - | 12x | |||||||
| remainder | 8x | - | 48 | ||||||||
| - divisor | * 8x0 | 8x | - | 48 | |||||||
| remainder | 0 |
Quotient : x3-5x2+2x+8 Remainder: 0
Polynomial Roots Calculator :
3.3 Find roots (zeroes) of : F(x) = x3-5x2+2x+8
See theory in step 3.1
In this case, the Leading Coefficient is 1 and the Trailing Constant is 8.
The factor(s) are:
of the Leading Coefficient : 1
of the Trailing Constant : 1 ,2 ,4 ,8
Let us test ....
| P | Q | P/Q | F(P/Q) | Divisor | |||||
|---|---|---|---|---|---|---|---|---|---|
| -1 | 1 | -1.00 | 0.00 | x+1 | |||||
| -2 | 1 | -2.00 | -24.00 | ||||||
| -4 | 1 | -4.00 | -144.00 | ||||||
| -8 | 1 | -8.00 | -840.00 | ||||||
| 1 | 1 | 1.00 | 6.00 | ||||||
| 2 | 1 | 2.00 | 0.00 | x-2 | |||||
| 4 | 1 | 4.00 | 0.00 | x-4 | |||||
| 8 | 1 | 8.00 | 216.00 |
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
x3-5x2+2x+8
can be divided by 3 different polynomials,including by x-4
Polynomial Long Division :
3.4 Polynomial Long Division
Dividing : x3-5x2+2x+8
("Dividend")
By : x-4 ("Divisor")
| dividend | x3 | - | 5x2 | + | 2x | + | 8 | ||
| - divisor | * x2 | x3 | - | 4x2 | |||||
| remainder | - | x2 | + | 2x | + | 8 | |||
| - divisor | * -x1 | - | x2 | + | 4x | ||||
| remainder | - | 2x | + | 8 | |||||
| - divisor | * -2x0 | - | 2x | + | 8 | ||||
| remainder | 0 |
Quotient : x2-x-2 Remainder: 0
Trying to factor by splitting the middle term
3.5 Factoring x2-x-2
The first term is, x2 its coefficient is 1 .
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 1 • -2 = -2
Step-2 : Find two factors of -2 whose sum equals the coefficient of the middle term, which is -1 .
| -2 | + | 1 | = | -1 | That's it |
Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -2 and 1
x2 - 2x + 1x - 2
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 :
1 • (x-2)
Step-5 : Add up the four terms of step 4 :
(x+1) • (x-2)
Which is the desired factorization
Final result :
(x + 1) • (x - 2) • (x - 4) • (x - 6)
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