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 :
12 ((((t2)-7t)+————)-(2•3t2))+9 (t3)Step 2 :
12 Simplify —— t3
Equation at the end of step 2 :
12
((((t2) - 7t) + ——) - (2•3t2)) + 9
t3
Step 3 :
Rewriting the whole as an Equivalent Fraction :
3.1 Adding a fraction to a whole
Rewrite the whole as a fraction using t3 as the denominator :
t2 - 7t (t2 - 7t) • t3
t2 - 7t = ——————— = ——————————————
1 t3
Equivalent fraction : The fraction thus generated looks different but has the same value as the whole
Common denominator : The equivalent fraction and the other fraction involved in the calculation share the same denominator
Step 4 :
Pulling out like terms :
4.1 Pull out like factors :
t2 - 7t = t • (t - 7)
Adding fractions that have a common denominator :
4.2 Adding up the two equivalent fractions
Add the two equivalent fractions which now have a common denominator
Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:
t • (t-7) • t3 + 12 t5 - 7t4 + 12
——————————————————— = —————————————
t3 t3
Equation at the end of step 4 :
(t5 - 7t4 + 12)
(——————————————— - (2•3t2)) + 9
t3
Step 5 :
Rewriting the whole as an Equivalent Fraction :
5.1 Subtracting a whole from a fraction
Rewrite the whole as a fraction using t3 as the denominator :
(2•3t2) (2•3t2) • t3
(2•3t2) = ——————— = ————————————
1 t3
Polynomial Roots Calculator :
5.2 Find roots (zeroes) of : F(t) = t5 - 7t4 + 12
Polynomial Roots Calculator is a set of methods aimed at finding values of t for which F(t)=0
Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers t 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 12.
The factor(s) are:
of the Leading Coefficient : 1
of the Trailing Constant : 1 ,2 ,3 ,4 ,6 ,12
Let us test ....
P | Q | P/Q | F(P/Q) | Divisor | |||||
---|---|---|---|---|---|---|---|---|---|
-1 | 1 | -1.00 | 4.00 | ||||||
-2 | 1 | -2.00 | -132.00 | ||||||
-3 | 1 | -3.00 | -798.00 | ||||||
-4 | 1 | -4.00 | -2804.00 | ||||||
-6 | 1 | -6.00 | -16836.00 | ||||||
-12 | 1 | -12.00 | -393972.00 | ||||||
1 | 1 | 1.00 | 6.00 | ||||||
2 | 1 | 2.00 | -68.00 | ||||||
3 | 1 | 3.00 | -312.00 | ||||||
4 | 1 | 4.00 | -756.00 | ||||||
6 | 1 | 6.00 | -1284.00 | ||||||
12 | 1 | 12.00 | 103692.00 |
Polynomial Roots Calculator found no rational roots
Adding fractions that have a common denominator :
5.3 Adding up the two equivalent fractions
(t5-7t4+12) - ((2•3t2) • t3) -5t5 - 7t4 + 12
———————————————————————————— = ———————————————
t3 t3
Equation at the end of step 5 :
(-5t5 - 7t4 + 12)
————————————————— + 9
t3
Step 6 :
Rewriting the whole as an Equivalent Fraction :
6.1 Adding a whole to a fraction
Rewrite the whole as a fraction using t3 as the denominator :
9 9 • t3
9 = — = ——————
1 t3
Step 7 :
Pulling out like terms :
7.1 Pull out like factors :
-5t5 - 7t4 + 12 = -1 • (5t5 + 7t4 - 12)
Polynomial Roots Calculator :
7.2 Find roots (zeroes) of : F(t) = 5t5 + 7t4 - 12
See theory in step 5.2
In this case, the Leading Coefficient is 5 and the Trailing Constant is -12.
The factor(s) are:
of the Leading Coefficient : 1,5
of the Trailing Constant : 1 ,2 ,3 ,4 ,6 ,12
Let us test ....
P | Q | P/Q | F(P/Q) | Divisor | |||||
---|---|---|---|---|---|---|---|---|---|
-1 | 1 | -1.00 | -10.00 | ||||||
-1 | 5 | -0.20 | -11.99 | ||||||
-2 | 1 | -2.00 | -60.00 | ||||||
-2 | 5 | -0.40 | -11.87 | ||||||
-3 | 1 | -3.00 | -660.00 | ||||||
-3 | 5 | -0.60 | -11.48 | ||||||
-4 | 1 | -4.00 | -3340.00 | ||||||
-4 | 5 | -0.80 | -10.77 | ||||||
-6 | 1 | -6.00 | -29820.00 | ||||||
-6 | 5 | -1.20 | -9.93 | ||||||
-12 | 1 | -12.00 | -1099020.00 | ||||||
-12 | 5 | -2.40 | -177.89 | ||||||
1 | 1 | 1.00 | 0.00 | t - 1 | |||||
1 | 5 | 0.20 | -11.99 | ||||||
2 | 1 | 2.00 | 260.00 | ||||||
2 | 5 | 0.40 | -11.77 | ||||||
3 | 1 | 3.00 | 1770.00 | ||||||
3 | 5 | 0.60 | -10.70 | ||||||
4 | 1 | 4.00 | 6900.00 | ||||||
4 | 5 | 0.80 | -7.49 | ||||||
6 | 1 | 6.00 | 47940.00 | ||||||
6 | 5 | 1.20 | 14.96 | ||||||
12 | 1 | 12.00 | 1389300.00 | ||||||
12 | 5 | 2.40 | 618.37 |
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
5t5 + 7t4 - 12
can be divided with t - 1
Polynomial Long Division :
7.3 Polynomial Long Division
Dividing : 5t5 + 7t4 - 12
("Dividend")
By : t - 1 ("Divisor")
dividend | 5t5 | + | 7t4 | - | 12 | ||||||||
- divisor | * 5t4 | 5t5 | - | 5t4 | |||||||||
remainder | 12t4 | - | 12 | ||||||||||
- divisor | * 12t3 | 12t4 | - | 12t3 | |||||||||
remainder | 12t3 | - | 12 | ||||||||||
- divisor | * 12t2 | 12t3 | - | 12t2 | |||||||||
remainder | 12t2 | - | 12 | ||||||||||
- divisor | * 12t1 | 12t2 | - | 12t | |||||||||
remainder | 12t | - | 12 | ||||||||||
- divisor | * 12t0 | 12t | - | 12 | |||||||||
remainder | 0 |
Quotient : 5t4+12t3+12t2+12t+12 Remainder: 0
Polynomial Roots Calculator :
7.4 Find roots (zeroes) of : F(t) = 5t4+12t3+12t2+12t+12
See theory in step 5.2
In this case, the Leading Coefficient is 5 and the Trailing Constant is 12.
The factor(s) are:
of the Leading Coefficient : 1,5
of the Trailing Constant : 1 ,2 ,3 ,4 ,6 ,12
Let us test ....
P | Q | P/Q | F(P/Q) | Divisor | |||||
---|---|---|---|---|---|---|---|---|---|
-1 | 1 | -1.00 | 5.00 | ||||||
-1 | 5 | -0.20 | 9.99 | ||||||
-2 | 1 | -2.00 | 20.00 | ||||||
-2 | 5 | -0.40 | 8.48 | ||||||
-3 | 1 | -3.00 | 165.00 |
Note - For tidiness, printing of 19 checks which found no root was suppressed
Polynomial Roots Calculator found no rational roots
Adding fractions that have a common denominator :
7.5 Adding up the two equivalent fractions
(-5t4-12t3-12t2-12t-12) • (t-1) + 9 • t3 -5t5 - 7t4 + 9t3 + 12
———————————————————————————————————————— = —————————————————————
t3 t3
Checking for a perfect cube :
7.6 -5t5 - 7t4 + 9t3 + 12 is not a perfect cube
Trying to factor by pulling out :
7.7 Factoring: -5t5 - 7t4 + 9t3 + 12
Thoughtfully split the expression at hand into groups, each group having two terms :
Group 1: 9t3 + 12
Group 2: -5t5 - 7t4
Pull out from each group separately :
Group 1: (3t3 + 4) • (3)
Group 2: (5t + 7) • (-t4)
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 :
7.8 Find roots (zeroes) of : F(t) = -5t5 - 7t4 + 9t3 + 12
See theory in step 5.2
In this case, the Leading Coefficient is -5 and the Trailing Constant is 12.
The factor(s) are:
of the Leading Coefficient : 1,5
of the Trailing Constant : 1 ,2 ,3 ,4 ,6 ,12
Let us test ....
P | Q | P/Q | F(P/Q) | Divisor | |||||
---|---|---|---|---|---|---|---|---|---|
-1 | 1 | -1.00 | 1.00 | ||||||
-1 | 5 | -0.20 | 11.92 | ||||||
-2 | 1 | -2.00 | -12.00 | ||||||
-2 | 5 | -0.40 | 11.30 | ||||||
-3 | 1 | -3.00 | 417.00 |
Note - For tidiness, printing of 19 checks which found no root was suppressed
Polynomial Roots Calculator found no rational roots
Final result :
5t5 - 7t4 + 9t3 + 12 ————————————————————— t3
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