Finding the roots of polynomials
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This solution deals with finding the roots (zeroes) of polynomials.
Solution found
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Step by Step Solution
Step 1 :
Equation at the end of step 1 :
(((y3)+5y)+(6•(y2)))+6•(y+1)Step 2 :
Equation at the end of step 2 :
(((y3) + 5y) + (2•3y2)) + 6 • (y + 1)
Step 3 :
Checking for a perfect cube :
3.1 y3+6y2+11y+6 is not a perfect cube
Trying to factor by pulling out :
3.2 Factoring: y3+6y2+11y+6
Thoughtfully split the expression at hand into groups, each group having two terms :
Group 1: 11y+6
Group 2: 6y2+y3
Pull out from each group separately :
Group 1: (11y+6) • (1)
Group 2: (y+6) • (y2)
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 :
3.3 Find roots (zeroes) of : F(y) = y3+6y2+11y+6
Polynomial Roots Calculator is a set of methods aimed at finding values of y for which F(y)=0
Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers y 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 6.
The factor(s) are:
of the Leading Coefficient : 1
of the Trailing Constant : 1 ,2 ,3 ,6
Let us test ....
| P | Q | P/Q | F(P/Q) | Divisor | |||||
|---|---|---|---|---|---|---|---|---|---|
| -1 | 1 | -1.00 | 0.00 | y+1 | |||||
| -2 | 1 | -2.00 | 0.00 | y+2 | |||||
| -3 | 1 | -3.00 | 0.00 | y+3 | |||||
| -6 | 1 | -6.00 | -60.00 | ||||||
| 1 | 1 | 1.00 | 24.00 | ||||||
| 2 | 1 | 2.00 | 60.00 | ||||||
| 3 | 1 | 3.00 | 120.00 | ||||||
| 6 | 1 | 6.00 | 504.00 |
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
y3+6y2+11y+6
can be divided by 3 different polynomials,including by y+3
Polynomial Long Division :
3.4 Polynomial Long Division
Dividing : y3+6y2+11y+6
("Dividend")
By : y+3 ("Divisor")
| dividend | y3 | + | 6y2 | + | 11y | + | 6 | ||
| - divisor | * y2 | y3 | + | 3y2 | |||||
| remainder | 3y2 | + | 11y | + | 6 | ||||
| - divisor | * 3y1 | 3y2 | + | 9y | |||||
| remainder | 2y | + | 6 | ||||||
| - divisor | * 2y0 | 2y | + | 6 | |||||
| remainder | 0 |
Quotient : y2+3y+2 Remainder: 0
Trying to factor by splitting the middle term
3.5 Factoring y2+3y+2
The first term is, y2 its coefficient is 1 .
The middle term is, +3y its coefficient is 3 .
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 3 .
| -2 | + | -1 | = | -3 | ||
| -1 | + | -2 | = | -3 | ||
| 1 | + | 2 | = | 3 | That's it |
Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, 1 and 2
y2 + 1y + 2y + 2
Step-4 : Add up the first 2 terms, pulling out like factors :
y • (y+1)
Add up the last 2 terms, pulling out common factors :
2 • (y+1)
Step-5 : Add up the four terms of step 4 :
(y+2) • (y+1)
Which is the desired factorization
Final result :
(y + 2) • (y + 1) • (y + 3)
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