Solution - Polynomial long division
Other Ways to Solve
Polynomial long divisionStep by Step Solution
Step 1 :
Equation at the end of step 1 :
(((4•(x2))-25)2)-9•(2x-5)2Step 2 :
Equation at the end of step 2 :
((22x2 - 25)2) - 9 • (2x - 5)2
Step 3 :
3.1 Evaluate : (4x2-25)2 = 16x4-200x2+625 3.2 Evaluate : (2x-5)2 = 4x2-20x+25
Step 4 :
Pulling out like terms :
4.1 Pull out like factors :
16x4 - 236x2 + 180x + 400 =
4 • (4x4 - 59x2 + 45x + 100)
Checking for a perfect cube :
4.2 4x4 - 59x2 + 45x + 100 is not a perfect cube
Trying to factor by pulling out :
4.3 Factoring: 4x4 - 59x2 + 45x + 100
Thoughtfully split the expression at hand into groups, each group having two terms :
Group 1: 45x + 100
Group 2: 4x4 - 59x2
Pull out from each group separately :
Group 1: (9x + 20) • (5)
Group 2: (4x2 - 59) • (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 :
4.4 Find roots (zeroes) of : F(x) = 4x4 - 59x2 + 45x + 100
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 4 and the Trailing Constant is 100.
The factor(s) are:
of the Leading Coefficient : 1,2 ,4
of the Trailing Constant : 1 ,2 ,4 ,5 ,10 ,20 ,25 ,50 ,100
Let us test ....
| P | Q | P/Q | F(P/Q) | Divisor | |||||
|---|---|---|---|---|---|---|---|---|---|
| -1 | 1 | -1.00 | 0.00 | x + 1 | |||||
| -1 | 2 | -0.50 | 63.00 | ||||||
| -1 | 4 | -0.25 | 85.08 | ||||||
| -2 | 1 | -2.00 | -162.00 | ||||||
| -4 | 1 | -4.00 | 0.00 | x + 4 | |||||
| -5 | 1 | -5.00 | 900.00 | ||||||
| -5 | 2 | -2.50 | -225.00 | ||||||
| -5 | 4 | -1.25 | -38.67 | ||||||
| -10 | 1 | -10.00 | 33750.00 | ||||||
| -20 | 1 | -20.00 | 615600.00 | ||||||
| -25 | 1 | -25.00 | 1524600.00 | ||||||
| -25 | 2 | -12.50 | 87975.00 | ||||||
| -25 | 4 | -6.25 | 3617.58 | ||||||
| -50 | 1 | -50.00 | 24850350.00 | ||||||
| -100 | 1 | -100.00 | 399405600.00 | ||||||
| 1 | 1 | 1.00 | 90.00 | ||||||
| 1 | 2 | 0.50 | 108.00 | ||||||
| 1 | 4 | 0.25 | 107.58 | ||||||
| 2 | 1 | 2.00 | 18.00 | ||||||
| 4 | 1 | 4.00 | 360.00 | ||||||
| 5 | 1 | 5.00 | 1350.00 | ||||||
| 5 | 2 | 2.50 | 0.00 | 2x - 5 | |||||
| 5 | 4 | 1.25 | 73.83 | ||||||
| 10 | 1 | 10.00 | 34650.00 | ||||||
| 20 | 1 | 20.00 | 617400.00 | ||||||
| 25 | 1 | 25.00 | 1526850.00 | ||||||
| 25 | 2 | 12.50 | 89100.00 | ||||||
| 25 | 4 | 6.25 | 4180.08 | ||||||
| 50 | 1 | 50.00 | 24854850.00 | ||||||
| 100 | 1 | 100.00 | 399414600.00 |
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
4x4 - 59x2 + 45x + 100
can be divided by 3 different polynomials,including by 2x - 5
Polynomial Long Division :
4.5 Polynomial Long Division
Dividing : 4x4 - 59x2 + 45x + 100
("Dividend")
By : 2x - 5 ("Divisor")
| dividend | 4x4 | - | 59x2 | + | 45x | + | 100 | ||||
| - divisor | * 2x3 | 4x4 | - | 10x3 | |||||||
| remainder | 10x3 | - | 59x2 | + | 45x | + | 100 | ||||
| - divisor | * 5x2 | 10x3 | - | 25x2 | |||||||
| remainder | - | 34x2 | + | 45x | + | 100 | |||||
| - divisor | * -17x1 | - | 34x2 | + | 85x | ||||||
| remainder | - | 40x | + | 100 | |||||||
| - divisor | * -20x0 | - | 40x | + | 100 | ||||||
| remainder | 0 |
Quotient : 2x3+5x2-17x-20 Remainder: 0
Polynomial Roots Calculator :
4.6 Find roots (zeroes) of : F(x) = 2x3+5x2-17x-20
See theory in step 4.4
In this case, the Leading Coefficient is 2 and the Trailing Constant is -20.
The factor(s) are:
of the Leading Coefficient : 1,2
of the Trailing Constant : 1 ,2 ,4 ,5 ,10 ,20
Let us test ....
| P | Q | P/Q | F(P/Q) | Divisor | |||||
|---|---|---|---|---|---|---|---|---|---|
| -1 | 1 | -1.00 | 0.00 | x+1 | |||||
| -1 | 2 | -0.50 | -10.50 | ||||||
| -2 | 1 | -2.00 | 18.00 | ||||||
| -4 | 1 | -4.00 | 0.00 | x+4 | |||||
| -5 | 1 | -5.00 | -60.00 | ||||||
| -5 | 2 | -2.50 | 22.50 | ||||||
| -10 | 1 | -10.00 | -1350.00 | ||||||
| -20 | 1 | -20.00 | -13680.00 | ||||||
| 1 | 1 | 1.00 | -30.00 | ||||||
| 1 | 2 | 0.50 | -27.00 | ||||||
| 2 | 1 | 2.00 | -18.00 | ||||||
| 4 | 1 | 4.00 | 120.00 | ||||||
| 5 | 1 | 5.00 | 270.00 | ||||||
| 5 | 2 | 2.50 | 0.00 | 2x-5 | |||||
| 10 | 1 | 10.00 | 2310.00 | ||||||
| 20 | 1 | 20.00 | 17640.00 |
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
2x3+5x2-17x-20
can be divided by 3 different polynomials,including by 2x-5
Polynomial Long Division :
4.7 Polynomial Long Division
Dividing : 2x3+5x2-17x-20
("Dividend")
By : 2x-5 ("Divisor")
| dividend | 2x3 | + | 5x2 | - | 17x | - | 20 | ||
| - divisor | * x2 | 2x3 | - | 5x2 | |||||
| remainder | 10x2 | - | 17x | - | 20 | ||||
| - divisor | * 5x1 | 10x2 | - | 25x | |||||
| remainder | 8x | - | 20 | ||||||
| - divisor | * 4x0 | 8x | - | 20 | |||||
| remainder | 0 |
Quotient : x2+5x+4 Remainder: 0
Trying to factor by splitting the middle term
4.8 Factoring x2+5x+4
The first term is, x2 its coefficient is 1 .
The middle term is, +5x its coefficient is 5 .
The last term, "the constant", is +4
Step-1 : Multiply the coefficient of the first term by the constant 1 • 4 = 4
Step-2 : Find two factors of 4 whose sum equals the coefficient of the middle term, which is 5 .
| -4 | + | -1 | = | -5 | ||
| -2 | + | -2 | = | -4 | ||
| -1 | + | -4 | = | -5 | ||
| 1 | + | 4 | = | 5 | That's it |
Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, 1 and 4
x2 + 1x + 4x + 4
Step-4 : Add up the first 2 terms, pulling out like factors :
x • (x+1)
Add up the last 2 terms, pulling out common factors :
4 • (x+1)
Step-5 : Add up the four terms of step 4 :
(x+4) • (x+1)
Which is the desired factorization
Multiplying Exponential Expressions :
4.9 Multiply (2x-5) by (2x-5)
The rule says : To multiply exponential expressions which have the same base, add up their exponents.
In our case, the common base is (2x-5) and the exponents are :
1 , as (2x-5) is the same number as (2x-5)1
and 1 , as (2x-5) is the same number as (2x-5)1
The product is therefore, (2x-5)(1+1) = (2x-5)2
Final result :
4 • (x + 4) • (x + 1) • (2x - 5)2
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