Solution - Factoring binomials using the difference of squares
Step by Step Solution
Reformatting the input :
Changes made to your input should not affect the solution:
(1): "x8" was replaced by "x^8".
Step by step solution :
Step 1 :
Equation at the end of step 1 :
(25 • (x2)) - (2•3•5x8) = 0Step 2 :
Equation at the end of step 2 :
52x2 - (2•3•5x8) = 0
Step 3 :
Step 4 :
Pulling out like terms :
4.1 Pull out like factors :
25x2 - 30x8 = -5x2 • (6x6 - 5)
Trying to factor as a Difference of Squares :
4.2 Factoring: 6x6 - 5
Theory : A difference of two perfect squares, A2 - B2 can be factored into (A+B) • (A-B)
Proof : (A+B) • (A-B) =
A2 - AB + BA - B2 =
A2 - AB + AB - B2 =
A2 - B2
Note : AB = BA is the commutative property of multiplication.
Note : - AB + AB equals zero and is therefore eliminated from the expression.
Check : 6 is not a square !!
Ruling : Binomial can not be factored as the
difference of two perfect squares
Polynomial Roots Calculator :
4.3 Find roots (zeroes) of : F(x) = 6x6 - 5
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 6 and the Trailing Constant is -5.
The factor(s) are:
of the Leading Coefficient : 1,2 ,3 ,6
of the Trailing Constant : 1 ,5
Let us test ....
| P | Q | P/Q | F(P/Q) | Divisor | |||||
|---|---|---|---|---|---|---|---|---|---|
| -1 | 1 | -1.00 | 1.00 | ||||||
| -1 | 2 | -0.50 | -4.91 | ||||||
| -1 | 3 | -0.33 | -4.99 | ||||||
| -1 | 6 | -0.17 | -5.00 | ||||||
| -5 | 1 | -5.00 | 93745.00 | ||||||
| -5 | 2 | -2.50 | 1459.84 | ||||||
| -5 | 3 | -1.67 | 123.60 | ||||||
| -5 | 6 | -0.83 | -2.99 | ||||||
| 1 | 1 | 1.00 | 1.00 | ||||||
| 1 | 2 | 0.50 | -4.91 | ||||||
| 1 | 3 | 0.33 | -4.99 | ||||||
| 1 | 6 | 0.17 | -5.00 | ||||||
| 5 | 1 | 5.00 | 93745.00 | ||||||
| 5 | 2 | 2.50 | 1459.84 | ||||||
| 5 | 3 | 1.67 | 123.60 | ||||||
| 5 | 6 | 0.83 | -2.99 |
Polynomial Roots Calculator found no rational roots
Equation at the end of step 4 :
-5x2 • (6x6 - 5) = 0
Step 5 :
Theory - Roots of a product :
5.1 A product of several terms equals zero.
When a product of two or more terms equals zero, then at least one of the terms must be zero.
We shall now solve each term = 0 separately
In other words, we are going to solve as many equations as there are terms in the product
Any solution of term = 0 solves product = 0 as well.
Solving a Single Variable Equation :
5.2 Solve : -5x2 = 0
Multiply both sides of the equation by (-1) : 5x2 = 0
Divide both sides of the equation by 5:
x2 = 0
When two things are equal, their square roots are equal. Taking the square root of the two sides of the equation we get:
x = ± √ 0
Any root of zero is zero. This equation has one solution which is x = 0
Solving a Single Variable Equation :
5.3 Solve : 6x6-5 = 0
Add 5 to both sides of the equation :
6x6 = 5
Divide both sides of the equation by 6:
x6 = 5/6 = 0.833
x = 6th root of (5/6)
The equation has two real solutions
These solutions are x = 6th root of ( 0.833) = ± 0.97007
Three solutions were found :
- x = 6th root of ( 0.833) = ± 0.97007
- x = 0
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