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): "x6" was replaced by "x^6".
Step by step solution :
Step 1 :
Equation at the end of step 1 :
(x2) - 22x6 = 0Step 2 :
Step 3 :
Pulling out like terms :
3.1 Pull out like factors :
x2 - 4x6 = -x2 • (4x4 - 1)
Trying to factor as a Difference of Squares :
3.2 Factoring: 4x4 - 1
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 : 4 is the square of 2
Check : 1 is the square of 1
Check : x4 is the square of x2
Factorization is : (2x2 + 1) • (2x2 - 1)
Polynomial Roots Calculator :
3.3 Find roots (zeroes) of : F(x) = 2x2 + 1
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 2 and the Trailing Constant is 1.
The factor(s) are:
of the Leading Coefficient : 1,2
of the Trailing Constant : 1
Let us test ....
| P | Q | P/Q | F(P/Q) | Divisor | |||||
|---|---|---|---|---|---|---|---|---|---|
| -1 | 1 | -1.00 | 3.00 | ||||||
| -1 | 2 | -0.50 | 1.50 | ||||||
| 1 | 1 | 1.00 | 3.00 | ||||||
| 1 | 2 | 0.50 | 1.50 |
Polynomial Roots Calculator found no rational roots
Trying to factor as a Difference of Squares :
3.4 Factoring: 2x2 - 1
Check : 2 is not a square !!
Ruling : Binomial can not be factored as the
difference of two perfect squares
Equation at the end of step 3 :
-x2 • (2x2 + 1) • (2x2 - 1) = 0
Step 4 :
Theory - Roots of a product :
4.1 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 :
4.2 Solve : -x2 = 0
Multiply both sides of the equation by (-1) : 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 :
4.3 Solve : 2x2+1 = 0
Subtract 1 from both sides of the equation :
2x2 = -1
Divide both sides of the equation by 2:
x2 = -1/2 = -0.500
When two things are equal, their square roots are equal. Taking the square root of the two sides of the equation we get:
x = ± √ -1/2
In Math, i is called the imaginary unit. It satisfies i2 =-1. Both i and -i are the square roots of -1
Accordingly, √ -1/2 =
√ -1• 1/2 =
√ -1 •√ 1/2 =
i • √ 1/2
The equation has no real solutions. It has 2 imaginary, or complex solutions.
x= 0.0000 + 0.7071 i
x= 0.0000 - 0.7071 i
Solving a Single Variable Equation :
4.4 Solve : 2x2-1 = 0
Add 1 to both sides of the equation :
2x2 = 1
Divide both sides of the equation by 2:
x2 = 1/2 = 0.500
When two things are equal, their square roots are equal. Taking the square root of the two sides of the equation we get:
x = ± √ 1/2
The equation has two real solutions
These solutions are x = ±√ 0.500 = ± 0.70711
5 solutions were found :
- x = ±√ 0.500 = ± 0.70711
- x= 0.0000 - 0.7071 i
- x= 0.0000 + 0.7071 i
- x = 0
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