Solution - Factoring binomials using the difference of squares
Other Ways to Solve
Factoring binomials using the difference of squaresStep by Step Solution
Reformatting the input :
Changes made to your input should not affect the solution:
(1): "x8" was replaced by "x^8".
Rearrange:
Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :
6*(2*x^4)-(3*(4*x^8))=0
Step by step solution :
Step 1 :
Equation at the end of step 1 :
(6 • (2 • (x4))) - (3 • 22x8) = 0
Step 2 :
Equation at the end of step 2 :
(6 • (2 • (x4))) - (3•22x8) = 0Step 3 :
Equation at the end of step 3 :
(6 • 2x4) - (3•22x8) = 0
Step 4 :
Multiplying exponents :
4.1 21 multiplied by 21 = 2(1 + 1) = 22
Equation at the end of step 4 :
(22•3x4) - (3•22x8) = 0
Step 5 :
Step 6 :
Pulling out like terms :
6.1 Pull out like factors :
12x4 - 12x8 = -12x4 • (x4 - 1)
Trying to factor as a Difference of Squares :
6.2 Factoring: x4 - 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 : 1 is the square of 1
Check : x4 is the square of x2
Factorization is : (x2 + 1) • (x2 - 1)
Polynomial Roots Calculator :
6.3 Find roots (zeroes) of : F(x) = x2 + 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 1 and the Trailing Constant is 1.
The factor(s) are:
of the Leading Coefficient : 1
of the Trailing Constant : 1
Let us test ....
P | Q | P/Q | F(P/Q) | Divisor | |||||
---|---|---|---|---|---|---|---|---|---|
-1 | 1 | -1.00 | 2.00 | ||||||
1 | 1 | 1.00 | 2.00 |
Polynomial Roots Calculator found no rational roots
Trying to factor as a Difference of Squares :
6.4 Factoring: x2 - 1
Check : 1 is the square of 1
Check : x2 is the square of x1
Factorization is : (x + 1) • (x - 1)
Equation at the end of step 6 :
-12x4 • (x2 + 1) • (x + 1) • (x - 1) = 0
Step 7 :
Theory - Roots of a product :
7.1 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 :
7.2 Solve : -12x4 = 0
Multiply both sides of the equation by (-1) : 12x4 = 0
Divide both sides of the equation by 12:
x4 = 0
x = ∜ 0
Any root of zero is zero. This equation has one solution which is x = 0
Solving a Single Variable Equation :
7.3 Solve : x2+1 = 0
Subtract 1 from both sides of the equation :
x2 = -1
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
In Math, i is called the imaginary unit. It satisfies i2 =-1. Both i and -i are the square roots of -1
The equation has no real solutions. It has 2 imaginary, or complex solutions.
x= 0.0000 + 1.0000 i
x= 0.0000 - 1.0000 i
Solving a Single Variable Equation :
7.4 Solve : x+1 = 0
Subtract 1 from both sides of the equation :
x = -1
Solving a Single Variable Equation :
7.5 Solve : x-1 = 0
Add 1 to both sides of the equation :
x = 1
5 solutions were found :
- x = 1
- x = -1
- x= 0.0000 - 1.0000 i
- x= 0.0000 + 1.0000 i
- x = 0
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