X^2-1 Difference Of Squares

[Marilina Crawn]

Thedifference of squares is a term used to describe a polynomial that can be factored into two perfect squares with a subtraction sign between them. In the case of x2 +1 and x2 -1, neither of these expressions can be factored into two perfect squares. Therefore, they are not considered difference of squares.

Understanding the difference between these two expressions is important in algebra because it can help with factoring and solving equations. Knowing that x2 +1 and x2 -1 cannot be factored as difference of squares can save time and prevent errors when trying to factor them.

This is a factoring calculator if specifically for the factorization of the difference of two squares. If the input equation can be put in the form of a2 - b2 it will be factored. The work for the solution will be shown for factoring out any greatest common factors then calculating a difference of 2 squares using the idenity:

If a is negative and we have addition such that we have -a2 + b2 the equation can be rearranged to the form of b2 - a2which is the correct equation only the letters a and b are switched; we can just rename our terms.

A binomial is factorable only if it is one of three things a Difference of Squares, a Difference of Cubes, or a Sum of Cubes. A binomial is a Difference of Squares if both terms are perfect squares. Recall we may have to factor out a common factor first.

If we determine that a binomial is a difference of squares, we factor it into two binomials. The first being the square root of the first term minus the square root of the second term. The second being the square root of the first term plus the square root of the second term, as in the following formula:

If you weren't able to come up with the insight, then you could just notice that the answer is divisible by , and . We can then use Fermat's Little Theorem for on the answer choices to determine which of the answer choices are divisible by both and . This is .

After expanding the first few terms, the result after each term appears to be where is the number of terms expanded. We can prove this using mathematical induction. The base step is trivial. When expanding another term, all of the previous terms multiplied by would give , and all the previous terms multiplied by would give . Their sum is equal to , so the proof is complete. Since is equal to , the answer is .

We notice that the first term is equal to . If we multiply this by the second term, then we will get , and we can simplify by using difference of squares to obtain . If we multiply this by the third term and simplify using difference of squares again, we get . We can continue down the line until we multiply by the last term, , and get .

Whenever you have a binomial with each term being squared (having an exponent of [latex]2[/latex]), and they have subtraction as the middle sign, you are guaranteed to have the case of difference of two squares.

These are other ways to write the formula of the difference of two squares using variables. Learn to recognize them in various appearances so that you know exactly how to handle them.

The first term of the binomial is definitely a perfect square because the variable [latex]x[/latex] is being raised to the second power. However, the second term of the binomial is not written as a square. So we need to rewrite it in such a way that [latex]9[/latex] is expressed as some number with a power of [latex]2[/latex]. I hope you can see that [latex]9 = \left( 3 \right)^2[/latex]. Clearly, we have a difference of two squares because the sign between the two squared terms is subtraction.

For this example, the solution is broken down in just a few steps to highlight the procedure. Once you get comfortable with the process, you can skip a lot of steps. In fact, you can go straight from the difference of two squares to its factors.

Now, we can truly rewrite this binomial as the difference of two squares with distinct terms that are being raised to the second power; where [latex]16y^4 = \left( 4y^2 \right)^2[/latex] and [latex]81 = \left( 9 \right)^2[/latex]

Are we done already? Well, examine carefully the binomials you factored out. The second parenthesis is possibly a case of difference of two squares as well since [latex]4y^2 = \left( 2y \right)\left( 2y \right)[/latex] and clearly, [latex]9 = \left( 3 \right)\left( 3 \right)[/latex].

Now we can deal with the binomial inside the parenthesis. It is actually a difference of two squares because we can express each term of the binomial as an expression with a power of [latex]2[/latex].

You may keep it in that form as your final answer. But the best answer is to combine like terms by adding or subtracting the constants. This also simplifies the answer by getting rid of the inner parenthesis.

An alternative method that can be used to more easily solve this equation comes from first isolating \(x^2\) and then taking the square root of both sides of the equation. Besides its simplicity, this method allows us to solve equations that do not factor. Continuing the above example to illustrate this alternative approach:

In summary, when there is no linear term in a quadratic equation, one method to solve it is to use the square root property. In this approach, the \(x^2\) term (or more generally the squared term) is isolated first, and then the square root of both sides of the equal sign is taken.

\[\beginalign* 25x^4-9&= 7 && \text The Square Root Property can be used\\

25x^4&= 16 && \textIsolate the squared term\\

(x^2)^2&= \frac1625 \\

x^2 &= \pm \dfrac1625 && \textSquare root both sides. Remember the \pm \text !!!\\

x^2 &= \pm \dfrac45 &&\textSimplify. \endalign*\]

Not all quadratic equations can be factored or can be solved in their original form using the square root property. In these cases, we may use a method for solving a quadratic equation known as completing the square. Using this method, we add or subtract terms to both sides of the equation until we have a perfect square trinomial on one side of the equal sign. We then apply the square root property. To complete the square, the leading coefficient, \(a\), must equal \(1\). If it is not, then divide the entire equation by \(a\) before beginning the complete the square process. Then, we can use the following procedures to solve a quadratic equation by completing the square.

So far, all of the examples have had a leading coefficient of \(1\). If this is not the case, remove it. This can be done by dividing both sides of the equal sign by the leading coefficient before completing the square.

The fourth method of solving a quadratic equation is by using the quadratic formula, a formula that will solve all quadratic equations. Although the quadratic formula works on any quadratic equation in standard form, it is easy to make errors in substituting the values into the formula. Pay close attention when substituting, and use parentheses when inserting a negative number.

When an expression is both a difference of squares and a sum or difference of cubes, if factoring as a difference of squares is done first, a more complete factorization is obtained. For example, if given the equation \( 64x^6-1=0\) to solve, when it is first factored as a difference of squares as \( (8x^3-1) (8x^3+1) \), and then as a difference of cubes, the solutions obtained are \( \pm \frac12\) and \( \pm \frac14(1 \pm i\sqrt3 ) \). In contrast, if it is first factored as a difference of cubes as \( (4x^2-1) (16x^4 +4x^2+1) \), the solutions eventually obtained are \( \pm \frac12\) and \( \pm \frac14\sqrt -2 \pm 2i\sqrt3 \).

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The LINEST function calculates the statistics for a line by using the "least squares" method to calculate a straight line that best fits your data, and then returns an array that describes the line. You can also combine LINEST with other functions to calculate the statistics for other types of models that are linear in the unknown parameters, including polynomial, logarithmic, exponential, and power series. Because this function returns an array of values, it must be entered as an array formula. Instructions follow the examples in this article.

if there are multiple ranges of x-values, where the dependent y-values are a function of the independent x-values. The m-values are coefficients corresponding to each x-value, and b is a constant value. Note that y, x, and m can be vectors. The array that the LINEST function returns is mn,mn-1,...,m1,b. LINEST can also return additional regression statistics.

The degrees of freedom. Use the degrees of freedom to help you find F-critical values in a statistical table. Compare the values you find in the table to the F statistic returned by LINEST to determine a confidence level for the model. For information about how df is calculated, see "Remarks," later in this topic. Example 4 shows use of F and df.

The equation of a straight line is y = mx + b. Once you know the values of m and b, you can calculate any point on the line by plugging the y- or x-value into that equation. You can also use the TREND function.

The accuracy of the line calculated by the LINEST function depends on the degree of scatter in your data. The more linear the data, the more accurate the LINEST model. LINEST uses the method of least squares for determining the best fit for the data. When you have only one independent x-variable, the calculations for m and b are based on the following formulas:

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