Difference Of Squares X^2-4

[Daisy Hughlett]

Thisis 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.

We now have several methods in our factoring arsenal. We should always look for a greatest common factor of all terms first, then how we proceed is primarily determined by the number of terms we have. If we have 4 terms, we should look at factoring by grouping. When we have a trinomial function of the form [latex]f(x)=ax^2+bx+c[/latex] trial and error or the ac-method are well suited for factoring. In this section, we are going to consider binomial functions of a special type; the difference of squares, and the sum or difference of cubes.

When we multiply two binomials, we usually end up with a trinomial. For example, [latex](x+3)(x-4)=x^2-x-12[/latex]. That is why we try to factor trinomials back into the product of two binomials. However, there is an occasion when the coefficients of the [latex]x[/latex]-terms end up with opposite signs and cancel each other out. For example, [latex](2x-3)(2x+3)=4x^2\colorblue+6x-6x-9=4x^2-9[/latex]. When this happens we end up with a difference of two squares. [latex]4x^2=(2x)^2[/latex] and [latex]9=3^2[/latex].

Perfect squares always have an exponent that is divisible by 2. For example, [latex](x^2)^2=x^4[/latex]; [latex](x^3)^2=x^6[/latex]; [latex](x^5)^2=x^10[/latex]; 4, 6, and 10 are all divisible by 2. So the difference of two squares can be applied when exponents are even.

Another binomial form that can be factored is the sum or difference of cubes. When we multiply a binomial and a trinomial we usually end up with a polynomial with 4 terms. Occasionally, some of the terms get cancelled out and we end up with the sum or difference of two cubes, [latex]a^3+b^3[/latex] or [latex]a^3-b^3[/latex]. Although the sum of two squares cannot be factored, the sum of two cubes can be factored into a binomial and a trinomial.

One way to factor an expression is to use the difference of two squares. Writing a binomial as the difference of two squares simply means you rewrite a binomial as the product of two sets of parentheses multiplied by each other. For example, \(a^2-b^2=(a+b)(a-b)\). The binomial \(a^2-b^2\) can be factored into two sets of parentheses multiplied by each other. \((a+b)(a-b)\) will produce \(a^2-b^2\) when multiplied.

Not all expressions can be factored using this method. There are a few clues to look for when determining whether an expression can be factored using the difference of squares. Notice in the previous example \(a^2-b^2\) that each term is a perfect square, and there is a subtraction symbol between each term. These are two helpful clues to look for when determining if a binomial can be factored using the difference of two squares. If these two clues are present, then the expression can be factored using the difference of squares.

The original subtraction symbol between the terms is what allows these middle terms to cancel out. If the original expression was \(x^2+9\), with addition between the two terms, it would not be possible to factor this using the difference of squares. If we tried to break this apart into \((x+3)(x+3)\), the result would be \(x^2+3x+3x+9\), which simplifies to \(x^2+6x+9\), which does not match the original expression. The important thing to remember is that subtraction between the two terms is required for an expression to be factored using the difference of squares.

Consider the binomial \(4x^2-49\). Can this be factored using the difference of squares? Subtraction is the symbol between both terms, which is a good start. The first term is raised to the power of two which is also good. When we look at the second term, \(49\), we notice that this can be written as \(72\). Now we have \(4x^2-7^2\). All clues indicate that the expression \(4x^2-49\) can be rewritten as the difference of two squares. \(4x^2-49\) becomes \((2x+7)(2x-7)\). We can check this by (FOIL)ing and checking that the product is in fact \(4x^2-49\).

To review, we know that a binomial can be written as the difference of two squares if both terms are perfect squares and they are separated by subtraction. The difference of two squares is a useful theorem because it tells us if a quadratic equation can be written as the product of two binomials.

The expression \(9x^2-16y^2\) can be factored using the difference of two squares because both terms are perfect squares, and the squared terms are separated by the subtraction symbol. \(9x^2\) can be expressed as \(3x\) times \(3x\), and \(16y^2\) can be expressed as \(4y\) times \(4y\). \(9x^2-16y^2\) becomes \((3x+4y)(3x-4y)\).

\(x^2-2\) is separated by subtraction, and the first term is a perfect square, however \(2\) is not a perfect square. There is no number that can be squared, with a result of \(2\). This expression cannot be factored using the difference of squares.

If both terms in a binomial are perfect squares, and the terms are separated by subtraction, then the binomial can be factored using the difference of squares. For example, \(x^2-4\) can be factored using the difference of squares because both terms are perfect squares, and they are separated by a subtraction sign.

The binomial \(9x^2-49y^2\) can be factored using the difference of squares because both terms are perfect squares and they are separated by a subtraction symbol. The first term \(9x^2\) can be split into \(3x\) times \(3x\). The second term \(49y^2\) can be split into \(7y\) times \(7y\). One set of parentheses needs to be addition and the other needs to be subtraction so that the middle term cancels out when FOILing.

(\(21xy\) and \(-21xy\) cancel out)

\(9x^2-49y^2\) becomes \((3x+7y)(3x-7y)\).

This is risky, as could be a viable answer too. Do not use this method unless you're sure about the answer. In other words, this solution is less reliable than the others, so only use it if you can't do the other methods.

Doing this we get Now you might be tempted to expand, but notice how we have two perfect squares in our equation: and which motivates us to take advantage of the difference of squares. For simplicity, we suppose that and Thus our equation is now We manipulate this equation by getting the LHS in a form that lets us do difference of squares, so we can rewrite as

Doing that, we get Also, when we set this equation up we not only do we make our fact useful, but we also can see that after expanding the RHS, we get on the RHS too, which makes it even a better move that we added both equations because remember our goal is to find

The calculator will try to factor any expression (polynomial, binomial, trinomial, quadratic, rational, irrational, exponential, trigonometric, or a mix of them), with steps shown. To do this, some substitutions are first applied to convert the expression into a polynomial, and then the following techniques are used: factoring monomials (common factor), factoring quadratics, grouping and regrouping, square of sum/difference, cube of sum/difference, difference of squares, sum/difference of cubes, and the rational zeros theorem.

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.

The main difference between factoring x^6 - y^6 as a difference of squares and as a difference of cubes is the number of terms in the final factored expression. When factoring as a difference of squares, the final expression will have two terms, while factoring as a difference of cubes will result in three terms.

To factor x^6 - y^6 as a difference of squares, you need to use the formula (a^2 - b^2) = (a + b)(a - b). In this case, a = x^3 and b = y^3, so the factored expression would be (x^3 + y^3)(x^3 - y^3).

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