How do you factor squares of binomials?

Answer 1

You should factor a2 - 9 and 4a2 - 4a + 1 by using certain algebraic properties listed below. First: (x + y)(x - y)=x2 - y2 Second: (x - y)(x - y)=x2 - 2xy + y2 So to factor a2 - 9 you use the first property to determine it is (a + 3)(a - 3). To factor 4a2 - 4a + 1 you use the second property to get (2a - 1)(2a - 1).

Answer 2

One of the most common ways we factor polynomials is through the use of the grouping method.

Let's take a look at three examples. Example one will contain a greatest common factor other than one with no remaining variables. Example two will contain a greatest common factor other than one, with remaining variables. Example three will contain a greatest common factor of one.

Example 1:

Please factor x² + 3x + 4x + 12x

When factoring, our first step should be to check if any of the terms (x², 3x, 4x, and 12x) share a common factor. Do they? Yes, the variable x is found in each of the terms.

If you find a common factor shared among all the terms, we need to pull it out. What ever we pull out goes out front of brackets.

x[ ................... ]

Since we have found that all four terms share the common factor of "x" we need to remove an x from each of the terms. Remember, we pulled it out from the problem so we need to restore equilibrium to our polynomial.

x² - x = x

3x - x = 3

4x - x = 4

12x - x = 12

Now that we removed an x from each of the terms, we can go ahead and plug back in the operations from the problem.

x + 3 + 4 + 12

Also remember you have that "x" you pulled from inside.

x[x + 3 + 4 + 12]

Now we take a look inside our brackets, do we have any like terms? Yes, 3, 4, and 12 are like terms. We can go ahead and combine these.

x[x + 19]

Lastly in this example, we check to see if we have anymore like terms inside the brackets. Do we? No. Our polynomial has been fully factored.

So, the factorization of x² + 3x + 4x + 12x = x[x + 19]

Note: Brackets and parentheses can be used interchangeably, however in the following examples, you will see how the use of brackets and parentheses can help to keep a polynomial cleaner.

Example 2:Please factor the following by grouping: 3x2 - 15x + 6x - 30

As in our first example, we should first look at the problem to determine if each of the terms (3x2, -15x, 6x, and -30) share a common factor. Well if we look are the variables first, 3x², -15x, and 6x, all share an "x" but -30 doesn't. So we've ruled out that the variable is the greatest common factor.

Now we look at the coefficient, or numbers, in the problem. Can any number divide into 3, -15, 6, and -30? Yes, 3.

3 ÷ 3 = 1

-15 ÷ 3 = -5

6 ÷ 3 = 2

-30 ÷ 3 = -10

Once again, just like our first example. We need to "yank" that common factor of "3" out of the numbers and place it outside the brackets.

3[ ........... ]

So we've yanked that 3 out, we can go ahead and place the remainder of the numbers back into our brackets.

3[x² - 5x + 2x - 10]

When we factor polynomials we're looking for two terms that share a common factor. Which terms in the above polynomial share common factors? x² and 2x (the "x" is common) and -5x and -10 (the "-5" is common).

Here, is where we introduce the parentheses. Just like in the next example, we need to separate the terms so the ones with like terms are next to each other.

3[(x² + 2x) - (5x - 10)]

Let's take a look at our first two terms inside the parentheses. What common factor is shared between the two terms? The "x" variable. Just like we did with the common factor 3, we need to yank that "x" out of there. However, it stays inside the brackets, just goes out front of the parentheses.

3[x(x + 2) - (5x - 10)]

As we did for the first two terms, we need to repeat the same steps for the second two steps.

3[x(x + 2) - 5(x + 2)]

Now that we've pulled out common factors out of our polynomial we found that the two terms have something in common. The factor of (x + 2).

We go ahead and pull this out, remembering to keep the 3 we pulled out earlier outside our brackets. We call the common factors we find inside out parentheses the "inner numbers".

3[(x + 2) ..... ]

Now that we've pulled out common inner numbers out, we need to take a look at the numbers that are outside the parentheses. We have, "x" and "-5". These are our "outer numbers". The outer numbers are the second pair of factors we're looking for.

3[(x + 2)(x - 5)]

You've now fully factored the original problem - 3x2 - 15x + 6x - 30 = 3[(x + 2)(x -5)

Example 3:

Please factor the following problem by grouping: x² + 3x + 2x + 6

Just like in the previous two examples. Our first goal should be to determine if the entire problem shares a common factor. Look at your variable first, the first three terms have an "x" variable in common, but 6 does not. The last three terms have no common factors what so ever. So since we've ruled out that the problem has a common factor, we say that the greatest common factor (GCF) is 1.

Now we proceed to factor like in the second term. Break the polynomial into two sets of two terms.

(x² + 3x) + (2x + 6)

In the first set of terms, are there are common factors shared between them? Let's take a look this time.

x² = x • x

3x = 3 • x

What is shared between the two? The variable "x". So like before, we yank that "x" out and place the "x' out side the parentheses while keeping the remainder of the stuff from the two terms inside the parentheses.

x(x + 3) + (2x + 6)

Just like the first set of terms, we need to take a look at the second set of terms. What is common between the second set of terms? Let's take a look.

2x = 2 • x

6 = 2 • 3

What is common between these two sets of terms? The "2" is. So like before with the x's in the earlier set of terms, we go ahead and pull those from the problem as well; placing them outside the parentheses.

x(x + 3) + 2(x + 3)

Similar to our earlier problem, look at our common "inner" factors. We can go ahead and write those down since they are common at this point.

(x + 3)( .... )

What about the "outer terms"? Exactly, we need to write those as the second set of parentheses.

(x + 3)(x + 2)

Now our problem has been factored to its fullest extent. x² + 3x + 2x + 6 = (x + 3)(x + 2).

Lastly, you may be wondering how to know if your factored answers are mathematically correct. Here is where we introduce the principal of distribution and the F.O.I.L method.

Example 1 Solution Check:

x² + 3x + 4x + 12x = x[x + 19]

If we distributed the x inside the brackets we would result in the following:

x • x = x²

x • 19 = 19x

x² + 19x

Now you're problem wondering why out redistributed answer looks different from our original answer. The answer is exactly the same as our original problem.

x² + 3x + 4x + 12x = x² + 7x + 12x = x² + 19x

Example 2 Solution Check:

3x² - 15x + 6x - 30 = 3[(x + 2)(x - 5)]

Ignoring the outside three for now, we want to take a look at our two factors inside the parentheses.

3x² - 15x + 6x - 30x = 3[(x + 2)(x - 5)]

We can turn this back into an original part of our equation by using a mathematical method called "Foiling" or "F.O.I.L" or First, Outer, Inner, and Last terms.

Now you may be asking, well what are the first, outer, inner, and last terms out of problem. I'll show you.

(x + 2)(x - 5)

First Terms: x • x = x²

(x + 2)(x - 5)

Outer Terms: x • (-5) = -5x

(x + 2)(x - 5)

Inner Terms: 2 • x = 2x

(x + 2)(x - 5)

Last Terms: 2 • (-5) = -10

Now we simply bring out outside three down and place in parentheses the newly foiled terms.

3[x² - 5x + 2x - 10]

Lastly, we redistribute that 3 we pulled out earlier.

3 • x2 = 3x2

3 • -5x = -15x

3 • 2x = 6x

3 • -10 = -30

If we rewrite our terms, what do we get? 3x2 - 15x + 6x - 30. Exactly our original problem.

Problem 3 Solution Check:

x² + 3x + 2x + 6 = (x + 3)(x + 2)

Exactly as we did with our last solution check, we need to F.O.I.L our answer.

x • x = x²

x • 2 = 2x

3 • x = 3x

3 • 2 = 6

Rewrite your new figures. You result in: x2 + 2x + 3x + 6. Exactly the same as our original problem.

Note: Even though our newly formed polynomial appeared in a different order than our original, it is still correct. As an example of this, imagine the commutative property. (a + b) = (b + a). So would 1 + 2 be any different than 2 + 1? No, they are just in a different, or reversed, order.

Answer 3

There are many different methods to factor polynomials in general; specifically for binomials, you can check:

• whether you can separate a common factor,
• whether the binomial is the difference of two squares,
• whether the binomial is the sum or difference of two cubes (or higher odd-numbered powers)

Answer 4

You will find more details about this in a high school algebra book - I suggest you study that. It isn't easy. But in summary, you try out several special cases.

* First, you should check whether there is any common factor among the different monomials. If there is, you take that out as a common factor.

* You should also watch out for special cases, such as the difference of two squares, the sum of two cubes, the difference of two cubes, or a trinomial that is a perfect square, and use the corresponding formulae. For example, x squared minus a squared is equal to (x + a) (x - a).

* A case that appears quite frequently is a degree-2 trinomial, which can be separated into two binomials. The details here are a bit complicated; better check your algebra textbook. But basically - and in the simplest case - you need to find two factors, (x + a)(x + b), such that the product ab is equal to the third coefficient, and the sum a + b is equal to the second coefficient.

Answer 5

(x2-3x-10) divided by (x+2)