Factoring Part 1

Numbers and algebraic expressions can be written as either sums or products. For example the number 12 can be written as 5 + 7 or as $$2 \cdot 6$$. In the product form, we call each multiplier a factor. Both 2 and 6 are called factors of 12. The prime factorization of a number is the number written as a product of prime factors (factors which can only be divided by 1 and themselves). So, the prime factorization of 12 is: $$2 \cdot 2 \cdot 3$$. We use prime factorization when we are looking for common factors between two or more numbers or two or more algebraic expressions.

It’s useful to be able to change a sum into a product, as well as to change a product into a sum. We use the distributive property to multiply a polynomial by a constant or a variable, thus changing a product into a sum:

Recall that you have to multiply every term in the polynomial by the number you are multiplying by. Well, factorization is the reverse process of this, in a sense: because we will change a sum into a product. In the sum, $$16x^2 + 24x - 40$$ we know that each term has factors. We imagine that each term might have a common factor. Notice that 8 is a factor of each term.

$$16x^2 + 24x - 40 = 8\left( {2x^2 } \right) + 8\left( {3x} \right) - 8\left( 5 \right)$$

$$16x^2 + 24x - 40 = 8\left( {2x^2 } \right) + 8\left( {3x} \right) - 8\left( 5 \right) =8\left( {2x^2 + 3x - 5} \right)$$

Use the same process when you are finding a common variable factor. For instance, in the expression, $$5x^2 - 4x^3 + 7x^5 $$, we notice that each term consists of products of x’s. Therefore there is a common factor of x. But what is the greatest common factor? In this expression, it’s the x with the largest power that is present in each term. To find this, we look at their exponents and find that the lowest exponent on x is 2. So, using the laws of exponents, we can rewrite each term as a multiple of $$x^2 $$:

$$5x^2 - 4x^3 + 7x^5 = x^2 \left( 5 \right) - x^2 \left( {4x} \right) + x^2 \left( {7x^3 } \right)$$

So, we can “factor out” the $$x^2 $$ by dividing it out of each term and writing the original sum as a product.

$$5x^2 - 4x^3 + 7x^5 = x^2 \left( 5 \right) - x^2 \left( {4x} \right) + x^2 \left( {7x^3 } \right) = x^2 \left( {5 - 4x + 7x^3 } \right)$$

Now, let’s look at an example in which we can factor out both constants and variables:

Write the following sum as a product by factoring out the least common factor from the following expression:

1. So, first let’s factor out any constants that we can. We look at each term in the polynomial and see that, first of all, they are all divisible by 2. So, let’s factor out a 2 by dividing it out of each term:

$$12xy^3 + 6x^3 y^4 + 60x^2 y^5 + 18y^3 = 2\left( {6xy^3 + 3x^3 y^4 + 30x^2 y^5 + 9y^3 } \right)$$

2. Now, we look at the terms inside the parentheses and see that they are each divisible by 3, so we can factor out a 3 by dividing each term by 3:

3. Each factor that we “factor out” can be multiplied together on the outside of the parentheses, since we divided them each out of the polynomial, so we can write this as:

$$2 \cdot 3\left( {2xy^3 + x^3 y^4 + 10x^2 y^5 + 3y^3 } \right) = 6\left( {2xy^3 + x^3 y^4 + 10x^2 y^5 + 3y^3 } \right)$$

Note that we could have factored out a 6 to begin with, if we had noticed that each term was divisible by 6 in the beginning.

4. Now, let’s look at the variables and see which ones are factors in all of the terms in the polynomial and what the greatest common factor would be. First, is x a factor in every term? No. It only appears in the first 3 terms, so there is no way that we can factor any x’s out of our expression.

5. How about the y’s? Each term has factors of y’s. What is the highest power of y that we can factor out of the expression? Well, the term with the smallest exponent on the y has an exponent of 3, so we can factor out $$y^3 $$ from each term:

$$6\left( {2xy^3 + x^3 y^4 + 10x^2 y^5 + 3y^3 } \right) = 6 \cdot y^3 \left( {2x + x^3 y + 10x^2 y^2 + 3} \right)$$

So, now we are done, since we have factored out the greatest common factor. There are no more numbers or variables that we can divide each term in the polynomial by evenly.

Rewrite each of the following sums as a product by factoring out the greatest common factor from the following expressions:

Solution: $$7\left( {y^4 - 2y^2 + 7} \right)$$

Solution: $$2x\left( {3x + x^2 - 6} \right)$$

Solution: $$xy^2 \left( {3y - 5x + 7} \right)$$

Solution: $$12ab^5 \left( {12 - 5a + 4b + 1} \right) = 12ab^5 \left( {13 - 5a + 4b} \right)$$

Solution: We cannot factor anything out of the expression because, even though the first 3 terms are divisible by a lot of the same factors, the last term, 3, is not.

Factoring Part 1

Expressing Sums as Products: the greatest common factor