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A polynomial is a mathematical expression containing multiple terms. For example, the following expressions are polynomials:
Factoring polynomials is a way to rewrite them in a simpler format, breaking them down into smaller components. We are answering this question: what do you have to multiply to get this expression? For example, x2 + 3x + 1 becomes (x+2)(x+1). 👍
Sometimes, this is the only thing you can do to factor. You could go further and try to get more factors out of an expression, but generally, you don’t need to; if you get to a point where decimal quantities are a part of your expression, like (x+3.4)(x-3.2), you should deem that your stopping point. Rarely will problems ever require a non-whole-number answer. 💘
Example: Factor the expression 8x^4 + 10x^3 + 6x.
What can we divide out of each coefficient (the number next to a variable, i.e. x) of each term in the expression? 2. What can we divide out of every term? x! Therefore, the greatest common factor of each term in the expression is 2x, and the factored solution is as follows: 2x (4x^2 + 5x + 3).
This method relies on a common factor, typically a binomial (two-term polynomial), within a polynomial once it is separated. Because not every polynomial has a common binomial once separated, this doesn’t always work, but it is worth trying if you can spot a common binomial! 🔭
Example: Factor the expression 4x^2 + 8x + 2x + 4.
First, let’s separate the polynomial into two expressions: (4x^2 + 2x) + (8x + 4).
We can use the same strategy of factoring out a GCF to simplify each part. (4x^2 + 2x) becomes x (4x+2) and (8x + 4) becomes 2 (4x+2). We multiplied 2 different quantities by the same factor, (4x+2): x and 2. Therefore, the expression simplified by grouping becomes (x+2)(4x+2).
This method works mainly for simpler 3-term AKA quadratic polynomials. Still, if you have times tables of larger numbers memorized or have practiced factoring polynomials more, you can use it on more complicated quadratics or even cubic polynomials.
Example 1: Factor x^2 - 3x - 2
While guess-and-checking, we must look for a term that adds to the middle term's coefficient and has a product of the last term of a trinomial/quadratic (three-term) expression. In this case, what numbers add to -3 and yield a product of -2? -2 and -1 do.
Those are the numbers we insert into the format of factors of a quadratic equation (x-(factor))(x-(factor)): (x-2)(x-1).
Example 2: Factor x^2 - x - 12.
What numbers add to -1 and yield a product of -12? -4 and 3, so we insert them into the aforementioned mold to get an answer of (x+3)(x-4)!
Special cases of polynomials have formulas that always work. Using them can save time:
One of the most common special cases is the difference of squares, the third formula above: a^2-b^2 = (a+b)(a-b). x^2-4 qualifies as a difference of squares.
Of course, the b-term in the difference of squares is not explicitly written; you can see that 4 = 2^2 and that 2 is, therefore, the b-term in this expression. Thus, the solution is (x+2)(x-2).
The formula for a difference of cubic expressions or “cubes” is a^3-b^3 = (a-b)(a^2+ab+b^2). For a sum of cubes, the formula isn’t much different: (a+b)(a^2-ab+b^2). One way to differentiate between the two formulas is to recall that the sum of cubes formula has alternating signs: +, -, +, whereas the difference of cubes is -, +, +.
The first special case listed in the image above is a^2+2ab+b^2 = (a+b)^2. For example, 9x^2+24x+16 rewritten as (3x)^2 + (23x4) + 4^2 leads us to (3x+4)^2.
The second special case listed is a^2-2ab-b^2 = (a-b)^2. The expression 16x^2+16x+4 fits this special case. Rewritten to fit the formula, it becomes (4x)^2 + (24x2) + (2^2). Our final answer is then (4x+2)^2!
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