分组：三次多项式

Grouping Cubics

We can break a polynomial into smaller groups with a common factor. This method is especially helpful when factoring cubic functions. This is called factoring by grouping. Rearranging the terms in descending exponent order helps. Here's an example: Let's say you need to factor $$3x^2+6+2x+x^3$$

First, let's rearrange the terms.

We can see the decreasing exponent order, where the cubic term is first, and the constant term is last. You'll see why the order is important in this next step.

We can factor this expression by literally boxing terms into groups, as seen below:

 We use these groups because each group has a common factor. The first two terms have a common factor of $x^2$. The last two terms have a common factor of $2$. Let's factor out each term.

Each group has a common factor of $x+3$. We can add these groups back together to get: $\boxed{(x^2 + 2)(x + 3)}.$

As you can see, we looked for the greatest common factor, GCF, between the first two terms and the last two terms. Then, we essentially factored out the $x+3$. You can use this pattern for nearly all polynomials.

Once you master these fundamentals, you will apply these concepts to more complex topics, like sums and differences of cubes.

What If We Don't Have a GCF?

Remember that the acronym GCF stands for greatest common factor. It is the largest possible term that can divide evenly into each term of an expression. Factoring the GCF is a good way to start factoring, it doesn't always work.

Some polynomials have four terms that don't share a factor. You can still factor these, but you need to start by grouping the terms.

We can use the following steps when there are four terms and no common factors:

Let's work through an example. Suppose that we want to factor the polynomial $2x^3+7x^2-8x-28$.

STEP 1: Put the terms in order of degree

STEPS 2-3: Group the first two terms and the second two terms

We use parentheses to group the first two terms and the last two terms.

$$(2x^3 + 7x^2) + (-8x-28)$$

STEP 4: Factor a GCF out of each newly grouped pair of terms

Right again!

Watch your negative signs. Oftentimes, people get messed up when factoring out negative signs. In this example, we factored out a $-4$, which means we have to divide the first term, $-8x$, by $-4$, which leaves us with $2$. We do the same thing with the second term. We factor out the $-4$, which means we divide the second term, $-28$, by $-4$. That equals 7.

Factoring out GCFs leaves us with the following expression:

$$x^2(2x+7) -4(2x+7)$$

Our expression contains a common factor, $2x + 7$. Factoring out that binomial leaves the factored form of the polynomial:

$$(x^2-4)(2x+7)$$

At this point, we have factored our original polynomial successfully. However, we have not yet fully factored it, and we're really close to doing so!

Take a closer look at the first binomial in our new expression, $\left(x^2-4\right)$.

If you didn't notice that pattern at first, don't worry! With enough practice, you'll start to recognize factoring patterns.

And we're finally done! The fully factored form of our original polynomial is:

$$2x^3+7x^2-8x-28 = (x-2)(x+2)(2x+7)$$

(Video) Factoring by Grouping

by mathman1024

This video by mathman1024 demonstrates the factor by grouping method.

$\large{\text{Summary}}$

The first polynomial our instructor factors is $x^3+3x^2+7x+21$. He uses parentheses to group his polynomial into two binomials:

$$(x^3+3x^2)+(7x+21)$$

Next, he factors a greatest common factor out of each binomial.

$$x^2(x+3)+7(x+3)$$

The polynomial now consists of two terms. Both are multiplied by the binomial $x_3$. He can factor out that binomial, which gives us the following:

$$(x+3)(x^2+7)$$

And there's our answer!

He goes on to factor one more example polynomial: $9x^3-12x^2-3x+4$. Try to factor this polynomial on your own. You can use the explanation in the video to check your answer.

$$\begin{align} -x^3+x^2-3x+3&=x^2(-x+1)+3(-x+1)\\ &=(x^2+3)(-x+1) \end{align}$$

A common factor is a term or expression that divides evenly into all of a polynomial's terms. For example, the polynomial $ab + ac$ has a common factor of $a$.

Now, when we are tasked with factoring an expression that contains a common factor, the first thing we want to do is factor out that common factor.

So, if we needed to factor the polynomial $ab + ac$, we would start by factoring out the polynomial's common factor of $a$, like so:

$$ab + ac$$ $$= a(b + c)$$

On a side note, if we were to multiply the term $a$ by the binomial $b + c$, we would get our original $ab + ac$ expression. You can utilize this technique of redistributing a term that you've just factored out to check your work when factoring other polynomials.

Example 1:

Factor $5x + 10xy + 25xz^2$.

Each of this expression's terms can be divided evenly by the term $5x$. Therefore, the expression's factored form is $5x(1 + 2y + 5z^2)$.

Example 2:

Factor $\frac{27x + 9y + 36z}{3}$.

There are two ways you could go about factoring this expression.

Method 1: Divide all of the numerator's terms by $3$ and then factor the result.

$$\frac{27x + 9y + 36z}{3}$$ $$=9x + 3y + 12z$$ $$=3(3x + y + 4z)$$

Method 2: Factor the numerator and then divide that factored expression by 3.

$$\frac{27x + 9y + 36z}{3}$$ $$=\frac{9(3x + y + 4z)}{3}$$ $$=3(3x + y + 4z)$$

Example 3:

Factor $2x + 8x^2 + 5y$.

While this expression does not contain any common factors, its first two terms can be divided evenly by the term $2x$. This means that we can group those first two terms using parentheses and factor a common factor out of that newly created binomial, like so:

$$2x + 8x^2 + 5y$$ $$= (2x + 8x^2) + 5y$$ $$= 2x(1 + 4x) + 5y$$

Test Yourself!

(Videos) Factoring Trinomials: Grouping and Trial and Error

by PatrickJMT

These videos by Patrick JMT show how to factor by grouping.

$\large{\text{Summary}}$

The grouping method is used quite often in factoring. Let's look at the polynomial $$12x^2+34x+10$$

We factor out a $2$ to get $2(6x^2+17x+5)$. Now, we factor $6x^2+17x+5$. First multiply $6$ and $5$ to get $30$. What are two numbers that multiply to get $30$ and add to $17$. How about $2$ and $15$?

Now, we can rewrite the middle term using $2$ and $15$. $$2(6x^2+2x+15x+5)$$

Now, we factor by grouping. Look at the first two terms. What is the GCF? $2x$. What about the second two terms? $5$. Now we can rewrite such that $$2[2x(3x+1)+5(3x+1)]$$

Now, we can write this like $$2(2x+5)(3x+1)$$