Спочатку винеси за дужки одночлен

We already know how to factor quadratic expressions with the general form: $$ax^2+bx+c=0$$

Sometimes, though, we're tasked with factoring expressions that look almost quadratic and factorable...but aren't.

Consider this polynomial: $x^3-2x^2-15x$. How are we supposed to factor this?

Factoring Out a Monomial First

Many of these "weird" polynomials share a monomial in each term. That monomial gets factored out to give us a nicer polynomial.

That's right! Despite looking a little different, these are all monomials.

Right! This is basically the opposite of multiplying a monomial and a polynomial. We can factor the monomial $x$ out from our polynomial, which leaves us with an expression consisting of a quadratic that is being multiplied by a single $x$ term: $$x^3-2x^2-15x=x(x^2-2x-15)$$

It looks like there are a lot of negative numbers here. Always watch your signs, whether you're working with positive or negative numbers.

You may need to know how to factor with a negative $b$ value!

The goal of factoring a non-quadratic is to factor out the greatest common factor. Factoring a monomial is not always necessary. However, it can make difficult polynomials much easier.

Let's go over a few more examples.

Right! All of those items are factors that we could pull out of our polynomial. However, we want to find our polynomial's GREATEST common factor.

Yep! We're left with thexpression $2a(a^2-5a+4)$ after factoring out $2a$. We can then factor the quadratic portion of that expression to get our polynomial's fully factored form.

$$\begin{align} 2a(a^2-5a+4)&=2a(a^2-a-4a+4)\\ &=2a(a(a-1)-4(a-1))\\ &=2a(a-4)(a-1) \end{align}$$

$$\begin{align} 3b^3+12b^2-36b&=b(3b^2+12b-36)\\ &=3b(b^2+4b-12)\\ &=3b(b+6)(b-2) \end{align}$$

Remember, if you want to check your work for any of these problems, all you have to do is multiply out all of your factors in the correct answer! You may need to FOIL, square a binomial, or find the product of a sum of difference. Basically, to check your work, you're doing the opposite of what you just did!

(Video) Solving Polynomial Equations

by yaymath

This video by yaymath covers factoring with various polynomials.

$\large{\text{Summary}}$

First, our instructor demonstrates how to factor a four-term polynomial. The steps he uses to do so are as follows:

1. Split the polynomial down the middle. This helps factor it in pieces (if needed).
2. Factor out the greatest common factor from both pieces. You may need to use prime factorization.
3. Use the GCF of each piece to build the polynomial's first factor.
4. The GCFs should have created two identical binomials. Use one of these binomials as our polynomial's second factor.

The second section covers a difference of squares. Remember that a difference of squares polynomial has the form $a^2-b^2$. The steps he uses to factor this kind of polynomial are as follows:

1. If able, factor out any greatest common factors.
2. Write down two sets of parentheses. One will have a plus sign, the other a minus.
3. Take the square root of the polynomial's first term. Write the result in each parenthesis to the left of the sign.
4. Take the square root of the polynomial's second term and write the result inside both sets of parentheses, to the right of each sign.

1. Split the polynomial down the middle. This helps factor it in pieces (if needed). $$2xy-16x\quad \quad +\quad \quad 5y-40$$
2. Factor out the greatest common factor from both pieces. $$2x(y-8)\quad \quad + \quad \quad 5(y-8)$$
3. Use the GCF of each piece to build the polynomial's first factor. $$(y-8)(2x+5)$$
4. The GCFs should have created two identical binomials. Use one of these binomials as our polynomial's second factor. $$\boxed{(y-8)(2x+5)}$$

Use the distributive property to redistribute the factors to double check your work!

Using Common Factors to...Factor!

Factoring a quadratic equation lets us brake it into smaller parts. This makes it easier for us to solve for $x$. Let's take a look at a quadratic in its general and factored form.

Multiplying the factored form with the FOIL method gives us the original expression.

Sometimes, you might come across a quadratic equation that looks a little...ugly.

Notice how, in each of the examples we've seen so far, the first term in our quadratic has been $x^2$.

So what are we supposed to do when that first term is, for instance, $3x^2$?

Quadratics like these are harder to factor, but it's usually not impossible to do so.

We can try to simplify the quadratic by factoring out common factors.

First, let's practice identifying common factors.

Right! The largest number that can be divided evenly into both $10$ and $15$ is $5$. This means that $5$ is the greatest common factor of those two terms.

Yup! Both $4ab$ and $6a^2b$ can be divided evenly by the term $2ab$ ($\frac{4ab}{2ab}=2$, and $\frac{6a^2b}{2ab}=3a$.)

Now, let's look at how identifying greatest common factors can help us factor polynomials more easily.

Once we've determined what a polynomial's greatest common factor is, we can pull that common factor OUT of the polynomial:

Note: if we re-distribute that $3x$, we end up with our original polynomial.

In this case, we've pulled (i.e. factored) out our polynomial's greatest common factor ($3x$, in green).

We have removed the GCF, 3x. This should make factoring the polynomial much easier.

Yup! This means that we can write our factored polynomial. The common factor will be outside of the parentheses.

$$3x(x-7)(x-2)$$

This is the factored form of the expression on the left-hand side of our quadratic equation!

If you'd like, you can now re-distribute the common factor (for us, it's $3x$) to the terms in parentheses:

$$(3x^2 - 21x)(3x^2 -3x)$$