factor each polynomial completely. to begin, state which method should be applied to the first step, given the guidelines of this section. Then continue the exercise and factor each polynomial completely.
2p - 6q + pq - 3q^2 Please help, I don't know how or where to begin.
I don't know what the "guidelines of this section" are.
You can factor a p out of two terms and a q out of the other two. This results in
p(2 + q) - 3q(2 + q)
Since 2+q (or q+2, which is the same thing) is a factor of both of these terms this an be be rewritten as
(p-3q)(q+2)
To factor the polynomial completely, you can apply the method of factoring by grouping. Here's how you can proceed:
Step 1: Group the terms in pairs:
(2p - 6q) + (pq - 3q^2)
Step 2: Factor out a common factor from each pair of terms:
2(p - 3q) + q(p - 3q)
Step 3: Notice that the terms (p - 3q) appear in both groups. Factor them out:
(p - 3q)(2 + q)
Therefore, the polynomial 2p - 6q + pq - 3q^2 can be factored as (p - 3q)(2 + q).
To factor the polynomial 2p - 6q + pq - 3q^2 completely, we can follow these steps:
Step 1: Grouping
Since we have four terms, we can try grouping the terms in pairs. Let's group the first two terms and last two terms together:
(2p - 6q) + (pq - 3q^2)
Step 2: Factoring out common factors
Now, let's look for common factors in each group. The first group (2p - 6q) has a common factor of 2, so we can factor it out:
2(p - 3q)
The second group (pq - 3q^2) has a common factor of q, so we can factor it out:
q(p - 3q)
Step 3: Common factor (p - 3q)
Now that we have the common factor (p - 3q) in both groups, we can rewrite the expression:
2(p - 3q) + q(p - 3q)
Step 4: Factor the common binomial (p - 3q)
The expression now has a common binomial factor of (p - 3q). We can factor it out:
(p - 3q)(2 + q)
Therefore, the polynomial 2p - 6q + pq - 3q^2 is completely factored as (p - 3q)(2 + q).
Remember, the first step in factoring a polynomial is to try grouping the terms. Then, look for common factors in each group, and finally factor out any common binomial factors.