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The greatest common factor (GCF) is the largest number that can divide each term of a polynomial without leaving a remainder. Finding the GCF is often the first step in factoring polynomials.

Let's look at part (a) of the exercise: the polynomial is written as 10q^2 + 50.

The terms in the polynomial are 10q² and 50. To find the GCF, start by identifying the numbers that can divide both 10 and 50. The number 10 is common to both terms:

• 10 divides 10q² to give q²
• 10 divides 50 to give 5

By factoring out the GCF, we write the polynomial as:

\[10(q^2 + 5)\]This simplifies the polynomial and makes it easier to work with.

Practice finding the GCF for different polynomials to get comfortable with this fundamental step.

Factoring trinomials, especially those that are quadratic, is a key skill. A quadratic trinomial has the form ax^2 + bx + c, where a, b, and c are constants.

Consider part (b) of the exercise: a^2 - 5a - 14.

To factor this trinomial, we need to find two numbers that multiply to give -14 (the constant term) and add to give -5 (the coefficient of the middle term). These numbers are -7 and 2:

• (-7) x 2 = -14
• (-7) + 2 = -5

We rewrite the trinomial using these numbers:

\[a^2 - 5a - 14 = a^2 - 7a + 2a - 14\]Then, we can factor by grouping:

\[(a^2 - 7a) + (2a - 14)\]We factor out the common factor in each group:

\[a(a - 7) + 2(a - 7)\]Finally, we factor out the common binomial factor:

\[(a - 7)(a + 2)\]And that's it! The trinomial is factored into two binomials.

Factoring by grouping is effective when you have a polynomial with four terms. It involves pairing terms into groups that have common factors.

Look at part (c) of the exercise: uv + 2u + 3v + 6.

First, group the terms:

\[(uv + 2u) + (3v + 6)\]Next, factor out the common factors from each group:

• u is common in the first group, so we factor it out: u(v + 2)
• 3 is common in the second group, so we factor it out: 3(v + 2)

Now we have:

\[u(v + 2) + 3(v + 2)\]Notice that (v + 2) is a common binomial factor, so we factor that out as well:

\[(u + 3)(v + 2)\]By grouping, we have factored the polynomial into the product of two binomials.

This method is handy when direct factoring is not evident.