For the first one:

This is a general trinomial. To figure out its factors, first you look at the first and last terms and figure out what THEIR factors are – what numbers you can multiply to get to them.

In the case of 6x^2 + 11x – 7,

What can you multiply to get to 6? That’s easy – 2 times 3. Since both these numbers are prime, we know that these are the only factors for 6. So to get to 6x^2, what do you multiply? 2x times 3x.

Now, for the last term. What do you multiply to get to 7? Since it’s prime, its only factors are 1 and itself.

So – your factors are 2x times 3x, and 1 times 7.

Now, since this is a general trinomial and not a perfect square trinomial, we have to use an educated system of trial and error to find the answer. We combine the factors in different arrangements and see which arrangement is correct.

For the sake of time and space, I’ll just tell you that the factors for this trinomial are (2x +1) (3x – 7).

The first number in each factor comes from the factors for 6x^2 – 2x and 3x. The second number in each factor comes from the 7 – 1 and 7.

Now, multiply it out.

(2x + 1) (3x) + (2x + 1) (-7)

(2x) (3x) + (1) (3x) + (2x) (-7) + (1) (-7)

6x^2 + 3x + -14x + -7

6x^2 – 11x – 7

Now, if you wondered how I knew which operational sign to use in the factors, here’s how you can tell. Look at the sign of the last term, 7 – it’s negative. This means that the signs will be different. One will be positive, and one will be negative. We know this because a negative times a negative is a positive, and a positive times a positive is a positive.

Now, as to which sign goes to which factor – that part’s trial and error. After you multiply it out, if the answer is correct except that it’s negative where it should be positive or vice versa, just switch the signs in the factors and multiply again!

Your questions number 3 and 4 will be solved in the same way as this one.

Number 2 will be a little different because it’s a four term polynomial. For this kind, you use grouping symbols and then factor the smaller problems you have separated.

So for step one – insert grouping symbols.

(x^3 + x^2) + (3x + 3)

Now, instead of a four term polynomial, you have two much more manageable binomials. Look at the first one and see if there is anything that has been distributed to all the terms of the binomial.

Yes, an x^2. Factor it out:

(x^2) (x + 1) To double check that you did this right, you can multiply it out and make sure it comes back to the original binomial.

OK, now look at your second binomial. Has anything been distributed to all of the terms?

(3x + 3) Yes, a 3. Undistribute it –

(3) (x + 1)

Now that we have completely factored both binomials, let’s look at the big picture again. Our problem now looks like this:

(x^2) (x + 1) + (3) (x + 1)

Is there a term that has been distributed to both sides of the expression? Yes – (x + 1). Undistribute it and use the other two terms to create a second binomial and you have your answer:

(x + 1) (x^2 + 3)

Again, to make sure there are no mistakes, you can multiply it out again. If it comes back to the original polynomial, you will be sure you have the right factors.

(x + 1) (x^2) + (x + 1) (3)

(x) (x^2) + (1) (x^2) + (x) (3) + (1) (3)

x^3 + x^2 + 3x + 3

Yep, we’re right back to where we started! Now we know that we did the problem right.

I hope I helped you understand how it works – good luck with the rest of them! 🙂