We have three responses for you.

Hi John.

It looks to my like it already is "solved". But perhaps you meant that that algebraic expression is equal to 0. With an equal sign, it becomes an equation that we can solve.

If (2x+3)(6x+8) = 0, then ask yourself this: given two numbers that multiply to give me zero, what can I say about the two factors? The answer is that at least one of them must be zero, because zero times
anything is zero.

The two factors you have are (2x+3) and (6x+8), so you know at least one of these is equal to zero. Either:
2x + 3 = 0

or

6x + 8 = 0

You need to solve each of these simpler questions independently and you will get two possible values of x that will solve the equation (2x+3)(6x+8) = 0.

I hope this helps,
Stephen La Rocque.

Hi John,

I had the same question as Sue about what you sent. You didn't say what was to be solved. My guess as to your question was different than Sue's. I thought you were asking to expand the expression (2x+3)(6x+8).

To expand this expression you need to use the distributive law of multiplication over addition. This is the fact that for any numbers a, b and c

a(b + c) = ab + ac

In your expression take a = (2x + 3), b = 6x and c = 8. The distributive law then gives

(2x + 3)(6x + 8) = (2x + 3)6x + (2x + 3)8

Since multiplication is commutative, that is ab = ba for any numbers a and b, you can write this as

(2x + 3)(6x + 8) = 6x(2x + 3) + 8(2x + 3)

Now apply the distributive law again to each of the two terms on the right. For example

6x(2x + 3) = (6x)(2x) + (6x)3 = 12x² + 18x

Apply the distributive law to the second term and then collect terms to arrive at a final expression.

If you practice this you will soon be able to do it more quickly and perform some of the intermediate steps in your head.

Penny

Hello, when looking at questions like this it is always good to look at the basis of where the process is coming from. If we have the question x(x+3)= what we need to do is distribute the first term, x in this case, to everything within the brackets, that is x(x+3) = (x)(x) + (x)(3) and then do our monomial multiplication to get x² + 3x. Now in this case we are dealing with two monomials, but the idea behind it is exactly the same. We want to distribute the first term to everything inside the brackets of the second term, that is (2x+3)(6x+8) = (2x+3)(6x) + (2x+3)(8). and then multiply from there. I the way I explained this to my students is that everything in the first set of brackets must be multiplied to everything in the second bracket so in the example you will get (2x+3)(6x+8) = (2x)(6x) + (2x)(8) + (3)(6x) + (3)(8) and then simplify from there. Hope that helps you. One thing to watch out for when doing these questions is what signs each term has. In this question everything was positive, however, if they are negative you will need to do multiply with positive and negative numbers.

Brennan