Hi, and welcome. My name is Anthony Varela. And today we're going to go through more challenging examples of quadratic factoring. So first, we're going to review our basic quadratic factoring strategies. And then we'll talk about how to factor quadratics when a our value is not equal to 1.
So that would be our leading coefficient, the coefficient of the x squared term. And we'll see two different methods to factor when that a value is a coefficient other than 1.
So let's review our basic factoring techniques. So if we have a quadratic expression, x squared minus 2x minus 8, and we'd like to factor this. And so we want to identify two integers. We can call them p and q.
And when we add p and q, that's going to give us the x term coefficient. So we'll be looking for negative 2 when we add p and q. When we multiply p and q, we're looking for the constant term. So that would be a negative 8 when we multiply p and q.
So we're going to start with our negative 8. That's our constant term. And we'll list two integers that are factors of negative 8. So for example 1 and negative 8 would multiply to give us negative 8.
When we add these, we get negative 7. And we're not looking for negative 7. We're looking for negative 2.
So let's choose two more integers p and q. Well, 2 times negative 4 would give us negative 8. And when we add these two numbers together, we get our negative 2. So we've identified p as 2 and q as negative 4.
So what does that mean then with factoring? Well, we can express this then as x plus 2 times x minus 4. Well, that's our basic factoring techniques.
What happens when that leading coefficient is a number other than 1? So here we have 2x squared plus 5x minus 3. We'll also write this out in general factored form. It looks very similar to our example before, but we have these x terms here being multiplied by some number. So I'm going to call them n and m.
And looking at our first step of FOIL to get back into expanded form, that first step is to multiply the two x terms. So the coefficients in front of the x squared term has to be 2. And we see that's going to be a product of these two numbers right here.
So I know that mn has to equal 2 in this case. Well, it's nice because 2 is a prime number. So I know that it's going to look like 2x plus some integer p times 1x times plus some integer q.
Well, looking at our next steps to FOIL then, we can see that q is going to be multiplied by 2x and p is going to be multiplied by 1x. So when a does not equal 1, we can see then that either p and/or q must be multiplied by a number before we add them together to get our x term coefficient.
So we're still going to start by listing out factors of the constant term. So we might have 1 times negative 3. But remember, because q is being multiplied by 2x, I'm going to double q before I add it to p. So really, I'm adding 1 plus 2 times negative 3. And we can see that that sums to negative 5.
Well, I'm looking for positive 5. So let's choose another two integers. How about negative 1 and positive 3?
Once again, I have to double q before I add it to p. So negative 1 plus 2 times 3 is 5. And that's what we're looking for. So I've identified p as negative 1 and q as positive 3. So in factored form, this would be 2x minus 1 times x plus 3.
And we can FOIL this to make sure that we've factored correctly. So 2x times x is 2x squared. Then we have 2x times 3, so plus 6x.
Then we have negative 1 times x. So I will subtract an x. And finally, negative 1 times 3. So I'll subtract a 3. Combining those 2x terms in the middle, I get 2x squared plus 5x minus 3, which matches what we started with.
Well, there's another method to these more challenging quadratic factoring problems. So let's go ahead and factor 6x squared plus 11x minus 10. Now what we're going to do to start off this problem is we're going to multiply our a value, so that would be 6, and multiply that by the constant term, negative 10. So 6 times negative 10 is negative 60.
And we're going to list out p and q values that would multiply 10 and negative 60 and then we'll add them together and see if we get our constant term. So one possibility might be 6 and negative 10. But when we add those together I get a negative 4. That's not what I'm looking for.
So let's try another two numbers-- 4 times negative 15. But when we add these, we get negative 11. I'm looking for positive 11. So I'll use 4 and 15 again, but I'll switch the signs, so negative 4 and positive 15.
Well, when we add these together, we get our positive 11. And that's what we're looking for-- positive 11. So how am I going to use then negative 4 and positive 15 to factor?
Well, this is also referred to as the box method because I'm going to create a two-by-two grid. And in two of these corners, I'm going to put our 6x squared and our negative 10, so the x squared term and our constant term.
Now in the other two boxes I'm going to put our x terms. And these are our coefficients. Now it doesn't matter if I were to put 15x here and negative 4x here. That's OK. So I'm going to write in in our other two corners then our x terms using these as the coefficients.
And now what I'm going to do is factor our rows and columns. So what I mean by that is looking at this first column right here I'm going to factor out a common factor between 6x squared and 15x. Well, I notice numerically they share a common factor of 3 and they share an x term. So I'm going to factor out 3x.
In this other column, they're both negative, so I can factor out a negative 1. I'm also going to factor out a 2. And I'm not going to factor an x because they both don't share an x term. So I'm going to be factoring out a negative 2.
Looking then at my rows, so these numbers, both of them are not negative. So I can't factor a negative number. I can factor out a 2 and I can factor out an x. So I'm going to factor out 2x.
And here, once again, one's a positive and one's a negative 1. So I can't factor out a negative. I can't factor out an x because this doesn't have an x term. But I can factor out a 5. So I have a positive 5 here.
And the interesting thing then about this box method is these are our two binomial rows that are multiplied together to get this expression. So the box method is another way to factor these more challenging problems where we have an x term coefficient that isn't equal to 1.
So we said that this quadratic expression is the same as this quadratic expression when we factored it. So let's go ahead and confirm by foiling. So 3x times 2x is 6x squared, 3x times 5 is 15x, negative 2 times 2x is negative 4x, and negative 2 times 5 is negative 10. And we can see then that positive 15x minus 4x is positive 11x. So we've factored this correctly.
Let's review our notes. With basic factoring, we're looking for two integers p and q such that the sum is the x term coefficient and the product is the constant term. But when a does not equal 1, p and q must be multiplied by a number before added together.
And we also saw this box method where we factored out, vertically and horizontally defined our two binomials that multiply together in factored form.
So thanks for watching this tutorial on more challenging quadratic factoring. Hope to see you next time.