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Video: Factor Quadratics with 𝑥² Coefficient Not 1

Lucy Murray

Learn how to factor nonmonic quadratics (with 𝑥² coefficient not equal to 1). We find the product of the 𝑥² coefficient and the constant, and we find factors that sum to the coefficient of the 𝑥 term and then complete the process of factoring.

05:20

Video Transcript

Factor Quadratics When 𝑎 Is Not Equal to One

We know that any quadratic comes in the form 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐. And to factor a quadratic when 𝑎 is not equal to one, we need to employ the first step of when it is equal to one, so we need to draw a table.

And in this table, we are looking for something that our two factors will add to get and something that they will multiply to get, so we’re looking for the sum and the product. Well the sum will always be the number in front of the 𝑥, or we could say the 𝑥-coefficient. And then the product will always be 𝑎 multiplied by 𝑐 to the coefficient of 𝑥 squared multiplied by the number by itself.

And then we’ll list the factor pairs of 𝑎𝑐 to find out which factor pair fits with our quadratic, so let’s have a go at an example. We must factor six 𝑥 squared minus five 𝑥 minus four. First thing we need to do, draw a table.

So we can see that the sum will be the coefficient of the 𝑥, so that will be negative five. And the product will be six multiplied by negative four, which is equal to negative twenty-four.

Now, the fact that our product is negative tells us that one of the factor pairs must be negative and one must be positive, and we’re looking for a factor pair that has a difference of five. So, what we now do is list the factor pairs of twenty-four. First of all, we’ve got one and twenty-four. They do not have a difference of five. Two and twelve, nope, they don’t either. Three and eight, aha, yes they do. So because we know that they add to give us a negative five, then that tells us that the larger number, eight, must be negative.

And the reason we found the factor pair in this case is basically what we’re doing is we’re splitting that middle 𝑥-value. As we know, that three 𝑥 minus eight 𝑥 is negative five 𝑥. Now, we’re going to take each set of two terms separately and factor them like they are a linear expression. So, taking the first pair, we can see that they’ve both got three in common and also 𝑥. So if we factor that, then we know that three multiplied by two is six and 𝑥 multiplied by 𝑥 is 𝑥 squared.

And then for the next term, we’re asking what do I have to multiply three 𝑥 by to get three 𝑥. Well that’s just one. Be careful not to miss that one because that’s something that students sometimes do. Now let’s take the next pair.

We can see that they have both got a common factor, a greatest common factor, of negative four. Now, this is a moment to be careful because us- each parenthesis must be exactly the same inside. So negative eight 𝑥 divided by negative four is two 𝑥. That’s okay so far, and then negative four divided by negative four is one. Here we go; we’re okay.

The reason that these must be the same is because, now, if we look at that expression, we can see that each term actually has a factor of two 𝑥 plus one, so we’re going to factor two 𝑥 plus one. And that will give us our first set of parentheses, and then we look at what’s left over. We can see we’ve got a three 𝑥 and a negative four. So our second parenthesis becomes three 𝑥 minus four. Now we have it.

So just going back over what we did, first of all, we need to find what the sum and what the product are. So in this case, the sum was negative five and the product was six times negative four, so negative twenty-four. We then needed to list the factor pairs of twenty-four to try and find two numbers that would satisfy this sum and product. The reason we were doing this was because we were splitting our middle 𝑥-value.

giving us three 𝑥 and minus eight 𝑥. Then we took each pair of terms individually and factored them and then factored the same parentheses, giving us two 𝑥 plus one all multiplied by all of three 𝑥 minus four.