# Question Video: Solving Quadratic Equations by Factorisation

Find the solution set of 3𝑧² − 7𝑧 = 6 in ℝ.

07:40

### Video Transcript

Find the solution set of three 𝑧 squared minus seven 𝑧 equals six in the real numbers.

The solution set means we’re looking to find the set of all values of the variable, in this case that’s 𝑧, which satisfy the given equation. And we’re told that we’re only interested in solutions that exist in the real numbers.

Now, looking at the equation we’ve been given, we can see that it is a quadratic equation because the highest power of our variable 𝑧 is two. The most general form of a quadratic equation when the variable is 𝑥 is 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 is equal to zero. We collect all the terms on the same side of the equation. So our first step will be to do this for our quadratic equation in 𝑧. We can achieve this in a single step by subtracting six from each side of the equation to give three 𝑧 squared minus seven 𝑧 minus six is equal to zero.

Now, there are a number of methods that we can use to solve a quadratic equation. The method we’re going to consider here though is the method of factoring. We need to be a little bit careful though because the coefficient of 𝑧 squared in this equation isn’t equal to one; it’s equal to three, which makes things slightly more complex.

We’re going to use the method of factoring by grouping. To do this, we look for two numbers which sum to the coefficient of 𝑧. In this case, that’s negative seven. And those same two numbers must have a product which is equal to the product of the coefficient of 𝑧 squared, that’s three, and the constant term negative six. So the two numbers must have a product of negative 18. In the general form, what we’re looking for is two numbers which sum to be the coefficient of 𝑥 and whose product is 𝑎𝑐.

Now, listing the factor pairs of 18 may help. We have one and 18, two and nine, and three and six. We want the product to be negative 18 though, which means we need one number to be negative and the other positive. With a little bit of thought, we see that if we choose the numbers negative nine and two, then these numbers have a product of negative 18 and they also have a sum of negative seven. So these are the two numbers we’re looking for.

Now, the next step is key. We’re not ready to start trying to put anything into parentheses yet. All we’re going to do is rewrite our quadratic equation. But we’re going to take that middle term of negative seven 𝑧 and rewrite it using these two numbers we’ve just found of negative nine and two. So we express our quadratic equation as three 𝑧 squared, there’s no change there, minus nine 𝑧 plus two 𝑧, that gives the original negative seven 𝑧, minus six, no change there, is equal to zero.

Now we’re ready to begin factoring our quadratic. And remember, the method we’re using is called factoring by grouping. We’re going to divide our quadratic equation in half so that we have two terms in each half. And we’re then going to factor each of these expressions individually.

Looking first at the binomial, three 𝑧 squared minus nine 𝑧, these two terms have a common factor of three 𝑧. Inside the parentheses, we need 𝑧 minus three so that when we expand, we get three 𝑧 squared minus nine 𝑧. Looking at our second binomial, positive two 𝑧 minus six, these two terms have a common factor of two. Inside the parentheses, we need 𝑧 minus three because then when we expand, two multiplied by 𝑧 gives two 𝑧 and two multiplied by negative three gives negative six.

What we should notice now is that the two halves of our expression have a common factor of 𝑧 minus three. And this is no coincidence. If a quadratic equation can be solved by factoring by grouping and if we performed all of the preceding steps correctly, then we should always find a shared factor. We can therefore factor our entire expression by 𝑧 minus three.

To complete the second set of parentheses, we look at what we have to multiply 𝑧 minus three by in each half of our expression. In the first half, we have to multiply by three 𝑧. And in the second half, we have to multiply by positive two. And so the expression inside our second set of parentheses is three 𝑧 plus two. And now we have a fully factored form of this quadratic equation. We have two factors multiplying to give zero. And we recall that the only way this can happen is if at least one of the factors is itself equal to zero. So we take each factor in turn, set it equal to zero, and then solve the resulting linear equation.

Our first equation, 𝑧 minus three equals zero, can be solved by adding three to each side to give 𝑧 equals three. Our second equation, three 𝑧 plus two equals zero, can be solved by first subtracting two and then dividing by three to give 𝑧 is equal to negative two-thirds. So we found the solution set of this quadratic equation. It’s the set containing the values negative two-thirds and three.

Now, it’s important to note at this stage here that once we’ve found those two numbers of negative nine and two, which had a sum of negative seven and a product of negative 18, we could’ve chosen to write our quadratic with these two coefficients in the opposite order. So we could’ve written it as three 𝑧 squared plus two 𝑧 minus nine 𝑧 minus six is equal to zero. You may wonder if this would lead to a different answer. So let’s just consider how the working would’ve developed if we’d done things this way.

As before, we would divide our quadratic into two halves and then factor each half separately. The two terms in the first half, three 𝑧 squared and positive two 𝑧, have a common factor of 𝑧. So this half factors as 𝑧 multiplied by three 𝑧 plus two. In the second half of our expression, the terms negative nine 𝑧 and negative six have a common factor of negative three. So this half factors as negative three multiplied by three 𝑧 plus two. We are adding these two halves of the expression together. But a positive and a negative together in this context will make a negative overall.

Adding negative three lots of three 𝑧 plus two is the same as subtracting three lots of three 𝑧 plus two. So we don’t need to worry about including this plus sign here. This time, we notice that our shared factor is three 𝑧 plus two. And so we can factor the entire expression by this set of parentheses. To complete our second set of parentheses, we have to multiply by 𝑧 and then negative three. So the factored form of our quadratic here is three 𝑧 plus two multiplied by 𝑧 minus three is equal to zero. Hopefully, you can see that these two factored forms of this quadratic are identical. The factors are just expressed in different orders. And so they’ll lead to the same solution from this point. It doesn’t matter then whether we choose to express that middle term as negative nine 𝑧 plus two 𝑧 or two 𝑧 minus nine 𝑧. We’ll end up with the same result in each case.

The solution set of the given quadratic equation is the set of values negative two-thirds, three.