Video: Determining Where the Graph of a Quadratic Equation Crosses the π‘₯-Axis by Factorization

Factor the equation 𝑦 = 4π‘₯Β² + 5π‘₯ βˆ’ 6. At which values of π‘₯ does the graph of 𝑦 = 4π‘₯Β² + 5π‘₯ βˆ’ 6 cross the π‘₯-axis?

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Video Transcript

Factor the equation 𝑦 equals four π‘₯ squared plus five π‘₯ minus six. At which values of π‘₯ does the graph of 𝑦 equals four π‘₯ squared plus five π‘₯ minus six cross the π‘₯-axis?

So, let’s begin by factoring the expression four π‘₯ squared plus five π‘₯ minus six. And there are usually more than one way to do these. We’ll consider two separate methods that you might have come across. The first is almost trial and error. We have a quadratic equation which consists of three coprime terms. That is, the terms four π‘₯ squared, five π‘₯, and negative six have no common factors aside from one. And that means we can factor this into two brackets.

So, what goes into each of these brackets? Well, we know that the first term in each binomial must have a product of four π‘₯ squared. So, this could be four π‘₯ and π‘₯ or two π‘₯ and two π‘₯. Similarly, the second term in each polynomial must have a product of negative six. So, forgetting about the negative for a moment, the numbers we could be interested in are one and six or two and three. We’re going to use a little bit of trial and error and see if we can figure out which of these numbers is correct.

Let’s begin by checking four π‘₯ and π‘₯ and one and six. And I’ve just chosen this in, in very much a random order. So, we have four π‘₯ and one and π‘₯ and six. What we’re going to do is imagine we’re redistributing the parentheses. We multiply the first term in each expression. So, four π‘₯ times π‘₯ gives us four π‘₯ squared, which is what we were hoping for. We multiply the outer terms in each expression. Four π‘₯ times six is 24π‘₯. We multiply the inner terms. One times π‘₯ is π‘₯. And when we multiply the last terms, we get six. Now, we didn’t put any signs in here, but it really doesn’t matter. There’s no way we can use 24π‘₯ and π‘₯ to make that central term five π‘₯.

So, let’s try something else. Let’s try swapping the six and the one around. We still get four π‘₯ squared. And actually, when we multiply the last term in each expression, we also still get six. When we multiply the outer terms, we get four π‘₯. And multiplying the inner terms gives us six π‘₯. Once again, there’s still no way that we can use four π‘₯ and six π‘₯ to make five π‘₯. So, we can disregard one and six from this option.

Let’s now try two and three. Again, let’s just put these in any order. Let’s try two in the first expression and three in the second. We get the four π‘₯ squared and the six we require. Four π‘₯ times three gives us 12π‘₯. And two times π‘₯ gives us two π‘₯. There’s no way we can use 12π‘₯ and two π‘₯ to make five π‘₯. So, once again, we disregard this set.

Before we try two π‘₯ and two π‘₯, let’s just swap the three and the two around and see if we can get this to work. This time, four π‘₯ times two is eight π‘₯. And three times π‘₯ is three π‘₯. So, we have four π‘₯ squared, eight π‘₯, three π‘₯, and six. Now, you might be able to spot that we can actually make five π‘₯ with eight π‘₯ and three π‘₯. We need plus eight π‘₯ minus three π‘₯. To achieve this, our two expressions are four π‘₯ minus three and π‘₯ plus two. And we go back to multiplying the last term in each expression. Negative three times positive two is negative six, as required.

And so, we factored the expression four π‘₯ squared plus five π‘₯ minus six. And we get four π‘₯ minus three times π‘₯ plus two. Now, that method works quite nicely, but it is a little bit long-winded. Let’s consider the alternative. In this method, we take the coefficient of π‘₯ squared and the final term, the constant, and we multiply them. Four times negative six is negative 24. We now need to find two numbers whose product is negative 24 and whose sum is the coefficient of π‘₯. So, their sum is five.

Well, negative three times eight is indeed negative 24. And the sum of negative three and eight is five. So, now, what we do is we leave the first term and the last term as they are. That’s four π‘₯ squared and negative six. And we take this five π‘₯ and we split it. And we use those two numbers to do so. We know that five π‘₯ can be written as negative three π‘₯ and eight π‘₯. We’re now going to individually factor the two halves of this expression.

So, we’re going to factor four π‘₯ squared minus three π‘₯ and eight π‘₯ minus six. The only common factor four π‘₯ squared and negative three π‘₯ have is π‘₯. So, we get π‘₯ times four π‘₯ minus three. Then, to factor eight π‘₯ minus six, their highest common factor is two. So, we get two times four π‘₯ minus three. We then factor this by four π‘₯ minus three. And when we do, we get four π‘₯ minus three times π‘₯ plus two. Notice that this looks exactly the same as the earlier result.

And so, the answer to the first part is 𝑦 equals four π‘₯ minus three times π‘₯ plus two. Now, it’s important to realize that because we did this using two different methods, we didn’t bother checking our second method. But it’s always sensible, if this is the method that you choose, to redistribute these parentheses and check. You do indeed get the original expression.

The second half of this question says, at which values of π‘₯ does the graph of 𝑦 equals four π‘₯ squared plus five π‘₯ minus six cross the π‘₯-axis?

We recall that the π‘₯-axis can be written using the equation 𝑦 equals zero. And so, to find the values of π‘₯ for which the graph crosses the π‘₯-axis, we set 𝑦 equal to zero and we solve for π‘₯. That is, zero equals four π‘₯ squared plus five π‘₯ minus six. Now, it isn’t a coincidence that the first part of this question asked us to factor the equation. We’re going to use that result and write zero equals four π‘₯ minus three times π‘₯ plus two.

And the reason we do this is cause it makes it a lot easier to solve. We now have two expressions. We have four π‘₯ minus three. And it’s multiplying another expression, π‘₯ plus two. And when we do, we get the result of zero. Now, the only way for the product of these two numbers to be equal to zero is if individually either four π‘₯ minus three is equal to zero or π‘₯ plus two is equal to zero. Let’s now individually solve each of these equations.

To solve this first equation, we’ll begin by adding three to both sides. So, we get four π‘₯ equals three. We then divide through by four. We see that π‘₯ is equal to three-quarters. To solve the second equation, all we need to do is subtract two from both sides. And that gives us a value of π‘₯ as negative two. And so, there are two solutions to the equation four π‘₯ squared plus five π‘₯ minus six equals zero. And these tell us the values of π‘₯ for which the graph of 𝑦 equals four π‘₯ squared plus five π‘₯ minus six crosses the π‘₯-axis. They are three-quarters and negative two.

Now, it’s often sensible to check our answers. Here, we can check by substituting π‘₯ equals three-quarters and π‘₯ equals negative two into the expression four π‘₯ squared plus five π‘₯ minus six. When we do, we get zero both times, which is what we’re expecting. So, we know these solutions are correct. They are three-quarters and negative two.

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