Video: Solving Quadratic Equations: Factoring

In this video, we will learn how to solve quadratic equations by factoring.

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Video Transcript

In this video, we’ll learn how to solve quadratic equations by factoring, sometimes called factorizing. We’ll consider what it actually means graphically to solve a quadratic equation and how this process looks for a variety of equations. Before approaching this topic, it’s important that you’re able to factor fully simple quadratic expressions of the form π‘₯ squared plus 𝑏π‘₯ plus 𝑐. And have techniques for factoring those of the form π‘Žπ‘₯ squared plus 𝑏π‘₯ plus 𝑐. And those which involve finding the difference of two squares.

To give us some context, we’re going to briefly turn our attention to a quadratic graph whose equation is 𝑦 equals π‘₯ squared plus four π‘₯ minus 12. We say that the roots of our equation are the values of π‘₯ where our graph crosses the π‘₯-axis. If we read those from our graph, we see that π‘₯ is equal to negative six and π‘₯ is equal to two are the roots of our equation. But what else does this tell us?

Well, the π‘₯-axis has the equation 𝑦 equals zero. So π‘₯ equals negative six and π‘₯ equals two are the solutions to the equation π‘₯ squared plus four π‘₯ minus 12 equals zero. We’ve simply replaced 𝑦 with zero in our equation. But we of course don’t always want to draw the graph of our equation. So what if instead we were to factor the quadratic expression π‘₯ squared plus four π‘₯ minus 12?

If we have a look at our expression, we see that all three terms are coprime. That is, they share no other factors than one. In the case of a quadratic, that’s a good indication to us that the expression will factor into two pairs of brackets or parentheses. We know though when we multiply or distribute these parentheses, we’re going to need π‘₯ squared. So the first term in each bracket must be π‘₯, since π‘₯ times π‘₯ gives us π‘₯ squared.

Recall next that the number in each bracket is found by finding a pair of numbers whose product is negative 12, they times to make negative 12, and whose sum is four. Well, the factor pairs of 12 are one and 12, two and six, and three and four. We know that the difference between two and six is four and that a positive times a negative is a negative. That is, six times negative two gives us negative 12 and six minus two gives us four.

The quadratic expression π‘₯ squared plus four π‘₯ minus 12 can therefore be written as π‘₯ plus six times π‘₯ minus two. Now, this is equal to zero. And we see that each expression in our brackets will simply give us a number. When we times two numbers together, we get zero. So what does that tell us about one or other of these numbers?

Well, the only way for this to happen is if either number is zero. In other words, either π‘₯ plus six is equal to zero or π‘₯ minus two is equal to zero. Let’s solve for π‘₯. We solve the first equation by subtracting six from both sides. So π‘₯ is negative six. We solve the second equation by adding two to both sides. So π‘₯ is equal to two. And we’ve solved the equation π‘₯ squared plus four π‘₯ minus 12 equals zero. Notice that these are the roots we identified earlier.

So we’ve learned that to solve a quadratic equation, we ensure it’s equal to zero and factor. But this also tells us the values of π‘₯ where our graph crosses the π‘₯-axis. Let’s have a look at an example of this.

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

We’re going to factor the quadratic expression six π‘₯ squared plus nine π‘₯. Notice that, in this expression, we have two terms who share a common factor. Both six π‘₯ squared and nine π‘₯ are divisible by three and π‘₯. So their highest common factor is three π‘₯. This means when we factor the expression, we take three π‘₯ to the outside of our parentheses. To find the terms inside the parentheses, we divide both six π‘₯ squared and nine π‘₯ by the greatest common factor. Six divided by three is two. So six π‘₯ squared divided by three π‘₯ is two π‘₯.

Then we see that nine π‘₯ divided by three π‘₯ is three. So when we factor the expression six π‘₯ squared plus nine π‘₯, we get three π‘₯ times two π‘₯ plus three. And so our equation is 𝑦 equals three π‘₯ times two π‘₯ plus three.

The second part of this question asks us to find the values of π‘₯ where the graph of our equation crosses the π‘₯-axis. These are the roots of our equation. And of course, we recall that the equation of the π‘₯-axis is 𝑦 equals zero. So our graph crosses the π‘₯-axis for values of π‘₯ such that 𝑦 is equal to zero. We set 𝑦 equal to zero in our original equation. And we get zero equals six π‘₯ squared plus nine π‘₯.

Remember though, we factor this expression and we got three π‘₯ times two π‘₯ plus three. So we replace six π‘₯ squared plus nine π‘₯ with its factored form. And we now see that we have two expressions, that’s three π‘₯ and two π‘₯ plus three, whose product is zero. Now, for this to be the case, either three π‘₯ must be equal to zero or two π‘₯ plus three must be equal to zero. The roots of our equation, that is, the values of π‘₯ where our graph crosses the π‘₯-axis, are the solutions to each of these equations.

And so we solve for π‘₯. We solve our first equation by dividing both sides by three. So π‘₯ is equal to zero. To solve our second equation, we begin by subtracting three from both sides. So two π‘₯ is negative three. Finally, we divide through by two. And we get π‘₯ is equal to negative three over two or negative 1.5. And so the values of π‘₯ where our graph crosses the π‘₯-axis is zero and negative three over two.

We’ll now consider an example of how to solve a monic quadratic, that’s one whose coefficient of π‘₯ squared is one, by factoring.

Solve the equation π‘₯ squared minus four π‘₯ plus four equals zero by factoring.

The question tells us to solve this equation by factoring. So we’re going to begin by factoring the quadratic expression π‘₯ squared minus four π‘₯ plus four. We see that we have a quadratic expression with three terms. And these terms are coprime. That is, they have only a factor of one in common. That tells us we can factor this expression into two brackets.

We know that the term in the front of each of these brackets must be π‘₯, since π‘₯ times π‘₯ gives us the π‘₯ squared we require. But to find the number in each bracket, we’re looking for two numbers whose product is four and whose sum is negative four. And so we simply begin by listing the factor pairs of four. They are one and four and two and two.

Well, the sum of two and two is four. But we also know that a negative multiplied by a negative is a positive. And so if we choose our numbers to be negative two and negative two, we find that negative two times negative two gives us the positive four we require. But their sum, that’s negative two plus negative two, gives us the negative four we require. And so when we factor π‘₯ squared minus four π‘₯ plus four, we get π‘₯ minus two times π‘₯ minus two.

There are now two routes we can take to solve this quadratic equation. The first is to recall that when we multiply a number by itself, we call that squaring. So π‘₯ minus two times π‘₯ minus two is π‘₯ minus two squared. And our equation becomes π‘₯ minus two squared equals zero.

We’re now going to take the square root of both sides of this equation. Usually, we would look to take both the positive and negative square root of whatever was on the right-hand side. But of course, the square root of zero is zero. So when we take the square root of both sides, we simply get π‘₯ minus two equals zero. We’ll solve this equation for π‘₯ by adding two to both sides. And we find π‘₯ is equal to two is the solution to our equation.

Now, it might seem unusual for a quadratic equation to have just one solution. So we’ll look at what the other method tells us. In the other method, we really think about what’s happening here. We have two numbers whose product is zero. Now, for this to be the case, either one or other of our numbers must be equal to zero. So either π‘₯ minus two is equal to zero or π‘₯ minus two is equal to zero. But these are the same equations, so they both have a solution of π‘₯ equals two. When this happens, we say that the equation has two equal roots. In graphical form, it means that the vertex or the turning point of our graph is the place where it intercepts the π‘₯-axis. And so it might look a little something like this.

We’ll now consider how to find the roots of a nonmonic quadratic equation.

Solve the equation nine π‘₯ squared plus 30π‘₯ plus 25 equals zero by factoring.

This is an equation that contains a nonmonic quadratic, a quadratic with a coefficient of π‘₯ squared is not equal to one. This means the quadratic expression is a little bit more difficult to factor than usual. We might notice that it’s a perfect square, with π‘Ž and 𝑐 being square numbers. But if we didn’t spot this, we could use trial and error or use the following method to factor.

In this method, the first thing that we do is we multiply the coefficient of π‘₯ squared and the constant. Nine times 25 is 225. And so we look for two numbers whose product is 225 and whose sum is 30. Well, 225 is a square number such that 15 times 15 is 225. And we also know that the sum of 15 and 15 is 30.

Our next step then is to break the 30π‘₯ into 15π‘₯ and 15π‘₯. And so our quadratic expression is nine π‘₯ squared plus 15π‘₯ plus 15π‘₯ plus 25. We now individually factor the first two terms and the last two terms. The greatest common factor of nine π‘₯ squared and 15π‘₯ is three π‘₯. So factoring these first two terms, we get three π‘₯ times three π‘₯ plus five. Then the greatest common factor of our last two terms is five. And so when we factor 15π‘₯ plus 25, we get five times three π‘₯ plus five.

Notice now that we have a common factor of three π‘₯ plus five. So we’re going to factor that. Three π‘₯ plus five is multiplied by three π‘₯ and five. So that’s the other binomial. And our expression becomes three π‘₯ plus five times three π‘₯ plus five.

Now, of course, we’re solving the equation nine π‘₯ squared plus 30π‘₯ plus 25 equals zero. So let’s set this equal to zero. And we know that for the product of these two numbers to be equal to zero, either one or other number must itself be equal to zero. So we see that three π‘₯ plus five is equal to zero or three π‘₯ plus five is equal to zero. In fact, these are the same equation and they’ll yield the same result.

So we’re just going to solve the equation three π‘₯ plus five equals zero. We subtract five from both sides. So three π‘₯ is negative five. And then we divide through by three. So π‘₯ is equal to negative five-thirds. And so we see that the equation nine π‘₯ squared plus 30π‘₯ plus 25 equals zero has the solution π‘₯ equals negative five-thirds. We can say that our equation has two equal roots or a repeated root.

And remember, we could actually check our working by substituting π‘₯ equals negative five-thirds into our original expression. And if we’d done that correctly, we’d find it’s equal to zero.

We’ll now consider the graph of a function where the coefficient of π‘₯ equals zero.

At which values of π‘₯ does the graph of 𝑦 equals π‘₯ squared minus seven cross the π‘₯-axis?

Remember, the equation of the π‘₯-axis is 𝑦 equals zero. So our graph crosses the π‘₯-axis when 𝑦 is equal to zero. We therefore set 𝑦 equal to zero and solve for π‘₯. So our equation becomes zero equals π‘₯ squared minus seven. Now, this is a quadratic equation. And so we might be thinking that we need to factor. However, the coefficient of π‘₯ here is zero. So we can solve simply by rearranging.

We’ll begin by adding seven to both sides of our equation so that seven equals π‘₯ squared. Our next job is to take the square root of both sides. But we remember that, in doing so, we have to take the positive and negative square root of seven. So π‘₯ is equal to the positive and negative square root of seven. This means that the roots of our equation, the values of π‘₯ where our graph crosses the π‘₯-axis, are root seven and negative root seven.

In our final example, we’ll look at how being given a root of an equation can help us to find the other root.

Given that negative 10 is a root of the equation two π‘₯ squared plus 13π‘₯ minus 70 equals zero, what is the other root?

We’re told that negative 10 is a root of our equation, which means that our quadratic must be equal to zero when π‘₯ is equal to negative 10. Essentially, it’s a solution to the equation two π‘₯ squared plus three π‘₯ [13π‘₯] minus 70 equals zero. Now, what this actually means is that π‘₯ plus 10 must be a factor of two π‘₯ squared plus 13π‘₯ minus 70. Two π‘₯ squared plus 13π‘₯ minus 70 can therefore be written as π‘₯ plus 10 times some other binomial.

Now, let’s give that binomial a form. Let’s say it’s in the form π‘Žπ‘₯ plus 𝑏, where π‘Ž and 𝑏 are real constants. What we’re going to do is distribute the parentheses on the right-hand side of this equation and see what we get. We begin by multiplying π‘₯ by π‘Žπ‘₯. That’s π‘Žπ‘₯ squared. We then multiply the outer terms. That’s π‘₯ times 𝑏, which is 𝑏π‘₯. Next, we multiply the inner terms. That’s 10 times π‘Žπ‘₯, which is 10π‘Žπ‘₯. And finally, we multiply 10 by 𝑏, to give us 10𝑏. So we find that this is equal to two π‘₯ squared plus 13π‘₯ minus 70.

And we now use a process called comparing coefficients. We look at the coefficients of our various terms. Let’s begin by comparing our coefficients of π‘₯ squared. On the left-hand side, we have a two. And on the right-hand side, the coefficient of π‘₯ squared is π‘Ž. So when we compare coefficients of π‘₯ squared, we find π‘Ž is equal to two.

We could next compare coefficients of π‘₯. In fact though, we’re going to compare constants. We might say these are the coefficients of the π‘₯ to the power of zero terms. On the left-hand side, our constant is negative 70. And on the right-hand side, we have 10𝑏. So negative 70 equals 10𝑏. And so we’re going to divide by 10 to solve for 𝑏. 𝑏 is therefore equal to negative seven. This means that our quadratic expression can be written as π‘₯ plus 10 times two π‘₯ minus seven. We’ve replaced π‘Ž and 𝑏 with their solutions.

But we know that we’re using this to solve the equation two π‘₯ squared plus 13π‘₯ minus 70 equals zero. We already know that we have one root of negative 10. That’s found by setting π‘₯ plus 10 equal to zero. We’re now going to set two π‘₯ minus seven equal to zero and solve for π‘₯. We’ll add seven to both sides of this equation so that two π‘₯ is equal to seven. And then we’ll divide through by two. So π‘₯ is equal to seven over two or 3.5. And we could check this solution by substituting π‘₯ equals seven over two into our original equation, making sure that it is indeed equal to zero.

In this video, we learned that the roots of a quadratic equation are the solutions to π‘Žπ‘₯ squared plus 𝑏π‘₯ plus 𝑐 equals zero. We saw that if we’re trying to solve a quadratic equation of this form, we begin by factoring the quadratic expression where possible. Once we’ve done that, we take the bit in each bracket, set that equal to zero, and solve for π‘₯. Of course, we can check any working by substituting the solutions back into the original expression.

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