### Video Transcript

In this video, weβll learn how to
solve quadratic equations by factoring, sometimes called factorizing. Weβll 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, itβs
important that youβre able to factor fully simple quadratic expressions of the form
π₯ squared plus ππ₯ plus π. And have techniques for factoring
those of the form ππ₯ squared plus ππ₯ plus π. And those which involve finding the
difference of two squares.

To give us some context, weβre
going to briefly turn our attention to a quadratic graph whose equation is π¦ 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
π¦ 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. Weβve simply replaced π¦ with zero
in our equation. But we of course donβt 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, thatβs
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, weβre 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. Letβs 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 weβve solved the equation π₯
squared plus four π₯ minus 12 equals zero. Notice that these are the roots we
identified earlier.

So weβve learned that to solve a
quadratic equation, we ensure itβs equal to zero and factor. But this also tells us the values
of π₯ where our graph crosses the π₯-axis. Letβs have a look at an example of
this.

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

Weβre 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 π¦ 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 π¦ equals zero. So our graph crosses the π₯-axis
for values of π₯ such that π¦ is equal to zero. We set π¦ 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, thatβs 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.

Weβll now consider an example of
how to solve a monic quadratic, thatβs 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 weβre 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, weβre 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, thatβs 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.

Weβre 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. Weβll 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 weβll look at what the other
method tells us. In the other method, we really
think about whatβs 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.

Weβll 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 itβs a perfect
square, with π and π being square numbers. But if we didnβt 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 weβre going to factor that. Three π₯ plus five is multiplied by
three π₯ and five. So thatβs the other binomial. And our expression becomes three π₯
plus five times three π₯ plus five.

Now, of course, weβre solving the
equation nine π₯ squared plus 30π₯ plus 25 equals zero. So letβs 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 theyβll yield the same result.

So weβre 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 weβd done that correctly,
weβd find itβs equal to zero.

Weβll now consider the graph of a
function where the coefficient of π₯ equals zero.

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

Remember, the equation of the
π₯-axis is π¦ equals zero. So our graph crosses the π₯-axis
when π¦ is equal to zero. We therefore set π¦ 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.

Weβll 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, weβll 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?

Weβre 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, itβs a solution to the
equation two π₯ squared plus 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, letβs give that binomial a
form. Letβs say itβs in the form ππ₯
plus π, where π and π are real constants. What weβre 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
ππ₯. Thatβs ππ₯ squared. We then multiply the outer
terms. Thatβs π₯ times π, which is
ππ₯. Next, we multiply the inner
terms. Thatβs 10 times ππ₯, which is
10ππ₯. And finally, we multiply 10 by π,
to give us 10π. 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. Letβs 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. So when we compare coefficients of
π₯ squared, we find π is equal to two.

We could next compare coefficients
of π₯. In fact though, weβre 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
10π. So negative 70 equals 10π. And so weβre going to divide by 10
to solve for π. π is therefore equal to negative
seven. This means that our quadratic
expression can be written as π₯ plus 10 times two π₯ minus seven. Weβve replaced π and π with their
solutions.

But we know that weβre 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. Thatβs found by setting π₯ plus 10
equal to zero. Weβre now going to set two π₯ minus
seven equal to zero and solve for π₯. Weβll add seven to both sides of
this equation so that two π₯ is equal to seven. And then weβll 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 ππ₯ squared plus ππ₯ plus π
equals zero. We saw that if weβre trying to
solve a quadratic equation of this form, we begin by factoring the quadratic
expression where possible. Once weβve 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.