In this video, we are going to look at how to complete the square.

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For example:
If we are given
$(x+3)^2$
we can multiply it out to get
$x^2+6x+9$

Here we have
$x^2+4x+c$
and we want c to be some number so that when factored, both binomials are exactly the same. To figure out what this c value will be, we will do the following:
$(\frac{b}{2})^2$
So in this case,
$(\frac{4}{2})^2$ is the same as
$2^2$ or $4$
In this case, c needs to be 4.

$x^2+3x+c$
then
$(\frac{3}{2})^2=\frac{9}{4}$
so
$x^2+3x+\frac{9}{4}$
If we had to factor from here, then the two terms are whatever $\frac{b}{2}$ was. When we factor this we get
$(x+\frac{3}{2})(x+\frac{3}{2})$

Video-Lesson Transcript

Let’s go over solving equations by factoring.

The first thing to note here is that we have a quadratic and it’s equal to zero.

$x^2 + 5x + 6 = 0$

So we have to make sure that the right side is zero.

Now, let’s factor this.

$x^2 + 5x + 6 = 0$

Draw two parenthesis and write $x$ on each.

Then find two numbers that will add up to $5$ and when multiplied will be equal to $6$.

$(x + 2) (x + 3) = 0$

When can two numbers when multiplied be equal to zero?

One of them has to be zero. Because of zero product property.

So we have to come up with one of these terms be zero.

Meaning $x + 2 = 0$ or $x + 3 = 0$

Either of the two should be zero.

$x + 2 = 0$,

$x + 2 - 2 = 0 - 2$,

$x = -2$

and

$x + 3 = 0$,

$x + 3 - 3 = 0 - 3$,

$x = -3$

So we have two solutions:

{$-2, -3$}

We usually use the brackets not the parenthesis.

Let’s have another one

$x^2 + 4x - 12 = 0$

Again, let’s factor this.

Draw two parenthesis and $x$ in each parenthesis.

Then think of two numbers that when you add up will equal $4$ and when multiplied will equal $-12$.

$x^2 + 4x - 12 = 0$,

$(x + 6) (x - 2) = 0$,

$x + 6 = 0$,

$x + 6 - 6 = 0 - 6$,

$x = -6$

and

$x - 2 = 0$,

$x - 2 + 2 = 0 + 2$,

$x = 2$

Our two solutions are

{$-6, 2$}

Take a look at this one:

$x^2 + 3x + 4 = 14$

Notice that the right side is not zero.

So we have to make it zero.

$x^2 + 3x + 4 - 14 = 14 - 14$,

$x^2 + 3x - 10 = 0$

Now we can factor this out.

$x^2 + 3x - 10 = 0$,

$(x + 5) (x - 2) = 0$

Let’s solve for $x$

$x + 5 = 0$,

$x + 5 - 5 = 0 - 5$,

$x = -5$

and

$x - 2 = 0$,

$x - 2 + 2 = 0 + 2$,

$x = 2$

So our solution set is

{$-5, 2$}