Hey guys! Welcome to this video on factoring quadratic equations.

To start, let’s review what a quadratic equation actually is. The term quadratic is derived from quad, which means “square,” and we call it this because the variable is squared.

Now, the standard form of a quadratic equation is this:

$$ax^2 +bx + c = 0$$

In order to factor a quadratic, you just need to find what you would multiply by in order to get the quadratic. The actual quadratic equation is the expanded, or multiplied out version, of your two factors that are being multiplied.

For example, $$(x + 2)$$ and $$(x + 6)$$ are my factors that are being multiplied together. Once you multiply together you get $$x^2+ 8x + 12$$.

So, again we have our factors $$(x + 2)(x + 6)$$ on the left, and when you multiply that you get the expanded version: $$(x^2+ 8x + 12)$$.

Now, expanding can be pretty easy; we know exactly what to do to expand them when given our factors, but figuring out how to factor our expanded version can be a little harder.

The easiest way to do this is to find the common factor.

Let’s look at how to do that. Say we have the equation $$8x^2 + 16x = 0$$

What are the common factors of $$8x^2 + 16x = 0$$?

Well, 8 and 16 share a common factor of 8. So, we can go ahead and factor out that 8.

Now, we have $$8(x^2 + 2x)$$ being multiplied by everything on the inside. But, we still have something that can be factored out. $$x^2$$ and $$x$$ share a common factor of $$x$$. So, now we have $$8x(x + 2)$$. And we’ve got it. Our factors of the equation $$8x^2 + 16x = 0$$ are $$8x$$ and $$x + 2$$.

Maybe you’re asking, why on earth do I even need to factor? Why can’t I just leave it as it is?

Well, if you recall a quadratic equation is always a parabola (or U-shaped graph). Well, factoring the quadratic equation then sets us up to be able to find out where exactly our roots are, and our roots just mean where our graph is equal to zero. To do that, we would set our factors equal to zero and solve.

So, $$8x$$ is $$0$$ when $$x = 0$$, and $$x + 2$$ is $$0$$ when $$x = -2$$.

When we look at the graph for $$8x^2+ 16x = 0$$, we can see that it’s zero at $$x = 0$$ and $$x = -2$$.

Alright, so that is kind of a side note to answer the question “why does factoring matter?”

But now, let’s look back at how to actually factor. In our last example, it was relatively simple to find a common factor for the two numbers 8 and 16. However, our numbers aren’t always so straightforward. So, it may be we don’t just know the factors off the top of our head, and guessing might not be as quick as we would hope. So, then what would we do?

Well, let’s look at an example of one of these equations that may be a little trickier to factor.

$$2x^2+ x – 6$$

Okay, so I don’t know about you, but I am not at the point where I can just look at this equation and just know off the top of my head what the factors are. So, I have to follow a kind of step-by-step method that works when we have a quadratic equation in standard form.

So, here are the steps that you should follow:

1. Identify which two numbers will multiply to get $$a\times c$$, and add together to get $$b$$.
2. Replace the middle of the equation with the two factors that you found
3. Group together the first two terms and the last two terms, then factor them individually
4. The two terms, at this point, should now have an obvious common factor

Let’s look at how to factor the equation $$2x^2 + x – 6$$ using these four steps.

### Step 1

So, first, we need to identify which two numbers multiply together to get $$a\times c$$, which is -12, and add to get $$b$$ (positive 1). So, let’s go ahead and list all the possible factors of 12. So, obviously there is 1 and 12, then 2 and 6, and lastly 3 and 4. Remember, we are actually dealing with a -12, so we need to consider that when choosing which two factors work. So, we need these factors to multiply to get -12, which each of them will if we through a negative sign in front of one of the factors, but we also need for it to add to get positive one. So, the only way that can happen is if we take the two factors 3 and 4, and throw a negative sign in front of 3. That makes our factors -3 and 4, which multiplies to give us -12 and add to give us positive 1. So, on to step 2.

### Step 2

Step two tells us to replace the middle of the equation with the two factors that we found. Doing that gives us $$2x^2 -3x + 4x -6$$.

### Step 3

Step 3 tells us to group together the first two terms, and the last two terms, then factor them individually. So, we would group together $$(2x^2 – 3x) + (4x-6)$$, then factor. From the first two terms, we can factor out an $$x$$, which would give us $$x(2x – 3)$$. From the last two terms, we can factor out a 2, which would give us $$2(2x – 3)$$. So, now we have $$x(2x – 3) + 2(2x – 3)$$.

### Step 4

Now, lastly, step number four tells us that we should be able to see an obvious common factor, which we have. $$(2x – 3)$$ is common to both terms, so we can write this as $$(x + 2)(2x – 3)$$, and we have our answer. To check this, we can multiply this out to see if you get our original expanded quadratic equation.

So, when we expand we get $$2x^2-3x + 4x – 6$$, and when we simplify that we get $$2x^2+ x – 6$$, which is our original equation. Great work, guys!

I hope that this video on how to factor quadratic equations was helpful.

See you guys next time!

## Factoring Quadratic Equations Practice Questions

Question #1:

$$x^2+5x+6$$

$$(x+6)(x+7)$$
$$(x+2)(x+9)$$
$$(x+2)(x+3)$$
$$(x+8)(x+4)$$

The correct answer is $$(x+2)(x+3)$$. The quadratic equation is currently in expanded form, with $$a=1$$, $$b=5$$, and $$c=6$$. First, find two numbers that multiply to $$a×c$$ and add to get b. In this case, the two numbers need to multiply to 6 and add to 5. The only two numbers that can do this are 2 and 3. Then, replace the middle of the equation with these two factors to get $$x^2+2x+3x+6$$. Group the first two terms and last two terms and factor them individually.

$$(x^2+2)+(3x+6)$$
$$x(x+2)+3(x+2)$$

Now, look for the common factor that they both share. In this case, it’s $$(x+2)$$ and $$(x+3)$$.

Question #2:

$$x^2-2x-35$$

$$(x+3)(x-2)$$
$$(x-7)(x+5)$$
$$(x+4)(x+9)$$
$$(x-3)(x-6)$$

The correct answer is $$(x-7)(x+5)$$. The quadratic equation is currently in expanded form, with $$a=1$$, $$b=-2$$, and $$c=-35$$. First, find two numbers that multiply to $$a×c$$ and add to get b. In this case, the two numbers need to multiply to -35 and add to -2. The only two numbers that can do this are -7 and 5. Then, replace the middle of the equation with these two factors to get $$x^2-7x+5x-35$$. Group the first two terms and last two terms and factor them individually.

$$(x^2-7x)+(5x-35)$$
$$x(x-7)+5(x-7)$$

Now, look for the common factor that they both share. In this case, it’s $$(x-7)$$ and $$(x+5)$$.

Question #3:

$$x^2-2x-15$$

$$(x-5)(x+3)$$
$$(x-6)(x-4)$$
$$(x-5)(x+4)$$
$$(x-6)(x-2)$$

The correct answer is $$(x-5)(x+3)$$. The quadratic equation is currently in expanded form, with $$a=1$$, $$b=-2$$, and $$c=-15$$. First, find two numbers that multiply to $$a×c$$ and add to get b. In this case, the two numbers need to multiply to -15 and add to -2. The only two numbers that can do this are -5 and 3. These two factors multiply to the original equation that is in expanded form, so the equation has been successfully factored.

Question #4:

$$3x^2-2x-5$$

$$(x+2)(x-5)$$
$$(4x+4)(5x-1)$$
$$(3x-5)(x+1)$$
$$(3x+2)(x-5)$$

The correct answer is $$(3x-5)(x+1)$$. The quadratic equation is currently in expanded form, with $$a=3$$, $$b=-2$$, and $$c=-5$$. First, find two numbers that multiply to $$a×c$$ and add to get b. In this case, the two numbers need to multiply to -15 and add to -2. The only two numbers that can do this are -5 and 3. Then, replace the middle of the equation with these two factors to get $$3x^2-5x+3x-5$$. Group the first two terms and last two terms and factor them individually.

$$(3x^2-5x)+(3x-5)$$
$$x(3x-5)+1(3x-5)$$

Now, look for the common factor that they both share. In this case, it’s $$(3x-5)$$ and $$(x+1)$$.

Question #5:

$$3x^2-x-4$$

$$(6x-3)(x+1)$$
$$(3x-4)(x+1)$$
$$(2x-1)(x+8)$$
$$(x+4)(x-1)$$

The correct answer is $$(3x-4)(x+1)$$. The quadratic equation is currently in expanded form, with $$a=3$$, $$b=-1$$, and $$c=-4$$. First, find two numbers that multiply to $$a×c$$ and add to get b. In this case, the two numbers need to multiply to -12 and add to -1. The only two numbers that can do this are -4 and 3. Then, replace the middle of the equation with these two factors to get $$3x^2-4x+3x-4$$. Group the first two terms and last two terms and factor them individually.

$$(3x^2-4x)+(3x-4)$$
$$x(3x-4)+1(3x-4)$$

Now, look for the common factor that they both share. In this case, it’s $$(3x-4)$$ and $$(x+1)$$.