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INSTRUCTORS Carleen Eaton Grant Fraser Eric Smith
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Lecture Comments (9)

1 answer

Last reply by: Professor Eric Smith
Wed Jul 16, 2014 8:12 PM

Post by patrick guerin on July 15, 2014

Thanks for the lecture!

1 answer

Last reply by: Professor Eric Smith
Wed Jul 16, 2014 8:12 PM

Post by patrick guerin on July 15, 2014

On example 3(3m2-9m=30), when you got to the zero factor property, could you possibly have it where you have 3(m+2) and 3(m+5) also with that problem?

1 answer

Last reply by: Professor Eric Smith
Wed Jul 16, 2014 8:04 PM

Post by patrick guerin on July 15, 2014

You said that some equations can be changed into the quadratic form. How would you know easily if they can be changed into a quadratic equation or not?

1 answer

Last reply by: Professor Eric Smith
Wed Apr 30, 2014 6:30 PM

Post by Fahad Chandia on April 30, 2014

In Example 5 you subtract x square from 2x square ,So reminder should b 2 not x Square .Can you explain to me ?

0 answers

Post by Fahad Chandia on April 30, 2014

if you put minus x 2

Solving Quadratic Equations by Factoring

  • A quadratic equation is an equation where the largest power present on a variable is two.
  • Factoring can help solve quadratic equations because of the zero factor property. If the quadratic is set equal to zero, and then factored, then each factor can be set equal to zero and solved separately.
  • Any of the previous factoring techniques for trinomials can be used such as reverse FOIL or the AC method.

Solving Quadratic Equations by Factoring

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Objectives 0:08
  • Solving Quadratic Equations by Factoring 0:19
    • Quadratic Equations
    • Zero Factor Property
    • Zero Factor Property Example
  • Example 1 4:00
  • Solving Quadratic Equations by Factoring Cont. 5:54
  • Example 2 7:28
  • Example 3 11:09
  • Example 4 14:22
  • Solving Quadratic Equations by Factoring Cont. 18:17
    • Higher Degree Polynomial Equations
  • Example 5 20:22

Transcription: Solving Quadratic Equations by Factoring

Welcome back to

In this lesson we are going to take a look at how you can solve quadratic equations using the factoring process.0002

Specifically we will focus on that factoring process and look at how factoring can help us to solve a few other types of equations.0010

Recall that when I'm talking about a quadratic equation and talking about any equation of the form ax2 + bx + c = 0.0022

We do not want our a to be a 0 value. 0032

We do not want to get rid of that x2 term.0036

Some examples of quadratic equations are below x2 + 5x +6 that is definitely a quadratic equation.0040

You can see that it fits the form pretty good. 0048

A would be 1, b would be 5 and c would be 6.0051

It also applies to other types of equations like this one.0055

Notice how this does not quite look like the form, 0059

but if you manipulate it just little bit and move the 3 to the other side it actually does fit the form quite well. 0061

One key feature of these quadratic is that you will have your squared variable in there somewhere.0071

Even this last one is a good example of a quadratic equation. 0078

This one now you can move the 4 to the other side, and put in a 0 placeholder to show that it does fit the form of our quadratic equation.0084

Keep looking for that squared term to be able to hunt these out. 0093

The way we are going to solve these quadratic equations is we are going to use a very special property known as the 0 factor property.0101

What the property says is that if you have two numbers call them a and b.0109

If those numbers multiply together and get you 0 in, either a or b must be 0.0114

The way I like to think of this is much in the context of a game.0121

Pretend that I have two numbers in my head, and they multiply together and get 0.0125

If I would ask you what you know about these numbers you would probably tell me that one of them better be a 0.0131

Because that is the only way we are going to multiply it and get that 0.0135

That is exactly what we are trying to say.0139

If the product of two numbers is 0, then one of the numbers or maybe both of them are 0. 0142

Watch how we use the 0 factor property because they get to our solutions for these quadratics.0147

We are going to use that to solve something like x +4 × x - 5 = 0. 0157

The reason why the 0 factor property is going to come in to play is because we treat each of these factors like one of those numbers. 0163

If I'm multiplying them together and I get 0 then I know that one of these factors or may be both of them are equal to 0.0173

Let us take these factors and create new equations for them.0181

I know that x + 4 = 0 or x – 5 = 0. 0188

The reason why that helps us out is notice how these new equations we form down here, they are much simpler than the original.0197

In fact, they are nice linear equations and I can easily solve them just by subtracting 4 from both sides of this one.0205

I can solve the other one by simply adding 5 to both sides.0217

What that 0 factor property helps us to do is take each of the factors that are in our quadratic and set them both equal to 0. 0226

That way we have some much simpler equations that we can solve from there.0236

Let us try some other ones that is already been factored out. 0243

I want to solve 2x + 3 × 5x + 7 = 0.0245

We will take each of the factors here and we will set them equal to 0 individually.0251

This one is equal to 0 and 5x + 7 will make that one equal to 0 as well.0258

It is just a matter of just solving these individually.0265

We will subtract 3 from both sides and we will divide by 2.0268

I know that one of my solutions is x = -3/2.0279

Let us solve the other one, -7 – 7 that will give us 5x = -7, divide both sides by 5 that would give us x = -7/5.0284

We have two solutions.0307

The 0 factor property works out good your factors are not that big.0312

In this one I have an x × x + 6.0317

We will take each of these factors and we will set them equal to 0.0323

There is not a whole lot of solving to be done with it each of these.0332

In fact, one of them are already solved, x could be 0.0334

On the other one, I will just subtract 6 on both sides, x = -6.0339

I have both my solutions for that one. 0350

We rarely have a quadratic equation that is already been factored out for us.0356

In fact, these first few examples here are very nice and simple ones to solve since they were already factored.0361

Our goal is to take any type of quadratic equation and factor it on our own using many of those different factoring techniques that we have learned.0367

Once we do have it factored then we will use our 0 factor property on it so that we can find our solutions.0377

Here is a good outline of what that might look like. 0383

Maybe I will start with equation like x2 - 3x -10=0.0386

I could use reverse foil or the AC method that actually factor that polynomial.0391

Once I have my factors I could set them each equal to 0 and solve and then get my solutions.0397

It is a good idea to just check your solutions by substituting them back into the original.0403

If you do this, you want to put it in for all copies of that variable. 0408

If I'm checking to see if 5 is a solution, I will put in for my x that has been squared.0414

I will put that in for the x right next to the 3 just so I can make sure that it works out in the entire equation. 0420

I get 25 – 15 – 10, does that equal 0? 0429

Sure enough, it does because of 25 – 25 that is how I would know that a solution checks out.0437

If we are going to do our factoring properly it means we always have to make sure that it set equal to 0 first.0453

Make sure that it is one of your first few steps when working on quadratic equation.0458

Let us give these two a try.0463

The first one is x2 + 2x = 8. 0465

I know it is one of my quadratics, I can see my squared term right there.0469

What I’m going to do is I'm going to get it equal to 0, x2 + 2x.0474

Let us go ahead and subtract 8 from both sides -8 = 0.0481

Let us use our factoring techniques to see if we can break it down. 0490

This one was not that big, I'm going to use just the reverse foil method to see if I can figure out what is going on here.0495

Two numbers that will multiply and give me x2, that better be an x and another x.0504

I need two things that will multiply to give me -8 but somehow add to give me 2.0513

4 and 2 will work as long as my 4 is positive and my 2 is negative.0520

I have both of my factors x + 4 and x - 2.0529

I will take each of these and set them equal to 0. 0535

Solving this one over here I will subtract 4 and solving the other one I would add 2.0545

Giving me the solutions x = -4 and x = 2.0553

You may have already notice that you can shortcut that last few steps just a little bit.0566

This number here will always be the opposite of this one right here in the factor.0572

I recommend going and least showing those steps for the first 3 times 0579

so you can be assured that you are using the 0 factor property in the background.0584

If you want to get a little bit more familiar with it, then feel free to use that short cut.0588

The next one is x2 = x + 30.0595

We will start this one off by getting everything over to one side.0599

I'm subtracting the x over and subtracting the 30 now it is equal to 0.0603

This was not that big either, so let us try reverse foil on that.0611

Two terms that would multiply to give me x2 would have to be x and x.0616

And two things that would multiply to give me 30, but add to be -1, I think we are looking at -6 and 5.0620

That will definitely do it.0632

We can take each of these and set them equal to 0.0636

Solving them I might have to add 6 to both sides of this one and subtract 5 from both sides of this one.0646

Leaving me with two solutions that x = 6 and x = -5.0652

Remember that if you are ever unsure of these answers here 0660

feel free to put them back into the original just to make sure that they do work out.0664

Let us try some more that are just a little bit more complicated.0672

This next one is 3m2 – 9m = 30.0676

I’m going to get it equal to 0 first.0681

That looks pretty good. 0692

Now I'm going to try and factor it.0694

One of the very first things I like to check for factoring is they all have something in common. 0697

Unfortunately, it looks like these ones do.0703

Everything in here is divisible by 3.0705

We will take out that common 3 before we get too far, that way we can make our factoring process much easier.0708

m2 is the left 9 ÷ 3 = 3m and 30 ÷ 3 =-10.0716

We just have to factor this.0725

m2 – 3m -10.0727

Two numbers that would multiply and give me m2 would have to be m and m.0737

Two numbers that would multiply to give me -10, but add to be -3, that will be - 5 and 2.0746

We have our factors, it is time to take all of them and set them equal to 0.0758

3 is that equal to 0? m -5 does that equal 0? and m + 2 is that equal 0?0764

Notice how I even took that 3, one common mistake is just to say that it is one of your solutions, 0773

but from 0 factor property you are checking to see if it equal 0. 0779

Notice how 3 is not equal to 0, that does not make any sense.0784

We will not even consider that.0787

The other ones I can go ahead and continue solving.0790

Add 5, add 5 and – 2m and -2.0793

m = -5 and m = -2.0799

I have my solutions for this quadratic equation.0805

By looking for that greatest common factory you can often save yourself quite a bit of work.0811

In this one I’m going to get my 3x to the other side. 0816

Then I immediately noticed that both of these have an x in common, so I can actually pull that x out.0822

I have an x and x -3, those are my two factors that I will set equal to 0.0834

It is nice having one of our factors equal to 0 already it means that we do not have to do much of the solving process.0841

It is already done.0847

I will go ahead and add 3 to the other equation that we formed that way that one is solved.0850

x = 3.0855

We have both of our solutions for that one. 0857

A tricky part of this entire thing is to make sure that it is set equal to 0 first.0864

Sometimes you might have to do some simplifying or combine things together before you can set it equal to 0. 0869

In this one we have x × 4x + 7 and all of that is equal to 2.0876

It is not that difficult to get it equal to 0 since we would simply subtract 2 from both sides.0882

But in order to have factors I need everything to be multiplied together and I still have some subtraction in here. 0892

What I'm going to do is distribute with my x here and then I will go ahead and try factoring.0899

4x2 + 7x – 2 = 0. 0906

This one is a little bit more complicated, it is not very obvious what the reverse foil method should do on it.0916

Let us go ahead and dig up our AC method and give that a try. 0924

The AC method we will multiply a and c together 4 × -2 = -8.0929

I'm looking for two numbers that will multiply to give me -8 but add to be 7. 0939

What do we have for possibilities, it could be 1, 8, it could be 2 and 4. 0946

It could be any of those in reverse.0951

We want them to multiply to give us -8 and we want it to add to be 7.0955

It looks like I have to use that 1 and 8.0965

The way this will work out as it will be -1 and 8.0971

That will help me split up my middle term.0976

We have done that we can go ahead and continue using factor by grouping and I will take out a common 8 from these first two terms. 0993

The next two, I can see a common 2 that I can pull out.1010

And now that we have that I can say what my factors are.1027

We have 4x - 1 and x + 2. 1035

That is quite a lengthy process but recognize we are not done yet.1042

All that was just done to factor it.1044

We need to take each of the factors and set them equal to 0.1047

4x - 1 = 0 and x + 2 = 0.1053

We will add one to both sides and divide by 4.1062

We have one of our solutions that x could equal ¼.1071

The other one let us go ahead and subtract 2.1078

Our other solutions is x =-2.1083

The factoring process can be lengthy, but it is just one step along the way to finding the solutions.1090

Let us try another one.1099

Some higher degree polynomials can also be solved using this factoring process.1101

Usually the ones that come to mind are the ones that have a greatest common factor that you can go ahead and dry out first.1106

Notice how this one is not a quadratic, it actually has x3 in there.1112

I can still use some of my factoring techniques because they both have 2x in common.1118

Let us pullout that 2x.1126

I would have x2 - 25. 1130

It looks like that what is left over happens to be a quadratic, but it is actually even more special than that. 1138

This right here is the difference of squares which means I can use one of my short cut formulas to help me out.1146

This would be x + 5 and x – 5.1162

I was able to factor it down completely and now you can take all of the pieces and set each one to 0.1169

2x, that could equal to 0, x + 5 that could equal 0, and x - 5 that could equal 0 as well.1178

Divide both sides by 2 for this one. 1191

That will give us x could be 0.1195

-5 on this one x could equal -5, add 5 to both sides of this one, x= 5.1200

This one has 3 solutions that I could go back and check.1214

Let us try out everything we have learned with factoring to see if we can tackle out one more problem.1225

This one is quite large. 1229

This one involves x -1 × 2x – 1 = x +12.1231

It looks like it might be quadratic after I do have something squared in there, but I have to get it equal to 0 first.1239

I need to factor it from there and I actually have to do some multiplying before actually get to that factoring process.1248

We will start over here on the left side. 1257

Let us go ahead and do some foiling okay.1258

My first terms would be 2x2, outside –x, inside - 2x, and my last terms 1.1262

Over on the right side of this, this is the same as x +1 × x +1.1281

We are going to use foil to help us spread this out.1289

Our first terms on that side x × x = x2, outside terms x, inside terms x, and last terms 1×1 =1.1293

I have better freedom on combining things together. 1306

Let us combine together these x's and these x's.1310

2x2 - 3x + 1= x2 + 2x + 1.1318

You can see that this is definitely quadratic. 1332

I got lots of x2 and I have a much better task of getting it all to one side and getting it equal to 0. 1335

I'm going to subtract the x2 from both sides. 1343

That will combine those ones and subtract 2x and we will subtract a 1.1346

This will give us x2 - 5x = 0.1356

This seems like a much nicer problem to solve. 1364

It even has a greatest common factor of x that I can take out from both of these terms.1366

I can take each of these factors and set them equal to 0.1378

Looks like one of those are already solved for us, so we will just leave that as it is.1389

The other one we will add 5 to both sides and then we will get our second solution that x must equal 5.1393

Some quadratics you can factor them using many of the factoring techniques that you learn before.1405

Take each of those factors and set them equal to 0 so you can find your solutions.1411

Thank you for watching