WEBVTT mathematics/algebra-1/smith
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Welcome back to www.educator.com.
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In this lesson we are going to take a look at how you can solve quadratic equations using the factoring process.
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Specifically we will focus on that factoring process and look at how factoring can help us to solve a few other types of equations.
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Recall that when I'm talking about a quadratic equation and talking about any equation of the form ax² + bx + c = 0.
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We do not want our a to be a 0 value.
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We do not want to get rid of that x² term.
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Some examples of quadratic equations are below x² + 5x +6 that is definitely a quadratic equation.
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You can see that it fits the form pretty good.
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A would be 1, b would be 5 and c would be 6.
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It also applies to other types of equations like this one.
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Notice how this does not quite look like the form,
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but if you manipulate it just little bit and move the 3 to the other side it actually does fit the form quite well.
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One key feature of these quadratic is that you will have your squared variable in there somewhere.
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Even this last one is a good example of a quadratic equation.
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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.
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Keep looking for that squared term to be able to hunt these out.
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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.
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What the property says is that if you have two numbers call them a and b.
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If those numbers multiply together and get you 0 in, either a or b must be 0.
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The way I like to think of this is much in the context of a game.
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Pretend that I have two numbers in my head, and they multiply together and get 0.
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If I would ask you what you know about these numbers you would probably tell me that one of them better be a 0.
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Because that is the only way we are going to multiply it and get that 0.
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That is exactly what we are trying to say.
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If the product of two numbers is 0, then one of the numbers or maybe both of them are 0.
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Watch how we use the 0 factor property because they get to our solutions for these quadratics.
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We are going to use that to solve something like x +4 × x - 5 = 0.
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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.
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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.
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Let us take these factors and create new equations for them.
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I know that x + 4 = 0 or x – 5 = 0.
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The reason why that helps us out is notice how these new equations we form down here, they are much simpler than the original.
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In fact, they are nice linear equations and I can easily solve them just by subtracting 4 from both sides of this one.
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I can solve the other one by simply adding 5 to both sides.
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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.
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That way we have some much simpler equations that we can solve from there.
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Let us try some other ones that is already been factored out.
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I want to solve 2x + 3 × 5x + 7 = 0.
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We will take each of the factors here and we will set them equal to 0 individually.
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This one is equal to 0 and 5x + 7 will make that one equal to 0 as well.
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It is just a matter of just solving these individually.
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We will subtract 3 from both sides and we will divide by 2.
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I know that one of my solutions is x = -3/2.
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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.
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We have two solutions.
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The 0 factor property works out good your factors are not that big.
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In this one I have an x × x + 6.
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We will take each of these factors and we will set them equal to 0.
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There is not a whole lot of solving to be done with it each of these.
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In fact, one of them are already solved, x could be 0.
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On the other one, I will just subtract 6 on both sides, x = -6.
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I have both my solutions for that one.
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We rarely have a quadratic equation that is already been factored out for us.
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In fact, these first few examples here are very nice and simple ones to solve since they were already factored.
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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.
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Once we do have it factored then we will use our 0 factor property on it so that we can find our solutions.
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Here is a good outline of what that might look like.
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Maybe I will start with equation like x² - 3x -10=0.
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I could use reverse foil or the AC method that actually factor that polynomial.
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Once I have my factors I could set them each equal to 0 and solve and then get my solutions.
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It is a good idea to just check your solutions by substituting them back into the original.
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If you do this, you want to put it in for all copies of that variable.
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If I'm checking to see if 5 is a solution, I will put in for my x that has been squared.
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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.
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I get 25 – 15 – 10, does that equal 0?
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Sure enough, it does because of 25 – 25 that is how I would know that a solution checks out.
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If we are going to do our factoring properly it means we always have to make sure that it set equal to 0 first.
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Make sure that it is one of your first few steps when working on quadratic equation.
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Let us give these two a try.
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The first one is x² + 2x = 8.
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I know it is one of my quadratics, I can see my squared term right there.
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What I’m going to do is I'm going to get it equal to 0, x² + 2x.
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Let us go ahead and subtract 8 from both sides -8 = 0.
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Let us use our factoring techniques to see if we can break it down.
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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.
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Two numbers that will multiply and give me x², that better be an x and another x.
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I need two things that will multiply to give me -8 but somehow add to give me 2.
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4 and 2 will work as long as my 4 is positive and my 2 is negative.
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I have both of my factors x + 4 and x - 2.
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I will take each of these and set them equal to 0.
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Solving this one over here I will subtract 4 and solving the other one I would add 2.
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Giving me the solutions x = -4 and x = 2.
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You may have already notice that you can shortcut that last few steps just a little bit.
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This number here will always be the opposite of this one right here in the factor.
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I recommend going and least showing those steps for the first 3 times
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so you can be assured that you are using the 0 factor property in the background.
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If you want to get a little bit more familiar with it, then feel free to use that short cut.
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The next one is x² = x + 30.
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We will start this one off by getting everything over to one side.
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I'm subtracting the x over and subtracting the 30 now it is equal to 0.
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This was not that big either, so let us try reverse foil on that.
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Two terms that would multiply to give me x² would have to be x and x.
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And two things that would multiply to give me 30, but add to be -1, I think we are looking at -6 and 5.
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That will definitely do it.
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We can take each of these and set them equal to 0.
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Solving them I might have to add 6 to both sides of this one and subtract 5 from both sides of this one.
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Leaving me with two solutions that x = 6 and x = -5.
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Remember that if you are ever unsure of these answers here
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feel free to put them back into the original just to make sure that they do work out.
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Let us try some more that are just a little bit more complicated.
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This next one is 3m² – 9m = 30.
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I’m going to get it equal to 0 first.
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That looks pretty good.
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Now I'm going to try and factor it.
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One of the very first things I like to check for factoring is they all have something in common.
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Unfortunately, it looks like these ones do.
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Everything in here is divisible by 3.
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We will take out that common 3 before we get too far, that way we can make our factoring process much easier.
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m² is the left 9 ÷ 3 = 3m and 30 ÷ 3 =-10.
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We just have to factor this.
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m² – 3m -10.
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Two numbers that would multiply and give me m² would have to be m and m.
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Two numbers that would multiply to give me -10, but add to be -3, that will be - 5 and 2.
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We have our factors, it is time to take all of them and set them equal to 0.
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3 is that equal to 0? m -5 does that equal 0? and m + 2 is that equal 0?
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Notice how I even took that 3, one common mistake is just to say that it is one of your solutions,
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but from 0 factor property you are checking to see if it equal 0.
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Notice how 3 is not equal to 0, that does not make any sense.
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We will not even consider that.
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The other ones I can go ahead and continue solving.
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Add 5, add 5 and – 2m and -2.
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m = -5 and m = -2.
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I have my solutions for this quadratic equation.
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By looking for that greatest common factory you can often save yourself quite a bit of work.
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In this one I’m going to get my 3x to the other side.
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Then I immediately noticed that both of these have an x in common, so I can actually pull that x out.
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I have an x and x -3, those are my two factors that I will set equal to 0.
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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.
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It is already done.
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I will go ahead and add 3 to the other equation that we formed that way that one is solved.
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x = 3.
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We have both of our solutions for that one.
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A tricky part of this entire thing is to make sure that it is set equal to 0 first.
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Sometimes you might have to do some simplifying or combine things together before you can set it equal to 0.
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In this one we have x × 4x + 7 and all of that is equal to 2.
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It is not that difficult to get it equal to 0 since we would simply subtract 2 from both sides.
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But in order to have factors I need everything to be multiplied together and I still have some subtraction in here.
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What I'm going to do is distribute with my x here and then I will go ahead and try factoring.
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4x² + 7x – 2 = 0.
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This one is a little bit more complicated, it is not very obvious what the reverse foil method should do on it.
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Let us go ahead and dig up our AC method and give that a try.
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The AC method we will multiply a and c together 4 × -2 = -8.
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I'm looking for two numbers that will multiply to give me -8 but add to be 7.
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What do we have for possibilities, it could be 1, 8, it could be 2 and 4.
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It could be any of those in reverse.
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We want them to multiply to give us -8 and we want it to add to be 7.
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It looks like I have to use that 1 and 8.
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The way this will work out as it will be -1 and 8.
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That will help me split up my middle term.
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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.
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The next two, I can see a common 2 that I can pull out.
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And now that we have that I can say what my factors are.
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We have 4x - 1 and x + 2.
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That is quite a lengthy process but recognize we are not done yet.
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All that was just done to factor it.
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We need to take each of the factors and set them equal to 0.
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4x - 1 = 0 and x + 2 = 0.
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We will add one to both sides and divide by 4.
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We have one of our solutions that x could equal ¼.
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The other one let us go ahead and subtract 2.
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Our other solutions is x =-2.
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The factoring process can be lengthy, but it is just one step along the way to finding the solutions.
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Let us try another one.
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Some higher degree polynomials can also be solved using this factoring process.
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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.
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Notice how this one is not a quadratic, it actually has x³ in there.
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I can still use some of my factoring techniques because they both have 2x in common.
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Let us pullout that 2x.
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I would have x² - 25.
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It looks like that what is left over happens to be a quadratic, but it is actually even more special than that.
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This right here is the difference of squares which means I can use one of my short cut formulas to help me out.
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This would be x + 5 and x – 5.
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I was able to factor it down completely and now you can take all of the pieces and set each one to 0.
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2x, that could equal to 0, x + 5 that could equal 0, and x - 5 that could equal 0 as well.
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Divide both sides by 2 for this one.
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That will give us x could be 0.
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-5 on this one x could equal -5, add 5 to both sides of this one, x= 5.
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This one has 3 solutions that I could go back and check.
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Let us try out everything we have learned with factoring to see if we can tackle out one more problem.
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This one is quite large.
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This one involves x -1 × 2x – 1 = x +1².
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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.
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I need to factor it from there and I actually have to do some multiplying before actually get to that factoring process.
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We will start over here on the left side.
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Let us go ahead and do some foiling okay.
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My first terms would be 2x², outside –x, inside - 2x, and my last terms 1.
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Over on the right side of this, this is the same as x +1 × x +1.
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We are going to use foil to help us spread this out.
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Our first terms on that side x × x = x², outside terms x, inside terms x, and last terms 1×1 =1.
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I have better freedom on combining things together.
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Let us combine together these x's and these x's.
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2x² - 3x + 1= x² + 2x + 1.
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You can see that this is definitely quadratic.
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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.
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I'm going to subtract the x² from both sides.
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That will combine those ones and subtract 2x and we will subtract a 1.
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This will give us x² - 5x = 0.
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This seems like a much nicer problem to solve.
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It even has a greatest common factor of x that I can take out from both of these terms.
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I can take each of these factors and set them equal to 0.
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Looks like one of those are already solved for us, so we will just leave that as it is.
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The other one we will add 5 to both sides and then we will get our second solution that x must equal 5.
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Some quadratics you can factor them using many of the factoring techniques that you learn before.
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Take each of those factors and set them equal to 0 so you can find your solutions.
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