Solving Quadratic Equations: The Zero-Factor Property Pt. 1 [fbt] (The Zero-Product Property)

Solving Quadratic Equations: The Zero-Factor Property Pt. 1 [fbt] (The Zero-Product Property)



do you feel it because you don't know numbers you don't know which one go which way and they got all these marks and things on them I know honey is hard but there is a solution for then tutoring and now here go mr. Witt explained math to us mr. Witt hello ladies and gentleman this is mr. Witt with Fort Bend tutoring FBT and today's lesson is going to be about solving quadratic equations dealing with the zero factor property oh yeah and basically you're just looking at the equation ladies and gentlemen in the form of ax squared plus BX plus C equals zero a quadratic equation preferably in standard form or could already be factored out like in X minus B times X minus C equals to zero in other words you want to have it factored set equal to zero and in that case you can set each factor equal to zero and solve for it so in a problem like this X would equal to B and then X would equal to C and basically ladies and gentlemen that's the zero factor property but that means nothing to you until you see some examples right so let's look at problem number one I have it right here right underneath it let's check it out in problem number one we have the quantity of X plus three times the quantity of X minus four equal to zero so notice that we have two sets of parentheses multiplying against each other and they're set equal to zero we'll see that's what you want you want things factored and always set equal to zero when using the zero factor property so once it's factored and set equal to zero all you would need to do is set each factor equal to zero so I'll have it re-written as X plus 3 equal to 0 and X minus 4 equals zero and then you just solve for each equation so in the first equation I'll be subtracting 3 to both sides which gives me x equals to negative 3 and then the second equation I'll be adding 4 to both sides equal sign and this gives me x equals 2 positive 4 once I have these two solutions and those are my answers by the way ladies and gentlemen I can write them as a solution set so if I can drop oh Lord I cannot draw a brace what can I draw braces ladies and gentlemen alright so if I take my time there you go there you go that's a that's as good as it's going to get for me I cannot draw braces it kills me so what I'm going to do is I'm going to write my solutions in ascending order as though it's on a number so in other words I'll be started with the lowest number the smallest solution which is going to be negative 3 and then my largest number which is going to be positive 4 and then there you go alright so that's your solution set those are your answers done and done and I got my red box right there so that's the zero factor property you start out with a factored quadratic equation you set each factor equal to zero and you write your answer in descending order yeah and that's it done and done ok so here we go moving on to the next problem let's check it out and problem number two I have the following problem hmm I have x squared plus 8x plus 15 equals zero now the method that we're using the zero factor property is best used when you have an expression that can be factored so I would be looking to see if I can find two numbers at a multiply to give me 15 and add to give me eight and so those two numbers would be well very good it would be five and three that's right now if you have any difficulty factoring quadratic trinomials such as this please check out our videos on factoring quadratic trinomials part one part two and part three yeah they're very helpful so check those out all right so here I'm going to open up two sets of parentheses I'm going to make sure it's still set equal to zero I have my variable X because X times X gives me x squared and we said that five and three would multiply to give us 15 and this last sign here will tell us to add to get eight so both of these will need to be positive and there you go it's factored okay but since we're solving the equation we'll need to set each factor equal to zero and then solve for them separately so I have X plus 5 equal to zero I also have X plus 3 equal to zero and then I'll be subtracting 5 to both sides yep okay and then X equal to negative 5 and then I'll also subtract 3 to both sides and I'll end up with x equals to negative 3 so here are my solutions and once again I'll be attempting to draw a brace so pray for me please let's see oh my goodness this is just is ridiculous there you go there you go nice brace so we'll have negative 5 and negative three okay and we're going to deal with that brace right there okay all right send in any strategies that you want to teach me how to draw a brace I cannot do it all right so here's the solution negative five negative three so what do we do we started out with our equation x squared plus 8x plus 15 equal to zero we factor the trinomial we set each factor equal to zero we had our two solutions negative five and negative three and we wrote them in a solution set writing the solutions in ascending order from least to greatest now depending on your teacher depending on whatever online website you may be using they may not care what order your solutions are in in other words you could put the negative three first and it just doesn't matter what order it is I'm just letting you know as a mathematician and as a math student it would be good practice for you to always write your solutions in ascending order from least to greatest all right so there you go moving on mm-hmm the next problem we have pride number three we have 5x squared minus 30x equal to zero so in this particular problem I would like to let you know about a couple of strategies that you can use for one you always start by looking for the greatest common factor when you're factoring your trinomials or binomials or whatever terms you may have when you're trying to use the zero factor property always look for the GCF first another tip I would like to inform you of is that anytime you have a coefficient many a number that they all have in common all right then you can just simply divide that away only when you're dealing with an equation okay so this is different when you're just asked to factor an expression completely here because we have an equation we have something on either side of equal sign we can literally just divide everything by five because each and every term can be divided by five evenly and we can simplify this equation yeah I can just simplify this into x squared minus 6x equal to zero okay notice how the x squared minus 6x equal to zero looks a lot less intimidating than our original problem right okay that's because we simplified it so you can always simplify an equation by a number but never a variable for instance these two terms here they also had X in common but I didn't divide away and so never divide away a variable because that's dividing away a solution so don't ever do that you can simplify it by dividing everything by a number however okay so keep that difference in mind please so continuing on these two terms have X in common so I'm going to factor out an X that'll leave me with X minus 6 equals 0 so here we have it completely factored out and you could have just simply factored out 5x in the beginning and that would have been a ok but I prefer simplified equations it's just a personal preference so here I'm going to set my first factor x equal to 0 and then I'll set my second factor X minus 6 equal to 0 and then I'll solve for each there's nothing left to do because X is already isolated then over here in our second equation we can add 6 to both sides yep and we'll have x equals 2 positive 6 as well so drawing the solution set all right so we have 0 6 as our solution and notice the ugly braces that there you go I can't do anything about this all right so 0 6 that's the answer to problem number 3 let's continue on problem number 4 in problem before we have 5x squared minus 11x equals 2 negative 2 so what I want to do in this problem here is I want to first of all make sure that the entire equation is set equal to zero and right now it's not so I'm going to add 2 to both sides of the equal sign and rewrite this equation all right so I'll be rewriting this as 5x squared minus 11x plus 2 equals 0 from there I want to factor the trinomial on the Left I want to factor that 5x squared minus 11x plus 2 if it can be factored well I can find that out by multiplying 5 times 2 well 5 times 2 will give me 10 so I would be looking for two numbers out of multiply to give me 10 and add to give me 11 well that would be 10 and 1 so I'm going to open up two sets of parentheses I have my variable X I have 10 and 1 and both of these will need to be negative so these two numbers will multiply to give me positive 10 and also add to give me a negative 11 once I have this initial framework set up I'm going to divide both of these values by that leading coefficient of five all right now this technique ladies and gentlemen I have plenty of examples on in my factoring quadratic trinomials part two videos so check that out if you've never seen it done this way so you'll have X minus two times this denominator will go in front and that will be 5x minus 1 and that's all now set equal to 0 so I have a completely factored now mm-hmm once I have the equation factored I can then set each factor equal to 0 and therefore solve the equation so I'll have X minus 2 equal to 0 I have 5x minus 1 equal to 0 and then I'll solve for X in each of the equations alright so I'll be adding 2 to both sides of the equation as well as I'll be adding 1 to the second equation alright that gives me 5x equals to 1 then I'll divide both sides of the equation by 5 so I'll end up with x equals to 1/5 all right so as a solution set let's see all right that's going to have to do for that brace I'll end up with 1/5 and 2 as my solutions all right let's see if I get this red box around here all right so to recap over that problem we started out by setting the initial equation equal to 0 by adding to the both sides we then factor the trinomial all right and you can use any preferred method you know of all right and then after that once it's factored you set each factor separately equal to 0 and solve for your variable and I'm writing my solutions in descending order all right from least to greatest so I have my 1/5 and 2 and that does it for problem number 4 all right let's check out the next problem then for prod number five we have C times the quantity of C minus 11 equals to negative 18 and at first glance a lot of students would assume this equation is already factored but it's not the reason for that ladies and gentlemen is because it's not set equal to zero and you have to have it set equal to zero in order to use the zero factor property so instead what you'll do is multiply in other words you want to distribute first in other words get those arrows popping all right my favorite probably is the distributive property so get those arrows popping you'll end up with C times C which is going to be C squared C times negative 11 is negative 11 C and this is going to be set equal to negative 18 you still want to have this set equal to zero right so let's add 18 to both sides of the equal sign and then we'll be able to rewrite this as C squared minus 11 C plus 18 equals to 0 all right you'll be looking for two numbers that multiply to give you 18 and add to give you 11 those two values would be 9 and 2 so I'm going to have C minus 9 times C minus 2 equal to 0 once again if you still have any difficulty fracturing trinomials check out our videos factoring quadratic trinomials so then after I have these factored you're going to set each factor equal to 0 you'll have C minus 9 equal to 0 you'll have C minus 2 equal to 0 and then all I have to do from there is just add 9 to both sides to isolate my variable seen mm-hmm C equals 2 9 and then I'll add 2 to both sides equal sign this gives me C equals 2 2 and from there all I have to do is write it in my solution set so let's see about those braces jacked up still still jacked up so you'll end up with 2 and alright that's it done and done just cannot draw a brace to save my life no next problem we have proud number six here the equation is 2x cubed equals to 9x squared minus 4x notice that all three terms have an X in common but you will not be dividing away an X remember that by dividing away a variable you're eliminating a solution and you do not want to do that however I do want to have this set equal to 0 first so I'm going to subtract 9x squared and add 4x to both sides of the equal sign so subtracting 9x squared adding 4x to both sides equal sign if you wanted to you can just break this first step down into two steps meaning in your first step subtract 9x squared to both sides and then add 4x and another step but I'm doing all at once because I might add a much room to work with here so I'm going to rewrite my equation as 2x cubed minus 9x squared plus 4x equals 0 yeah that's what's up then I'll be factoring out X as my GCF from each term so factoring out X I can rewrite this as x times 2x squared minus 9x plus 4 equal to 0 then looking at the trinomial you want to see if this can be factored so if I multiply 2 times 4 I get 8 and I know that 2 factors of 8 that will add to give me 9 it's going to be 8 and 1 so this could be factored so I'll be bringing down my variable X opening up 2 sets of parentheses I'll need an X and an X and I'll use a negative 8 and a negative 1 okay because those will be a 2 multiplied to give me 8 and add to give me 9 and then because I had a coefficient in front of my variable x squared I'll need to divide both the 8 and the 1 by 2 okay always by that leading coefficient when using this method simplifying I'll have x times X minus 4 times 2x minus 1 equals 0 I have everything factored out I'll be setting each of these three factors now equal to zero so I have X equal to zero X minus 4 equal to 0 and 2x minus 1 equals 0 and then you'll simply solve for each of the equations so x equals 0 is already solved for I'll be adding 4 to both sides this gives me x equals to 4 I'll also add 1 to both sides in this last equation this gives me 2x equals to 1 and then I'll divide both sides by 2 and this gives me x equals to 1/2 and lexo so my final solution set is going to be 0 1/2 + 4 ok and that's it that's the solution now there is something that you can use to gauge how many solutions you will normally end up with your highest exponent will always tell you the number of solutions that you'll have in an equation sometimes that means that your answers will be the same or that they may not even be real however you'll know the number of solutions to any equation by the highest exponent every single time all right so that's problem number 6 ok ladies and gentlemen here we have problem number 7 in prime number 7 I have negative 8 N squared minus 16 n minus 6 equals 0 so the first thing I notice is 1 my leading coefficient is negative and I never liked that but also that all of the terms can be divided by 2 so that means I'm going to divide every single thing by negative 2 remember it's okay to divide away a number just not a variable so I'll be rewriting this as 4n squared plus 8n plus 3 equals 0 so by dividing the original equation by negative 2 I was able to rewrite this original equation into something that looks a lot easier to solve right so here I'm going to start out by multiplying 4 times 3 and that gives me 12 so 2 factors of 12 that will add to give me 8 is going to be 2 and 6 so I'm going to open up two sets of parentheses here I'll end up with n plus six times n plus two and then because I had that four in front I'll divide both of these numbers by four so next I'll simplify and I'll end up with n plus three-halves times n plus one half equals zero now if you're pretty savvy ladies and gentlemen it since we're solving the equation you could set these equal to zero as they are but I'm going to be consistent and show you what the completed factored form looks like first so anytime you have any denominators left over when you're factoring using this method those denominators will go in front of the variable so you'll end up with 2n plus 3 times 2n plus 1 equals 0 next I'll be setting each one of my factors equal to 0 separately and then solving for my variable n so I have 2n plus 3 equal to 0 I have 2n plus 1 equals 0 and then I'll subtract 3 to both sides in this first equation and then divide by 2 like so so that I end up with a result of n equals to negative three-halves for that first equation and then i'll be subtracting one to both sides and this will be 2n equals the negative one and then dividing both sides by two this gives me n equals to negative one-half so our solution is going to be negative three-halves and negative one half and that's it okay dining done all right that was proud number seven all right our last and final problem ladies and gentleman's problem number eight gave you a lot of examples in this lesson here here in front of rate I have 63 M squared minus 31 M minus 10 and we're trying to find two factors of a master product of 63 times 10 that would subtract to give you 31 ladies and gentlemen I'm not going to be multiplying 63 times 10 and I was going to be huge so instead what I will use is the prime factorization of 63 and 10 to help me find factors that will subtract to give me 31 so looking at this the prime factorization of 63 is going to be 3 times 3 times 7 where as the prime factorization of 10 is 2 times 5 all right and once again ladies and gentlemen this process this particular process that I'm showing you here is in our video factoring quadratic trinomials part three all right so I'll be finding 2 factors out of these numbers here that will subtract to give me 31 so what I believe I'll use is the 9 here alright that 3 times 3 times this 5 and 9 times 5 is 45 right what's left over is 7 times 2 which is 14 and 45 minus 14 is 31 bingo that's what I need it so I'm going to open up two sets of parentheses I know I need my variable M and I'll be using 45 and 14 so remember I got these two factors using the prime numbers of the original first coefficient and the last coefficient I know that the 45 needs to be negative the sign of the middle term is always the sign of your largest factor and because I'm subtracting the 14 needs to be positive and I'll be dividing both of these numbers by that leading coefficient which is 63 yep remember all of this is still set equal to zero simplifying these first numbers by nine I'll end up with five sevens simplifying the 14 and the 63 by seven I'll end up with two nights remember any remaining denominators will go in front of your variable so you'll end up with this factor as 7m minus five times nine M plus two equals to zero setting each factor equal to zero using a zero factor property will have 7m minus 5 equal to zero as well as 9m plus two equals to zero and then solving for each one of these I'll add five to both sides in the first equation they're giving 7m equal to five and then I'll divide both sides of the equal sign by seven to give me M equals two five seven so that's my first solution I got then in this second equation this 9m plus two equal to zero I'll subtract two to both sides of the equal sign I'll bring down nine M equals to negative two and then I'll divide both sides of the equal sign by nine leaving me with M equals to negative 2/9 all right so those are my two solutions as a solution set it will be negative 2/9 and positive 5/7 and that's your solution to problem number 8 alright ladies and gentleman that does it for this lesson of solving quadratic equations using the zero factor property we kicked out eight examples for you how about that please rate comment and subscribe ladies and gentlemen if you're able please donate we really appreciate it peace we certainly hope you enjoyed today presentation by foot been tutoring did you understand the program would you like to rate us or give us some feedback subscribe to us you could do all that on tutor me Mouse dotnet

Comments

  1. Emblem

    thank you so much for making this video, I love how you didn’t explain it and expected us to everything else besides this. It made my life a whole lot easier

  2. carnivalwrestler

    From the world of all martial arts, traditional and non-traditional, advice on how to draw better braces: "Slow makes smooth, and smooth makes quick" : ). Excellent video, btw.

  3. RB

    How come for problem 3 you say not to divide by x (which worked for me), yet on problem 6 you do divide by x? Also on problem 6, does it matter which side of the equation you make zero? Another words could you have used
    – 2x^3 + 9x^2 – 4x?

  4. Rights for all humans

    sir, sir, plz slow down, where I get confused is how you decide about the signs? sometimes, we can have 2 positive and other times it's choosen – & + , please SLOWLY explain, not 150 miles per hour… 🙂 Also, problem #4, how come you add +2 to the solution, and not subtract? (5×2-11x+2=0??? Oooo I am so confused!!! :'(

  5. john burens

    In problem #4 you added 2 to both sides -11x = -2…. Shouldn't the negative eleven have become a -9x? What am I missing? The process makes sense though so I like the way you're showing it because it simplifies to the next step. Thanks again Mr. Whitt!

  6. Nlrsuperstar

    AND I TELL EVERYONE WHO WILL LISTEN AND SOME PEOPLE WHO DON'T LISTEN ABOUT YOUR CHANNEL, I TELL THEM ANYWAY IM DOING THEM A FAVOR BY SHARING YOUR CHANNEL THEY WILL THANK ME LATER LOL

  7. Nlrsuperstar

    YOU MAKE THIS FUN, AND THAT IS SAYING SOMETHING, PLEASE KEEP DOING WHAT YOU ARE DOING, I WISH YOU ALL THE BLESSINGS IN THE WORLD YOU ARE HELPING A LOT OF PEOPLE (MEEEEEE!!) LOL

  8. Nlrsuperstar

    LOVE IS NOT A STRONG ENOUGH WORD FOR HOW I FEEL ABOUT YOUR VIDEOS, THANK YOU SOOOOOOOOO MUCH, IVE BEEN SUBSCRIBED FOR A WHILE AND EVEN TOLD MY COLLEGE INTERMEDIATE MATH TEACHER ABOUT YOUR AWESOMENESS, SHE BEING EQUALLY AWESOME QUICKLY WROTE DOWN YOUR YOUTUBE CHANNEL TO PASS ON TO OTHER STUDENTS.

  9. Jarima Alvarado

    So my professor solved an equation using the zero product property but I didn't understand how he did it. For one it's an online class, and two it looks complicated…

    The equation is

    x(2x+1)=15
    2x(squared)+x-15=0
    (2x-5)(x+3)=0
    x=5/2, -3

    {5/2, -3}

Leave a Reply

Your email address will not be published. Required fields are marked *