in this video you will learn how to solve quadratic equations by factoring the tutorial is divided into two parts in the first part we will solve two term quadratic equations using techniques like the greatest common factor and the difference of two squares formula in the second part we will solve quadratic trinomials both with a leading coefficient of one and different from one we will use techniques like the product sum method the AC method and factoring by grouping towards the end we will cover quadratic equations given in non-standard form and also address the most common mistake made when solving quadratic equations let's get started in our first three examples the constant term of each quadratic equation is missing in such cases our first step is to factor out the greatest common factor from each term in this example the GCF of x^2 and -10 X is X this is one of the factors of this equation the other factor is just the sum of what is left here which is x - 10 next we will apply the zero product property but first what is this property the zero product property states that if the product of two factors is equal to zero then at least one of the factors must be zero for instance If the product of a and b equals z then either a is Zer or b is z similarly for this product to be equal to zero either X must be zero or x - 10 must be zero the next step is to solve for x for the first equation x is already equal to zero for the second equation add 10 to both sides to isolate X these cancel out and we get x equal to 10 therefore the solutions are X = 0 or x = 10 we can check our solutions by substituting them back into the original equation substituting 0 0^ 2 is 0 10 * 0 is also 0 and the difference is also 0 likewise substituting 10 10 s is 100 10 * 10 is also 100 and the difference is zero as you can see both 0 and 10 check out confirming that they are indeed solutions to the quadratic equation let's work on another example like this as we did in the previous example first we factor out the GCF from each term the G G CF of the numbers is three and that of the variable is X so the GCF of the two terms is 3x the other factor is the sum of what remains here which is x + 5 Now using the zero product property for this product to be zero either 3x must be 0 or x + 5 must be zero then solve for x for the first equation divide both sides by 3 to isolate X these cancel out and we get x equal 0 for the second equation subtract five from both sides to isolate X therefore the solutions are X = 0 or X = -5 now is your turn please pause the video and give it a try first factor out the GCF from each term the GCF of the numbers is -4 and that of the variable is X right so the GCF of the two terms is -4x the other factor is 2 x + 3 next set each factor equal to 0 and solve for x from the first equation we get x = 0 for the second equation first subtract 3 from both sides to isolate 2x then divide both sides by two to isolate X therefore the solutions are X = 0 or -3 Hales in the next four examples the the middle term of each quadratic equation is missing in such cases we will use the difference of two squares formula to factor the equations the difference of two squared values is equal to the product of the sum and the difference of the values in this example 81 is a perfect square and can be written as 9^ squar right as you can see this is now the difference of two squares and we can Factor it as x + 9 * x - 9 now we have completed Factor factoring next using the zero product property we set each factor equal to zero then solve for x solving the first equation we get x = -9 solving the second equation we get x = 9 therefore the solutions are x = -9 or POS 9 in the next example the leading coefficient is negative and the constant term is positive in this case factor out the negative sign to the front making the equation suitable to be factored by using the difference of two squares formula now the expression inside the parentheses can be factored as x + 7 * x - 7 next apply the zero product property for this product to be zero either x + 7 must be 0 or x - 7 must be 0o from the first equation we get x = -7 from the second equation we get X = 7 therefore these are the Solutions in the next example the constant term is not a perfect square which might make it seem difficult to apply the difference of two squares formula however since 7 is a factor of 63 we can Factor it out making the constant term a perfect square in fact a general tip when solving quadratic equations with a leading coefficient different from one is to begin by factoring out the greatest common factor this simplifies the factoring process now apply the zero product property and solve for x subtract three from both sides of the first equation and add three to both sides of the second equation therefore these are the solutions now it's your turn please pause the video and give it a try in this example 9 and 25 do not have a common factor but both are perfect squares you can rewrite 9x2 as 3x^2 and 25 as 5^ 2 this is now a difference of two squares and it can be factored easily right next set each factor equal to zero and solve for x by the way these types of quadratic equations can also be easily solved using the square root property please make sure to check the link in the description after you finish watching this video now let's move on to solving quadratic trinomials we begin with equation that have a leading coefficient of one in such cases our first step is to find two numbers that multiply to give the constant term and at the same time add up to the coefficient of the middle term in this example the constant term is 24 and the coefficient of the middle term is 11 so we need to find two numbers that multiply to give 24 and add up to 11 to find these numbers easily we start by listing the factor pairs of 24 1 and 24 2 and 12 3 and 8 4 and 6 we also have the negative factor pairs which can be obtained by changing the sign of these factor pairs now we have a list of all pairs that multiply to give 24 next we need to figure out which pair adds up to 11 let's add them to find out 1 + 24 = 25 2 + 12 = 14 3 + 8 equal 11 we have found the numbers it is the pair 3 and 8 right there is no need to check the remaining pairs as only one pair satisfies both conditions therefore as three and 8 are the two numbers that multiply to give 24 and add up to 11 when we Factor this quadratic equation it becomes x + 3 * x + 8 which equals 0 at this point we have completed factoring the equation the next step is to apply the zero product property by setting each factor equal to zero then solve for x solving the first equation we find that x = -3 solving the second equation we find that x = 8 therefore the solutions are x = -3 or8 we can check our solutions by substituting them back into the original equation substituting -3 -32 is is 9 and 11 * -3 is -33 9 - 33 is -4 and this equals 0 right similarly substituting -8 8^2 is 64 and 11 * -8 is 88 64 - 88 is -24 and this equals 0 as you can see both -3 and 8 check out let's work on a another example like this similar to the previous example the leading coefficient of this quadratic equation is also one so the first step is to find two numbers that multiply to give the constant term and add up to the coefficient of the middle term to find these numbers easily first list the factor pairs of -36 1 and -36 2 and8 3 and -12 4 and 9 6 and -6 by changing the sign of these factor pairs you can obtain the other half of the factor pairs now you have a list of all pairs that multiply to give -36 next figure out which pair adds up to -5 simply add the pairs to find out which one it is it is the pair 4 and -9 right as I mentioned before there is no need to check the remaining pairs only one pair satisfies both conditions therefore as 4 and -9 are the two numbers that multiply to give -36 and add up to -5 when we Factor this quadratic equation it becomes x + 4 * x -9 which equals 0 next set each factor equal to 0 and solve for x solving the first equation we find that x = -4 solving the second equation we find that x = 9 therefore the solutions are x = -4 or 9 in the next example the leading coefficient is not one what should we do in such a case when the leading coefficient is not one our first step is to check if we can Factor it out in this case the leading coefficient is Nega one and it can be factored out easily simply change the sign of each term and place the negative sign in front of the parentheses now the expression inside the parentheses has a leading coefficient of one so you can follow the same steps as in the previous two examples please pause the video and try factoring the expression inside the parentheses find two numbers that multiply to give -42 and add up to 11 list the factor pairs of -42 1 and -42 2 and -21 3 and -14 6 and -7 changing the sign of these factor pairs gives you the other half now all these pairs multiply to give -42 but only one of them adds up to 11 which one is it it is only -3 and 14 right therefore these are two numbers we are looking for and the expression inside the parentheses can be factored as x - 3 * x + 14 now for this product to be zero either x - 3 must be 0 or x + 14 must be 0 solving the first equation we find that x = 3 solving the second equation we find that x = -14 therefore these are the solutions let's work on another example like this in this case the leading coefficient is three which is a common factor of both 27 and 60 therefore we can Factor it out now the expression inside the parentheses has a leading coefficient of one from our previous examples you know the process of factoring this right feel free to pause the video and give it a try find two numbers that multiply to give 20 and add up to -9 list the factor pairs of 20 1 and 20 2 and 10 four and five change their sign to obtain the other half now all these pairs multiply to give 20 but only -4 and -5 add up to -9 right therefore the expression inside the parentheses can be factored as x - 4 * x - 5 now for this product to be zero either x - 4 must be 0 or x - 5 must be 0 right from the first equation we find that x = 4 from the second equation we find that x equal 5 therefore these are the Solutions in this example factoring out three made the leading coefficient one however in the next example we cannot factor out five because it is not a common factor of both 18 and 9 in such cases start by multiplying the leading coefficient by the constant term in this example the leading coefficient is five the constant term is 9 and their product is 45 then find two numbers that multiply to give this product and add up to the coefficient of the middle term so you need to find two numbers that multiply to give 45 and add up to 18 list the factor pairs of 45 1 and 45 3 and 15 if you notice 3 and 15 add up to 18 right so these are the two numbers we looking for while listing the factor pairs if you notice a pair that works you don't need to continue listing the remaining pairs since there is only one factor pair that satisfies both conditions moreover if you can mentally figure out the two numbers there is no need to make a list you can proceed with that and by the way because we are multiplying a by C here this method is called the AC method the next step is to split the middle term using the two numbers we found we can rewrite 18x as the sum of three X and 15x if we add these terms back together we get the original 18x so we are not changing the value of the equation the other terms in the equation stay the same next we Factor by grouping group the first two terms together and the last two terms together then factor out the greatest common factor from each group in the first group the GCF is X which is one of the factors of this group the other factor of this group is simp simply the sum of what is left here which is 5x + 3 in the second group the GCF is three which is one of the factors of this group the other factor of this group is the sum of what remains here which is again 5x + 3 when you factor by grouping if the factors inside the parentheses are the same for both groups then you are on the right track otherwise you need to go back and check your previous steps next factor out the common factor for both groups this leaves behind x + 3 as the other Factor now we are done with factoring the next step is to set each factor equal to zero and solve for x solving the first equation we find that x equal -3 fths solving the second equation we get x = -3 therefore these are the Solutions in the next example the quadratic equation is not in standard form what should we do first in this case the first step is to rewrite the quadratic equation in standard form to do this subtract 43x from both sides of the equation now you have a quadratic equation in standard form with a leading coefficient different from one since 18 is not a common factor of both 43 and 5 it cannot be factored out therefore you need to use the AC method as in the previous example please feel free to pause the video and give it a try first multiply the leading coefficient by the constant term which equals 90 then find two numbers that multiply to give 90 and add up to -43 list the factor pairs of 90 1 and 90 2 and -45 notice that 2 and -45 add up to -43 right now as we have already found the two numbers we are looking for there is no need to list the remaining factor pairs so we can proceed to The Next Step which is to split the middle term using these two numbers you can rewrite - 43x as the sum of 2x and -45x the other terms stay the same next Factor by grouping group the first two terms together and the last two terms together and then factor out the GCF from each group in the first group the GCF of the numbers is two and the variable is X so the GCF is 2x notice that here 2x is the same thing as 2 x * 1 so the other factor of this group is 9x + 1 in the second group the GCF is -5 leaving 9x + 1 as the other Factor notice that the factors inside the parentheses are the same for both groups indicating that you are on the right track next factor out 9x + 1 which leaves 2 x - 5 as the other Factor finally apply the zero product property by setting each factor equal to zero and then solve for x solving the first equation you find that X = -19 solving the second equation you find that x = 5 Hales therefore these are the solutions let's solve our last equation which has only one solution when a quadratic equation is given in this form please avoid setting the factors equal to the number on the right side of the equation and solving for x this is a common mistake remember the zero product property can only be applied when the product of the factors is zero in this example the product of 4X and x - 5 is -25 which is not zero therefore the zero product property cannot be applied in this form so what should we do first to solve this equation because this quadratic equation is in non-standard form the first step is to rewrite it in standard form to do this start by Distributing 4X 4X * X is 4x² 4X * -5 is -2X then add 25 to both sides of the equation now you have a quadratic equation in standard form and your next step is to factor it since four is not a common factor of both 20 and 25 it cannot be factored out so you need to use the AC method first multiply 4 by 25 which equals 100 then find two numbers that multiply to give 100 and add up to -20 let's do this mentally try to think of two numbers that satisfy both conditions you know that 10 * 10 is 100 and 10 + 10 is 20 but if you make both 10 negative the sum becomes -20 and the product is still 100 right so you have found the two numbers the next step is to split -2X using these numbers the other terms stay the same next Factor by grouping please pause the video and give it a try in the first group the GCF of the numbers is two and that of the variable is X X so the GCF is 2x the other factor is 2x - 5 in the second group the GCF is -5 and the other factor is again 2x - 5 now factor out the common factor which is 2x - 5 as you can see the other factor is also 2x - 5 when we set each factor equal to zero and solve for x we find the same solution so this quadratic equation has only one solution in general when you factor a quadratic equation and obtain the same factors it implies there is only one solution this typically occurs when the trinomial is a perfect square thanks for watching I hope you found this video helpful please give it a thumbs up and consider subscribing if you have any questions please feel free to leave them in the comments below I will do my best to answer them