4/22/2024 0 Comments Factoring a quadratic equationGreatest common factor? So let's see, they'reĪll divisible by three, so you could factor out a three. Pause the video and see if youĬan factor this completely. So let's say that we had the expression negative three x squared And now I have actuallyįactored this completely. That as x minus three, times x plus one, x plus one. So I could re-write all of this as four times x plus negative three, or I could just write Plus one is negative two, and negative three times So let's see, A could beĮqual to negative three and B could be equal to one because negative three A plus B is equal to negative two, A times B needs to beĮqual to negative three. I get a negative value, one of the 'em is going to be positive and one of 'em is going to be negative. I get negative three, since when I multiply That add up to negative two, and when I multiply it Now am I done factoring? Well it looks like IĬould factor this thing a little bit more. And if I factor a four out of negative 12, negative 12 divided byįour is negative three. Of negative eight x, negative eight x dividedīy four is negative two, so I'm going to have negative two x. Out of four x squared, I'm just going to be So I could re-write this as four times, now what would it be, four times what? Well if I factor a four They're not all divisible by x, so I can't throw an x in there. So let's see, they'reĪll divisible by two, so two would be a common factor, but let's see, they'reĪlso all divisible by four, four is divisible by four,Įight is divisible by four, 12 is divisible by four, and that looks like the Try to find the greatest of the common factor, possible common factors There any common factor to all the terms, and I So the way that I like to think about it, I first try to see is So factor this completely, pause the video and have a go at that. ^ Sterling, Mary Jane (2010), Algebra I For Dummies, Wiley Publishing, p.See if we can try to factor the following expression completely.This can be a powerful tool for verifying that a quadratic expression of physical quantities has been set up correctly. Furthermore, by the same logic, the units of c must be equal to the units of b 2 / a, which can be verified without solving for x. If the constants a, b, and/or c are not unitless then the units of x must be equal to the units of b / a, due to the requirement that ax 2 and bx agree on their units. There will be no real values of x where the parabola crosses the x-axis. The complex roots will be complex conjugates, where the real part of the complex roots will be the value of the axis of symmetry. However, there is also the case where the discriminant is less than zero, and this indicates the distance will be imaginary – or some multiple of the complex unit i, where i = √ −1 – and the parabola's zeros will be complex numbers. If the discriminant is positive, the distance would be non-zero, and there will be two solutions. This is one of three cases, where the discriminant indicates how many zeros the parabola will have. Algebraically, this means that √ b 2 − 4 ac = 0, or simply b 2 − 4 ac = 0 (where the left-hand side is referred to as the discriminant). If this distance term were to decrease to zero, the value of the axis of symmetry would be the x value of the only zero that is, there is only one possible solution to the quadratic equation. The other term, √ b 2 − 4 ac / 2 a, gives the distance the zeros are away from the axis of symmetry, where the plus sign represents the distance to the right, and the minus sign represents the distance to the left. The axis of symmetry appears as the line x = − b / 2 a. X 1 = − b + b 2 − 4 a c 2 a and x 2 = − b − b 2 − 4 a c 2 a
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