How to factor a trinomial with the quadratic formula

[@bprpmathbasics]

Check out 5 easier ways: kzbin.info/www/bejne/e3LUm6CLpLClb5o

[@anuragguptamr.i.i.t.2329]

So, the quadratic formula is most useful, only when the Zero roots are imaginary or irrational. However, in beginning of a problem, we might not be able to know in advance, whether the roots would come out to be real/ imaginary/ rational/ irrational.

[@cyrusyeung8096]

This is basically an application of the factor theorem, which states that if k is a solution to f(x) = 0, then (x - k) is a factor of f(x)

[@savitatawade2403]

yep

[@notar2123]

This can also be generalized. P(x) = a₁ + a₂x² + ... + aₙxⁿ = aₙ(x-x₁)(x-x₂)...(x-xₙ), where x₁,x₂ etc. are roots of P(x).

[@DrFunkman]

This might be the first trinomial equation I’ve seen on the channel that you can solve by figuring it out in your head

[@enterlessguy]

Hey bprp, we actually don't always need to guess to be able to factor the expressions, heres an example: P(x)=2x^2+5x-7 Firstly, we will start by multiplying P(x) by a and writing our expression in this special way. (The expression can be written as (ax)^2+b(ax)+ca=> 2P(x)= (2x)^2+5(2x)-14 => Let 2x=u=>2P(x)=u^2+5u-14=> 2P(x)= (u+7)(u-2)==> 2P(x)=(2x+7)(2x-2) =>2P(x)=(2x+7)2(x-1)=> divide both sides by 2 (it would be a in general case) and done, => P(x)=(2x-7)(x-1) I think this is a cool method that people should know.

[@endru8]

How did you turn u²+5u-14 into (u+7)(u-2) without guessing?

[@MathNerd1729]

Nice vid! :) I agree; I'd much rather just factor it using a bit of guess-and-check. But on another note, the thought of factoring quadratics with the quadratic formula makes me think about factoring some cubics using the cubic formula 😂

[@ChavoMysterio]

6x²+7x-3 6x²+9x-2x-3 3x(2x+3)-1(2x+3) (3x-1)(2x+3) ❤

[@anuragguptamr.i.i.t.2329]

So, the quadratic formula is most useful, only when the Zero roots are imaginary or irrational. However, in beginning of a problem, we might not be able to know in advance, whether the roots would come out to be real/ imaginary/ rational/ irrational.

[@savitatawade2403]

you can...

[@Brid727]

if b^2 - 4ac is a perfect square number and greater than or equal to 0, then the entire quadratic expression is factorable and its roots are rational

[@wacdac1826]

didnt know you could do this

[@ZoidVERSE]

I thought everyone knew it though, since the concept of something being factored is finding it's roots, which is what the quadratic formula does

[@jeffw1267]

It took me about ten seconds to calculate this mentally. I'm a genius.

[@cloverisfan818]

Definitely easier if you have the quadratic formula built-in in your calculator

[@Aramlyric]

THAT'S MY LESSON FROM SCHOOL😂😂😂

[@kevinstreeter6943]

Why do you want to factor it if it is not equal to 0?

[@endru8]

you can use it for partial fraction decomposition when calculating an integral or something

[@NadiehFan]

I can see why you are not in favor of using the quadratic formula to factor a quadratic polynomial, particularly when this is a quadratic polynomial with integer coefficients that can be factored over the integers. What you are doing here is using the quadratic formula to find the zeros x₁ and x₂ of ax² + bx + c in order to write this as a(x − x₁)(x − x₂) and then distribute (factors of) a over (x − x₁) and (x − x₂) in order to obtain the required or desired factorization over the integers. However, this proces can be streamlined. First of all, a quadratic ax² + bx + c with _integer coefficients_ can be factored over the integers _if and only if_ its discriminant D = b² − 4ac is the square of an integer (this can easily be proved). Since we can check ahead of time by calculating the discriminant whether or not any given quadratic with integer coefficients can be factored over the integers we never have to waste time trying to factor a quadratic with integer coefficients that cannot be factored over the integers. But, calculating the discriminant of a quadratic ax² + bx + c with integer coefficients is not just useful to make sure the quadratic can actually be factored over the integers. Once we have established that D = b² − 4ac is the square of an integer and that √D is therefore an integer we can use this to take the trial and error out of factoring by grouping. As is well known, to factor a quadratic ax² + bx + c with integer coefficients over the integers using factoring by grouping we need to find two integers f and g with sum b and product ac. It is easy to see that if D = b² − 4ac is the square of an integer and therefore √D is an integer with the same parity as b then f = (b + √D)/2 and g = (b − √D)/2 are indeed two integers with sum b and product ac. To make the expressions for f and g easy to remember, just keep in mind that f and g are two integers with _sum_ b and _difference_ √D. To see how this works, let's factor your quadratic polynomial 6x² + 7x − 3 using factoring by grouping but _without_ any trial and error. First we calculate the discriminant which is D = 7² − 4·6·(− 3) = 49 + 72 = 121 = 11². So, D is the square of an integer which guarantees that this quadratic with integer coefficients can indeed be factored over the integers. We now calculate f = (7 + 11)/2 = 9 and g = (7 − 11)/2 = −2. Since b = f + g we can rewrite bx as fx + gx so we get 6x² + 9x − 2x − 3 3x(2x + 3) − (2x + 3) (3x − 1)(2x + 3) and we are done. No chance of wasting time trying to factor a quadratic with integer coefficients that cannot be factored over the integers, no trial and error to find two integers with a given sum and product, no messing with the quadratic formula and no messing with fractions. How about this?

[@Dominus-F6uZr]

I like your shirt

[@Raharepa]

Good

[@nvapisces7011]

Quadratic formula works if and only if there are real coefficients for the quadratic equation lol

[@notar2123]

Uhh, no.

[@Yilmaz4]

it works for imaginary coefficients too, test it yourself

[@Qermaq]

It works if the coefficients are parameters, it'll work with any complex values so long as the coefficient of x^2 is not 0.

[@hussainahmad27]

You are not allowed to use quadratic formula in expressions, it is derived only for equations. But with restriction x≠n, n can be any number; can you use quadratic formula. NO! An expression does not have any roots, and in equation, factorization is unnecessary, therefore this method is not a standard. Another theorem is required!!!

[@sumanbag729]

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