Factorising Quadratics


[tex]12x^{2} +x-15=(6x+?)(2x-?)[/tex]


Find the missing numbers.


I'm not entirely sure if this is possible, as my teacher is prone to making typos but I have no idea how to solve this; I've tried trial and error with factors of 15 and none of them seem to work!

Respuesta :

quadratic formula where:
[tex]ax ^{2} + bx + c[/tex]
[tex] \frac{ - b + { \sqrt{b^{2} - 4ac } }}{2a} [/tex]
[tex] \frac{ - b - \sqrt{b { }^{2} - 4ac } }{2a} [/tex]
do both do get the roots of the equation

Answer:


Step-by-step explanation:

A short answer is that you cannot factor this the way you are trying to do it. What to do to go on is a question. You could use the quadratic formula to get some sort of answer but if factoring is your only option, forget about it.

Here's how you can check.

  • a = 12
  • b = 1
  • c = -15

There is a part of the quadratic solution is the sqrt(b^2 - 4*a*c). In order to be factorable, this sqrt must result in a perfect square.

sqrt(1 - 4(12)(-15) ) = sqrt( 1 - - 720) = sqrt(721)

This is anything but a perfect square. (721 does factor, however, into 7 * 103). I03 is prime. So the long and short of it is, there's no way it will factor.

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