Question

How would you factor -(3x^2 - 4x + 15)?
Explain, please!

Answer 1

\(-(3x+a)(x+b) \) where \(ab=15\) and \(3b+a=-4\)

Answer 2

actually this one doesn't factor using integers

Answer 3

So wait, what. :c

Answer 4

\( -(3x^2 - 4x + 15)\) does not factor

Answer 5

The original problem was
\[ \frac{ (x^2 + 5) \times 3 - (3x + 2) \times 2x }{ (x^2 + 5)^2 }\]

Answer 6

So I made the numerator
(3x^2 + 15) - (6x^2 +4x)

Answer 7

\[\frac{ (x^2 + 5) \times 3 - (3x + 2) \times 2x }{ (x^2 + 5)^2 }\]\[=\frac{ 3x^2+15 -6x^2-4x}{ (x^2 + 5)^2 }\]

Answer 8

3x^2+15-6x^2-4x = -3x^2-4x+15 = -(3x^2+4x-15)

Answer 9

combine like terms get
\[\frac{-3x^2-4x+15}{(x^2+5)^2}\]

Answer 10

Which is what I thought I had to factor o:

Answer 11

-(3x^2+9x-5x-15) = -(3x(x+3)-5(x+3) = -(3x-5)(x+3)

Answer 12

\[-3x^2-4x+15=-(3x^2+4x+15)=-(x+3)(3x-5)\]

Answer 13

it was \(+4x\) not \(-4x\) when you factor out the \(-\)

Answer 14

Oh! Okay, thank you guys. (: