Question

6x^2 + x - 12 How to factor by splitting middle term?

Answer 1

try \( \bf 6x^2 + \color{red}{x} - 12\implies 6x^2 \color{red}{+ 9x-8x} - 12\implies (6x^2 + 9x)-(8x + 12)\)

Answer 2

Man do u know how to complete the square?

Answer 3

I will teach you how to solve any quadratic equation without having to factor anything:
\[ax^2+bx+c=0\]

It is called completing the square:
Lets do it on your quadratic:

\[6x^2+x-12=0\]

Divide everything by 6 to get it in a form with no a-term:
\[x^2+\frac{ 1 }{ 6 }x-2=0\]

Add the opposite of the c-term of the quadratic to both sides in this case (2) not (-2) but 2:

\[x^2+\frac{ 1 }{ 6 }x=2\]

Add half of the b-term squared to both sides so half of one-sixth or one twelfth squared or" (1/144)":
\[x^2+\frac{ 1 }{ 6 }x+\frac{ 1 }{ 144 }=2+\frac{ 1 }{ 144 }\]

Factor this by doing be (x+half of the b-term)^2=whatever

so for this quadratic:\[(x+\frac{ 1 }{ 12 })^2=2+\frac{ 1 }{ 144 }\]
The 1/12th is half of the b-term i was talking about

Take the square root of both sides:
\[x+\frac{ 1 }{ 12 }=\pm\sqrt{2+\frac{ 1 }{ 144 }}\]
Subtract 1/12 from both sides:

\[x=\pm \sqrt{2+\frac{ 1 }{ 144 }}-\frac{ 1 }{ 12 }\]