Question

How do you factor `6x^2+23x+6`?

Answer 1

Root1 `= - 0.282 (3dp)` & Root2 `= -3.552 (3dp)`

Explanation:

This is a quadratic equation of form `ax^2+bx+c` Here `a=6;b=23;c=6` Roots are `-b/(2a)+- sqrt(b^2-4ac)/(2a) or -23/12+-sqrt(529-144)/12 = -1.916 +- 19.62/12 = -1.916 +- 1.635 :.` Root1 `= - 0.282` & Root2 `= -3.552`[Ans]

Answer 2

`6x^2+23x+6=6(x+(23-sqrt(385))/12)(x+(23+sqrt(385))/12)`

Explanation:

Complete the square, then use the difference of squares identity:

`a^2-b^2=(a-b)(a+b)`

with `a=(x+23/12)` and `b=sqrt(385)/12` as follows:

`6x^2+23x+6`

`=6(x^2+23/6x+1)`

`=6((x+23/12)^2-23^2/12^2+1)`

`=6((x+23/12)^2-(529-144)/12^2)`

`=6((x+23/12)^2-385/12^2)`

`=6((x+23/12)-sqrt(385)/12)((x+23/12)+sqrt(385)/12)`

`=6(x+(23-sqrt(385))/12)(x+(23+sqrt(385))/12)`

Note that it is not possible to simplify `sqrt(385)` any further since `385 = 5*7*11` has no square factors.