Find factors of g(x)=2x^3+3x^2-23x-12 when (2x+1) is a factor

Step-by-step explanation:

STEP

1

:

Equation at the end of step 1

 (((2 • (x3)) -  3x2) -  23x) +  12

STEP  

2

:

Equation at the end of step

2

:

 ((2x3 -  3x2) -  23x) +  12

STEP

3

:

Checking for a perfect cube

3.1    2x3-3x2-23x+12  is not a perfect cube

Trying to factor by pulling out :

3.2      Factoring:  2x3-3x2-23x+12  

Thoughtfully split the expression at hand into groups, each group having two terms :

Group 1:  -23x+12  

Group 2:  2x3-3x2  

Pull out from each group separately :

Group 1:   (-23x+12) • (1) = (23x-12) • (-1)

Group 2:   (2x-3) • (x2)

Bad news !! Factoring by pulling out fails :

The groups have no common factor and can not be added up to form a multiplication.

if the eqn of the parabola is f(x) = (x-4)(x+2), we need to mult. this out to find the coordinates of the vertex.

then f(x) = x^2 + 2x - 4x - 8, and f(x) = x^2 - 2x - 8.

one way in which you can find the vertex is to calculate x = -b / (2a).   here, that comes to

        )

x = = 1.   f(1) = 1^2 - 2(1) - 8 = - 9.   vertex is at (1, -9).

        2(1) 

find another point on the curve:   let x=0, for which y=-8.   then (0, -8) is on the curve.   plot the vertex and this point and draw a smooth parabola thru both.

hope this

look a the sides they are perpendicular