Answer:
Step-by-step explanation:
6x² + 19 x +15 can be factor as (x-root1 )( x-root2)* coefficient of x²
use quadratic equation to fid the roots, x = (-b±√b²-4ac)/2a
6x² + 19 x +15 =0
x= (-19 ±√19²-4*6*15) / 2*6
x= (-19 ± √1)/ 12
x= (-19+ 1)/12 = -18/12 = -3/2
and
x= (-19-1)/12 = -20/12 = -5/3
6x² + 19 x +15 = 6*(x+(3/2)) (x+(5/3))
the length and with of the rectangle could be any combination of the factors of 6 and the (x+(3/2)) (x+(5/3))
if we consider 6 =2*3 we have length and width 2x+3 and 3x+5
because 2*[x+(3/2)]*3[ x+(5/3)]
if we consider 6 = 3*2 we have length and width 3x+(9/2) and 2x+(10/3)
because 3*[x+(3/2)]*2[ x+(5/3)]
if we consider 6= 6*1 we have length and width 6x+9 and x+(5/3)
because 6*[x+(3/2)]*1[ x+(5/3)]
if we consider 6= 1*6 we have length and width x+(3/2) and 6x+10
because 1*[x+(3/2)]*6[ x+(5/3)]
