How do you find values for #a# and #b# that would make the equation #-2y(y^2-6y-3) = ay^3+12y^2+by# true?

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Answer 1 · 2017-08-09T02:23:57

See a solution process below:

Expand the term on the left by multiplying each term within the parenthesis by the term outside the parenthesis:

#color(red)(-2y)(y^2 - 6y - 3 = ay^3 + 12y^2 + by#

#(color(red)(-2y) xx y^2) - (color(red)(-2y) xx 6y) - (color(red)(-2y) xx 3) = ay^3 + 12y^2 + by#

#-2y^3 - (-12y^2) - (-6y) = ay^3 + 12y^2 + by#

#-2y^3 + 12y^2 + 6y = ay^3 + 12y^2 + by#

Because the #y^2# term on each side of the equation has the same coefficient. We can solve for each of the other coefficients separately.

Solve For a:

#-2y^3 = ay^3#

#(-2y^3)/color(red)(y^3) = (ay^3)/color(red)(y^3)#

#-2 = a#

#a = -2#

Solve For b:

#6y = by#

#(6y)/color(red)(y) = (by)/color(red)(y)#

#6 = b#

#b = 6#