Question

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?

Answer 1

See a solution process below:

Explanation:

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`