Question

Solve the following polynomial equation by grouping and factoring.9y3−16y=0

Answer 1

Solve:

`9y^3-16y=0`

Factoring out y:

`y(9y^2-16)=0`

The expression in parentheses can be factored by using the special product:

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

Therefore, the equation becomes:

`y(3y-4)(3y+4)=0`

Now equate each factor to 0 and solve for y:

`\begin{gathered} y=0 \\ \\ 3y-4=0\therefore y=(4)/(3) \\ \\ 3y+4=0\therefore y=-(4)/(3) \end{gathered}`

We have 3 solutions:

y = 0, y = 4/3, y = -4/3