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:

$\begin{gathered} y=0 \\ \\ 3y-4=0\therefore y=\frac{4}{3} \\ \\ 3y+4=0\therefore y=-\frac{4}{3} \end{gathered}$

We have 3 solutions:

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