13 Assignment rematics 10C TT 2. (2 marks) 3. Brendan factored the trinomial $8 y^{2}-12 y-18$ in this way: \[ \begin{aligned} 8 y^{2}-12 y-18 & =2\left(4 y^{2}-6 y-9\right) \\ & =2(2 y-3)(2 y-3) \\ & =2(2 y-3)^{2} \end{aligned} \] Is his factoring correct? Explain.

Step 1 ：First, we check Brendan's factoring: $8y^2 - 12y - 18 = 2(4y^2 - 6y - 9)$

Step 2 ：Now, we try to factor the quadratic inside the parentheses: $4y^2 - 6y - 9$

Step 3 ：We notice that the quadratic is not a perfect square, so we cannot factor it as $(2y - 3)^2$

Step 4 ：Therefore, Brendan's factoring is incorrect, and the correct factoring should be $2(4y^2 - 6y - 9)$