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

Answer

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Answer

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

Steps

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

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

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

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