Problem

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.

Answer

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Answer

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

Steps

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 $2(4y^2 - 6y - 9)$

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