How do you factor $12 y^{2} - 4 y - 5$ $12y^(2)-4y-5$ ?

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How do you factor $12 y^{2} - 4 y - 5$ $12y^(2)-4y-5$ ?

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Answer 1 · 2017-02-03T00:17:34

$12 y^{2} - 4 y - 5 = (2 y + 1) (6 y - 5)$ $12y^2-4y-5 = (2y+1)(6y-5)$

Explanation:

Use an AC method:

Given:

$12 y^{2} - 4 y - 5$ $12y^2-4y-5$

Look for a pair of factors of $A C = 12 \cdot 5 = 60$ $AC = 12*5 = 60$ which differ by $B = 4$ $B=4$ .

The pair $10, 6$ $10, 6$ works.

Use this pair to split the middle term and factor by grouping:

$12 y^{2} - 4 y - 5 = (12 y^{2} - 10 y) + (6 y - 5)$ $12y^2-4y-5 = (12y^2-10y)+(6y-5)$

$12 y^{2} - 4 y - 5 = 2 y (6 y - 5) + 1 (6 y - 5)$ $color(white)(12y^2-4y-5) = 2y(6y-5)+1(6y-5)$

$12 y^{2} - 4 y - 5 = (2 y + 1) (6 y - 5)$ $color(white)(12y^2-4y-5) = (2y+1)(6y-5)$

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How do you factor $12 y^{2} - 4 y - 5$ $12y^(2)-4y-5$ ?

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