How do you factor $14 k^{2} - 9 k - 18$ $14k^(2)-9k-18$ ?

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How do you factor $14 k^{2} - 9 k - 18$ $14k^(2)-9k-18$ ?

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Answer 1 · 2016-04-01T20:44:45

Use an AC method to find:

$14 k^{2} - 9 k - 18 = (7 k + 6) (2 k - 3)$ $14k^2-9k-18=(7k+6)(2k-3)$

Explanation:

Look for a pair of factors of $A C = 14 \cdot 18 = 252$ $AC = 14*18 = 252$ which differ by $B = 9$ $B=9$ .

Since $252$ $252$ is close to $256 = 16^{2}$ $256 = 16^2$ , look for a pair close to $16 \pm \frac{9}{2}$ $16+-9/2$ . We find the pair $21, 12$ $21, 12$ works:

$21 \times 12 = 252$ $21 xx 12 = 252$

$21 - 12 = 9$ $21 - 12 = 9$

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

$14 k^{2} - 9 k - 18$ $14k^2-9k-18$

$= 14 k^{2} - 21 k + 12 k - 18$ $=14k^2-21k+12k-18$

$= 7 k (2 k - 3) + 6 (2 k - 3)$ $=7k(2k-3)+6(2k-3)$

$= (7 k + 6) (2 k - 3)$ $=(7k+6)(2k-3)$

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How do you factor $14 k^{2} - 9 k - 18$ $14k^(2)-9k-18$ ?

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