How do you factor completely $20y^2 + 3y - 2$ ?

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Answer 1 · 2017-01-09T00:47:34

$20y^2+3y-2 = (4y-1)(5y+2)$

Given:

$20y^2+3y-2$

Use an AC method:

Find a pair of factors of $AC=20*2 = 40$ which differ by $B=3$

The pair $8, 5$ works.

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

$20y^2+3y-2 = (20y^2+8y)-(5y+2)$

$color(white)(20y^2+3y-2) = 4y(5y+2)-1(5y+2)$

$color(white)(20y^2+3y-2) = (4y-1)(5y+2)$

How do you factor completely 20y^2 + 3y - 220y^2 + 3y - 2?