How do you factor $y= 2x^2 - 5x -3$ ?

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Answer 1 · 2015-12-14T23:08:47

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

This quadratic is of the form $ax^2+bx+c$ with $a=2$ , $b=-5$ and $c=-3$ . It has zeros given by the quadratic formula:

$x = (-b+-sqrt(b^2-4ac))/(2a) = (5+-sqrt(5^2-(4xx2xx-3)))/(2xx2)$

$=(5+-sqrt(49))/4 = (5+-7)/4$

That is $x = -1/2$ and $x = 3$

Hence our quadratic has linear factors $(2x+1)$ and $(x-3)$

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

Answer 2 · 2015-12-14T23:13:03

Find a suitable split for the middle term, then factor by grouping to find:

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

Look for a pair of factors of $AC = 2*3 = 6$ whose difference is $B = 5$

The pair $6, 1$ works. Use that to split the middle term then factor by grouping as follows:

$y = 2x^2-5x-3$

$=2x^2-6x+x-3$

$=(2x^2-6x)+(x-3)$

$=2x(x-3)+1(x-3)$

$=(2x+1)(x-3)$

How do you factor y= 2x^2 - 5x -3 y= 2x^2 - 5x -3 ?