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)#