How do you find the zeroes of #y= ( x - 2) ( 3x + 5)#?

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Answer 1 · 2018-05-31T12:52:53

See a solution process below:

To find the zeros for this equation, equate each term on the right to #0# and solve for #x#:

Solution 1:

#x - 2 = 0#

#x - 2 + color(red)(2) = 0 + color(red)(2)#

#x - 0 = 2#

#x = 2#

Solution 2:

#3x + 5 = 0#

#3x + 5 - color(red)(5) = 0 - color(red)(5)#

#3x + 0 = -5#

#3x = -5#

#(3x)/color(red)(3) = -5/color(red)(3)#

#(color(red)(cancel(color(black)(3)))x)/cancel(color(red)(3)) = -5/3#

#x = -5/3#

The Zeros for this problem are:

#x = {-5/3, 2}#