How do you factor #y=2x^2 - 5x – 3 #?

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Answer 1 · 2016-05-07T01:35:58

#y = (2x+1)(x-3)#

To find the factors, put #y=0#

#=>2x^2-5x-3=0#

This is a quadratic equation and will have 2 factors or 2 roots.

We first find two numbers that:

Let's list the factors of -6:

#color(blue)(1xx-6=-6)#
#2xx-3=-6#
#3xx-2=-6#
#6xx-1=-6#

From the above combinations, #1 + (-6) =1-6= -5#.

Now back to the quadratic equation:

#2x^2-5x-3=0#

Here, we write #-5# as #1-6#.
Then #-5x# will be #x-6x#

#=>2x^2color(red)(+1x-6x)-3=0#

Make 2 pairs:
#=>color(orange)(2x^2+x)color(blue)(-6x-3)=0#

Take out the common terms from each pair:

#=>color(orange)(x(2x+1))color(blue)(-3(2x+1))=0#

#color(red)(2x+1)# is common to the two terms in the equation. Take out the common terms, and write the remaining terms:

#=>color(red)((2x+1))color(green)((x-3))=0#

#y = (2x+1)(x-3)#

We can check our answer by working backwards:

i.e. solve: #y = (2x+1)(x-3)#

#y = 2x(x-3)+1(x-3)#

#y = 2x^2-6x+x-3#

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