Is it possible to factor #y=2x^2+7x+3 #? If so, what are the factors?

1 Answer
Dec 3, 2015

Answer:

Yes, it is possible.

#y = (2x^2 + 7x + 3) = (x + 1/2)(x + 3)#

Explanation:

To find a complete factorisation of #2x^2 + 7x + 3#, you need to find all solutions of the equation

#2x^2 + 7x + 3 = 0#.

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This can be done e.g. with "completion of the circle".
Let me show you how it's done!

... compute #-3# on both sides of the equation...

#<=> 2x^2 + 7x = -3#

... divide by #2# on both sides of the equation ...

#<=> x^2 + 7/2 x = - 3/2#

Now, the goal is to create a quadratic expression like #x^2 + 2ax + a^2# on the left side of the equation.

We already have #x^2# and #7/2 x = 2ax# which means that #a = 7/4#. So, to complete the circle, we need to add #(7/4)^2# to the expression on the left side.
However, as we don't want to destroy the equality, it is also needed to add the same expression on the right side, too.

#<=> x^2 + 7/2 x + (7/4)^2 = - 3/2 + (7/4)^2#

As next, due to the rule #x^2 + 2ax + a^2 = (x+a)^2#, you can transform the left side.

#<=> (x + 7/4)^2 = - 3/2 + (7/4)^2#

#<=> (x + 7/4)^2 = 25/16#

Now, you can draw the root on both sides.
Be careful though: you will have two solutions since both #(5/4)^2 = 25/16# and #(-5/4)^2 = 25/16# hold.

#x + 7/4 = 5/4 color(white)(xx)" or " color(white)(xx) x + 7/4 = - 5 / 4#

#x = - 1/2 color(white)(xxxiii)" or " color(white)(xx) x = -3#

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With your solutions #x = -1/2# and #x = -3# you have:

#color(white)(xxx) 2x^2 + 7x + 3 = 0#

#<=> (x- (-1/2))(x-(-3)) = 0#

#<=> (x + 1/2)(x+3) = 0#

Thus,

#2x^2 + 7x + 3 = (x + 1/2)(x+3)#.