Is it possible to factor y=2x2+7x+3? If so, what are the factors?

1 Answer
Lotusbluete
Dec 3, 2015

Yes, it is possible.

y=(2x2+7x+3)=(x+12)(x+3)

Explanation:

To find a complete factorisation of 2x2+7x+3, you need to find all solutions of the equation

2x2+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...

⇔2x2+7x=−3

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

⇔x2+72x=−32

Now, the goal is to create a quadratic expression like x2+2ax+a2 on the left side of the equation.

We already have x2 and 72x=2ax which means that a=74. So, to complete the circle, we need to add (74)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.

⇔x2+72x+(74)2=−32+(74)2

As next, due to the rule x2+2ax+a2=(x+a)2, you can transform the left side.

⇔(x+74)2=−32+(74)2

⇔(x+74)2=2516

Now, you can draw the root on both sides.
Be careful though: you will have two solutions since both (54)2=2516 and (−54)2=2516 hold.

x+74=54× or ×x+74=−54

x=−12×ξii or ×x=−3

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With your solutions x=−12 and x=−3 you have:

×x2x2+7x+3=0

⇔(x−(−12))(x−(−3))=0

⇔(x+12)(x+3)=0

Thus,

2x2+7x+3=(x+12)(x+3).

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