Solve Quadratic Trinomials w/Large Coeffs: Factoring Tips

[daigo]

I'm told that factoring is an important skill in calculus so I am avoiding using the quadratic formula. But for quadratic equations with large coefficients to factor, is there a better/faster way rather than guessing and checking every single combination?

 

[Char. Limit]

I personally prefer completing the square. It's rather more useful in calculus than even factoring, as it gives you the point where the derivative is zero without even ever having to take it.

EDIT:

Completing the square is also rather simple. Start with an equation a x^2 + b x + c = 0, right? Divide both sides by a so that you get this:

x^2 + \frac{b}{a} x + \frac{c}{a} = 0

Then add and subtract b^2 / 4 a^2 to the left side (effectively adding zero)

x^2 + \frac{b}{a} x + \frac{b^2}{4 a^2} + \frac{c}{a} - \frac{b^2}{4 a^2} = 0

Now, the first three terms in that are a perfect square, and the last two are a constant, so we can easily rearrange this to give...

\left( x + \frac{b}{2a}\right)^2 = \frac{b^2}{4 a^2} - \frac{c}{a}

And there we have it. This gives us the two roots with just a little algebraic manipulation (equivalent to the quadratic formula, as it happens) and also immediately gives us the vertex of the equation (hint, take the derivative of both sides).

 

[Number Nine]

Yes...the quadratic formula.
How do you intend to find irrational roots without using the quadratic formula?

 

[Char. Limit]

Yes...the quadratic formula.
How do you intend to find irrational roots without using the quadratic formula?

If you take a look at my post above, it can be done quite easily.

 

[Number Nine]

If you take a look at my post above, it can be done quite easily.

True, but at that point he's effectively utilizing the quadratic formula anyway.

 

[Char. Limit]

True, but at that point he's effectively utilizing the quadratic formula anyway.

Also true. I like completing the square better than a straight application, though, as it gives you more insight into why the quadratic formula works.

 

[mathman]

Also true. I like completing the square better than a straight application, though, as it gives you more insight into why the quadratic formula works.

After doing it a few times your way the concept should sink in. After that using the quadratic formula is faster.