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i need help with factoring quadratic equations

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2 Answers

Here is a general rule. In a general quadratic ax2+bx+c, if you can write b=m+n where m and n are integers such that a/m = n/c (equivalently, mn=ac), then you can factor easily by first rewriting the quadratic as ax2+mx+nx+c. This is because ax2+mx and nx+c will have a common factor. If such integers m and n don't exist, then the quadratic can't be factored.

For example, take 6x2 - 7x - 3.

We want m+n = -7 and 6/m = n/(-3).

We can choose m = 2 and n = -9. 

This gives us:

6x2 - 7x - 3 = 6x2 + 2x - 9x - 3 = 2x(3x + 1) - 3(3x + 1) = (2x - 3)(3x + 1).

Or equivalently, with m = -9 and n = 2:

6x2 - 7x - 3 = 6x2 - 9x + 2x - 3 = 3x(2x - 3) + 1(2x - 3) = (3x + 1)(2x - 3).

I think sometime it's best to work an example in reverse.  In a sense work backwards to get the idea.  If one were asked to multiply out the following,

(x-5)(x+2) [ and reach x2 - 3x -10],

the commonly used technique is called FOIL, for First Outer Inner Last.

We being by multiplying x by x, the First terms in each linear factor, giving x2.

Then multiply x by 2, the Outer terms, giving 2x.

Next mulitply -5 by x, the Inner terms, giving -5x.

Finally, muliply -5 by 2, the Last terms, giving -10.

Adding the results gives

x2 + 2x - 5x - 10 = x2 - 3x - 10.

Factoring quadratics is this same process, but in reverse.  We begin with a quadratic like,

x2 - 4x - 12,

and we seek to un-FOIL it, so to speak.  We know the result should resemble,

(x + A)(x + B),

but the task can sometimes be difficult for a while until one gets the hang of it.  But remembering that this FOILs to give x2 + Ax + Bx + A*B = x2 + (A + B)x + A*B we see a way to attack it.

In this example one seeks A and B to multiply to give -12, but add to give -4.  Through trial and error, one eventually finds A = 2 and B = -6 to satisfy these requirements.  Thus we find that our example factors as,

(x + 2)(x - 6).

How does this help when solving a quadratic equation?  Well let's see it in action.  Say we have the task of finding solutions to the following,

x2 + 6x - 40 = 0.

Here we factor the LHS using the same technique as above.

We eventually find (x - 4)(x + 10). We can check this by multiplying it out, i.e. FOIL.  After checking we continue rewriting the equation as

(x - 4)(x + 10) = 0.  Now our task is almost complete because to find our solutions we need only set each factor equal to 0, and solve.

x - 4 = 0 gives x = 4, and x + 10 = 0 gives x = -10.  Thus our two solutions, x = 4, -10.

Remember this tehnique requires a little care, making sure that the RHS is 0, before factoring.

This method can also be extended to more comlicated examples involving quadratics with a lead coeffectient (x2 coefficient) other than 1, like the examples above.  Those aren't too difficult once one has mastered the technique for simple examples like the ones worked in this post.

I hope this helps.  I may work some other examples soon, if you or others want to see more.

In the mean time, keep practicing.  This is an extremely necessary skill to have under one's belt before continuing on to more advanced math.  All it takes to master is lots of practice and patience.