Last time I introduced the method to factor some quadratic equations so that you can use the Null Factor Law to solve the equation. The example used was

\[{x}^{2}\hspace{0.33em}{-}\hspace{0.33em}{2}{x}\hspace{0.33em}{-}\hspace{0.33em}{35}\hspace{0.33em}{=}\hspace{0.33em}{0}

\]

which when factored became

\[{(}{x}\hspace{0.33em}{-}\hspace{0.33em}{7}{)(}{x}\hspace{0.33em}{+}\hspace{0.33em}{5}{)}\hspace{0.33em}{=}\hspace{0.33em}{0}

\]

where you can readily see the two solutions *x* = 7 and -5. Let’s do another example.

{x}^{2}\hspace{0.33em}{+}\hspace{0.33em}{2}{x}\hspace{0.33em}{-}\hspace{0.33em}{24}\hspace{0.33em}{=}\hspace{0.33em}{0}

\]

So we know the factors will look like \[

{(}{x}\hspace{0.33em}{+}\hspace{0.33em}{a}{)(}{x}\hspace{0.33em}{+}\hspace{0.33em}{b}{)}

\] and we need to find the *a* and* b* so that the factored form is equivalent to the left side of the equation. *a* and* b *must be numbers that multiply to equal 24 and add or subtract to equal +2, the coefficient in front of the *x*. 8 and 3 do not work as they do not add or subtract to equal 2. However, 6 and 4 look like contenders.

So the signs of 6 and 4 must be such that they add to equal +2 but multiply to equal -24. Looks like +6 and -4 work so

\[{x}^{2}\hspace{0.33em}{+}\hspace{0.33em}{2}{x}\hspace{0.33em}{-}{24}\hspace{0.33em}{=}\hspace{0.33em}{(}{x}\hspace{0.33em}{+}\hspace{0.33em}{6}{)(}{x}\hspace{0.33em}{-}\hspace{0.33em}{4}{)}\hspace{0.33em}{=}\hspace{0.33em}{0}

\]

where the two solution can be found by equating each factor to 0 which gives the two solutions *x* = 4 and -6.

Some quadratic equations may not be factorable but there is another method to solve these. I will show this method in my next post.

Now the generic quadratic equation looks like

\[{a}{x}^{2}\hspace{0.33em}{+}\hspace{0.33em}{b}{x}\hspace{0.33em}{+}\hspace{0.33em}{c}\hspace{0.33em}{=}\hspace{0.33em}{0}

\]

where *a*, *b*, and *c* are specific numbers. I have been merciful so far in that I have only looked at equations where *a* = 1. If it is some other value, it’s a little more difficult to factor, but it can be done. However, I will use the quadratic formula to work with these in my next post.