Quadratic Factoring, Part 2

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 ab, 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.

Quadratic Factoring

In my last post, I showed how to solve

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

Now this is really a quadratic equation in disguise. When I covered the Distributive Property, you saw how to distribute a factor within a set of brackets:

\[
{x}{(}{x}\hspace{0.33em}{+}\hspace{0.33em}{5}{)}\hspace{0.33em}{=}\hspace{0.33em}{x}\hspace{0.33em}\times\hspace{0.33em}{x}\hspace{0.33em}{+}\hspace{0.33em}{5}\hspace{0.33em}\times\hspace{0.33em}{x}\hspace{0.33em}{=}\hspace{0.33em}{x}^{2}\hspace{0.33em}{+}\hspace{0.33em}{5}{x}
\]

This can also be done in reverse by un-distributing the x to get the expression back into factored form so that you can take advantage of the Null Factor Law. What I didn’t cover, was how to take an expression like \[
{(}{x}\hspace{0.33em}{-}\hspace{0.33em}{7}{)(}{x}\hspace{0.33em}{+}\hspace{0.33em}{5}{)}
\] and un-factor it. To do this, you can distribute each term in the first set of brackets with each term in the second set:

\[
\begin{array}{l}
{{(}{x}\hspace{0.33em}{-}\hspace{0.33em}{7}{)}{(}{x}\hspace{0.33em}{+}\hspace{0.33em}{5}{)}\hspace{0.33em}{=}\hspace{0.33em}{x}{(}{x}{+}{5}{)}\hspace{0.33em}{-}\hspace{0.33em}{7}{(}{x}\hspace{0.33em}{+}\hspace{0.33em}{5}{)}\hspace{0.33em}{=}\hspace{0.33em}{x}^{2}\hspace{0.33em}{+}\hspace{0.33em}{5}{x}\hspace{0.33em}{-}{7}{x}\hspace{0.33em}{-}\hspace{0.33em}{35}}\\
{{=}\hspace{0.33em}{x}^{2}\hspace{0.33em}{-}\hspace{0.33em}{2}{x}\hspace{0.33em}{-}\hspace{0.33em}{35}}
\end{array}
\]

So this is now recognisable as a quadratic expression. However, if I originally had the equation

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

we could not use the Null Factor Law. So what can you do if given this equation? There is a way to solve this using something called the Quadratic Formula, but that will be covered later. Here I will show how to factor this equation back to the original form we started with so that we can use the Null Factor Law.

So the goal is to take \[
{x}^{2}\hspace{0.33em}{-}\hspace{0.33em}{2}{x}\hspace{0.33em}{-}\hspace{0.33em}{35}
\] and get it in the form

(something)(something else). If you look at how we unfactorised this, you can see that I can start with

\[
{(}{x}\hspace{0.33em}{+}\hspace{0.33em}{a}{)(}{x}\hspace{0.33em}{+}\hspace{0.33em}{b}{)}
\], where we need to find the a and the b so that the expressions are equivalent. I know this is the way to start as the two x‘s are needed to get \[
{x}^{2}
\] when they are multiplied together. Now to find the a and the b, including the sign of each, you need to look at all the possible factors of the known number in the quadratic, in this case -35, that add up to the coefficient of the middle term, in this case -2. Again, this is suggested if you look at how we expanded \[
{(}{x}\hspace{0.33em}{-}\hspace{0.33em}{7}{)}{(}{x}\hspace{0.33em}{+}\hspace{0.33em}{5}{)}
\]. The last number in the expansion (-35) is generated by multiplying the -7 and the +5. These two numbers also multiply the x‘s which are eventually added together to get the middle term, in this case -2x.

So to find the a and the b in \[
{(}{x}\hspace{0.33em}{+}\hspace{0.33em}{a}{)(}{x}\hspace{0.33em}{+}\hspace{0.33em}{b}{)}
\], let’s look at the possible factors of 35 (ignoring the sign for the moment): 35 and 1 or 7 and 5. 35 and 1 do indeed multiply to equal 35 but their addition or subtraction together do not equal 2. That leaves 7 and 5 which do satisfy both requirements: 7 × 5 = 35, 7 – 5 = 2. Now all that remains is to determine the signs. Since their multiplication has to equal -35, one of the numbers needs to be negative. And since the coefficient of the middle term is negative, the large number 7 needs to be negative as well. So in this case a = -7 and b = + 5 so that the factorisation we are looking for is \[
{(}{x}\hspace{0.33em}{+}\hspace{0.33em}{a}{)(}{x}\hspace{0.33em}{+}\hspace{0.33em}{b}{)}
\]. Now we can use the Null Factor Law to solve the equation, as we had done previously.

Not all quadratics can be factored like this, but this is a good skill to develop with practise. I will do several more examples in my next post.

Null Factor Law

I’d like to return to algebra and discuss the Null Factor Law which is useful in solving certain equations:

If two or more factors multiplied together equal zero, then the solutions can be found be equating each factor separately to zero.

This makes sense if you think of two numbers multiplied together equal 0:

\[
{ab}\hspace{0.33em}{=}\hspace{0.33em}{0}
\] can only be true if either a is zero, b is zero, or both are zero. No other non-zero numbers multiplied together can equal zero.

This is true for any algebraic expressions multiplied together. For example:

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

Can only be true if \[
{(}{x}{-}{7}{)\hspace{0.33em}=}\hspace{0.33em}{0}
\] or if \[
{(}{x}{+}{5}{)\hspace{0.33em}=}\hspace{0.33em}{0}
\]

Without formal algebra, you can see the two solutions to this equation are then x = 7 or -5.

Now this one was easy, but sometimes you are given an equation that is not a factored one:

\[
{x}^{2}\hspace{0.33em}{+}\hspace{0.33em}{5}{x}\hspace{0.33em}{=}\hspace{0.33em}{0}
\]

At first glance, it looks like the null factor law doesn’t apply here. But I did a post on the distributive property. Please review that if needed, but notice that there is a common factor of x in each of the terms on the left side of the equation. I can un-distribute this x to get:

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

Looks like the null factor law can be used as there are now two factors on the left side. So mentally setting each of these to zero, we get the two solutions x = 0 or -5.

I covered a similar example on my post about quadratic equations. You can click on the tags on the right or below (depending on the device you are viewing this on) to directly go to previous posts on the listed topics.

I will be covering more complex examples in my next post.