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 -2*x*.

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.