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.

## The Distributive Property and Like Terms

Now that you know what a factor is, we can use that knowledge to solve more interesting equations than x + 7 = 10. But first, another maths shortcut and a property that is used all the time in algebra.

Remember that multiplication is implied with two sets of brackets next to each other:

(2)(-7) = -14

Well the same thing applies when there are unknowns:

3 × x = 3x, a × b = ab

Also, notice that when you see just x, you can think of that as 1x where the “1” is left off as anything multiplied by “1” is the same anything. Now let’s look at one of the most used properties in maths: the distributive property.

If you had an expression like 3(2 + 4), which means 3 × (2 + 4), you could first add the numbers in the brackets then multiply: 3(2 + 4) = 3(6) = 18. But another way is to first “distribute” the 3 among the terms in the brackets:

3(2 + 4) = 3(2) + 3(4) = 6 + 12 = 18

This holds true for any numbers:

4(6 – 5) = 4(6) – 4(5) = 24 – 20 = 4

-3(5 – 3) = (-3)(5) – (-3)(3) = -15 + 9 = -6 (Remember the sign rules?)

This is called the distributive property. But it’s not that useful to use with known numbers as I would normally just do the math in the brackets first. Now let’s use this property with unknowns:

x(5 – 3) = 5x – 3x. Now you might want to write x5 – x3, but it is customary to write the known number first, then the unknown. You can do this because when you multiply numbers together, it doesn’t matter what order you do the multiplication (this is another property called the commutative property).

Now notice that if you first did the calculation in the brackets first, you would get the answer 2x. Well because of the distributive property, you can add things like 3x and 2x together to get 5x because you can un-distribute the x to get

3x + 2xx(3 + 2) = 5x

This will work for any combination of a number multiplying an unknown. When the unknown is the same letter, or combination of letters, these are called like terms and you can add them together because of the distributive property.

14a – 11a = 3a, because 14a and -11a are like terms, however,

3x + 4a cannot be added together because the unknowns are different.

More examples:

-2y + 3y = 1yy (remember the “1” doesn’t need to be shown)

5m – 2m = 3m

102x + 33x = 135x

20x – 5x + 2y = 15x + 2y (only the like terms can be added)