
A quadratic, or second order function of x is a function of the form y = ax² + bx + c, where a, b, and c are constants, and a is not zero. (If a were zero, this would be a linear, or first order, function.) It's called a "second order" function because the highest power of x is two.
A challenge presented by this equation is for any given constants a, b, and c, to find the values of x that make the equation true.
There can be as many as 2 such values, or "roots", of a 2nd order polynomial. The best way to find the roots is to "factor" the polynomial. That is, find lowerorder polynomial expressions that can be multiplied together to result in the original polynomial.
For example, consider x² + 4x  21 = 0. This equation is the same as (x3) (x+7) = 0. Multiply it out, and you'll see: x*x + x*7  3*x  21. Collect the terms x*7 and 3*x, which add up to 4x, and you'll see it multiplies out to x² + 4x  21.
Why is the factored form useful? Here's why: do you remember that zero times anything is zero? So (x3) (x+7) = 0 is true when either x3 is zero or x+7 is zero. In other words, the roots are 3 and 7. Make sure you see that before you continue reading.
Now you appreciate why it is handy to know how to factor a polynomial. But you still don't know how to do it. Hopefully, you are now motivated to learn how.
Take another look at our example, x² + 4x  21. The "21" comes from the product of 3 and 7. The "4" comes from the sum of 3 and 7. So to factor this polynomial, you need to look for a pair of numbers whose sum is 4 and whose product is 21. The numbers 3 and 7 just "jump out at you" when you think of it that way, right? OK, but what is a general method of finding the right pair of numbers if you know their sum and their product? (Make sure you understand why it's important to be able to do this before you read on!)
Now you know why it's important to be able to find a pair of numbers knowing only their product and their sum. Before we look at the general case, let's look at a few special cases:
What about the case where the sum is zero. For example, what two numbers add up to zero, and have a product of 16? Answer: 4 and 4. Do a few more examples like this, and you'll see how to solve this special case: a and b are the positive and negative square roots of the negative of the product. This helps us understand easily how to solve the special kind of 2nd order polynomial where the "x" coefficient is zero. For example, x²  16 is (x4) (x+4). This is an important rule, so be sure you understand it  it'll come up again in this lesson, and it will come up over and over in mathematics. The rule is this:
a²  b² is equal to (ab) (a+b).
Try some examples so you're certain you "get" it…
If you're a "visual thinker" then imagine this: a square array of dots with a square "bite" taken out of the upper right corner. Let's say the original square array of dots had 100 dots in it, and the "bite" was a small 4by4 square. 100  16 = 84 dots remain after the bite was taken. Now change your perspective. Instead of thinking of these 84 dots as a big square minus a small bite, think of it as a square and two rectangles added together. In your mind, "draw" a square around the largest number of dots you can starting from the lowerleft corner. In this case, this will be a 6by6 square  any bigger, and you'll be pushing into the area already bitten off. Now what are the two rectangles left over? There's a 6by4 rectangle on the right side of the 6by6 square, and another 6by4 rectangle on the top. Take one of those rectangles, and stack it next to the other one so the dots are now arranged into one big rectangle. How big is it? 6by14. You just showed geometrically that 10x10  4x4 equals 6x14. In other words, (104) (10+4).
I call 10 the "anchor" and 4 the "swing" because I picture swinging down from the anchor to get 6 and then swinging up to get 14. So I say this: the square of the "anchor" minus the square of the "swing" equals "swing down" times "swing up".
Are you sure you can factor the difference of two squares? Try some more examples now. Then read on.
Now another special case: the sum is two, and the product is the negative of one less than a perfect square. For example, what two numbers add up to 2, and have a product of 63? Answer: 9 and 7. Do you see how the "difference of squares" rule applies here? The anchor is 8, and the swing is 1. (8+1) (81) is equal to 64  1.
Are you ready for a more general case? Lets figure out the two numbers that add up to SUM and multiply out to PRODUCT. Using the Anchor and Swing approach, one of these two numbers is A+S, and the other is AS. You can tell the Anchor, A, right away: it's the average of A+S and AS, or half the sum of the two numbers. The Product is the difference between the square of A and the square of S  do you remember why? Because A²  S² equals (AS) (A+S). Since you know the Anchor, you can square it, and subtract the Product to get the square of the Swing.
Try some values for Sum and Product, below, and see how the numbers A+S and AS (that is, the numbers that have the given Sum and Product) are calculated:
As you can see, 7+3 is 10 and 7*3 is 21. You can change any one of the values in the table, above, to recalculate all the other values. Try a bunch of examples to convince yourself the formula works.
What would happen if the Swing were larger than the Anchor? That means the Anchor is so close to zero that when you "Swing low" you're in negative territory, and when you "Swing high" you're in positive territory. Well, for one thing, the Product will be negative, because the Product of a negative number and a positive number is a negative number. Amazingly, the formula still works. Try a Sum of 4, and a Product of 21, but first  Can you figure it out in your head? OK, now try it.
Putting it together:
x² + 4x  21 can be factored as
(x + 7) (x  3) because the Sum of 7 and 3 is 4, and their Product is 21.
Now we have almost the complete picture; there's just one little wrinkle that needs to be discussed: What if the x² coefficient is not one. For example, what about:
15x²  26x + 8 = 0
If you think about it, you'll see this is the same as
x²  (26/15)x + (8/15) = 0
because you can divide both sides by 15. Now I'll express this same equation a little differently  hang in there, because you'll see why I do this a little later!
x²  (26/15)x + (120/15²) = 0
Now we're looking for two numbers that add up to 26/15 and multiply out to 120/15². Let's suppose the two numbers are a/15 and b/15. Now we can see the Sum of a/15 and b/15 is 26/15, so the Sum of a and b is 26. The Product of a/15 and b/15 is 120/15², so the Product of a and b is 120. Now the problem boils down to finding two numbers, a and b, whose
Sum is 26
Product is 120
Use the calculator, above, or your own pencil and paper to figure out that the two numbers are 6 and 20. So a/15 is 6/15, and b/15 is 20/15, which are the two roots of the equation.
Putting it together:
15x²  26x + 8 = 0 can be divided by 15 to give
x²  (26/15)x + 120/15² = 0 and factored as
(x  6/15) (x  20/15) = 0
because the Sum of 6 and 20 is 26, and their Product is 8*15, or 120.
Simplifying further
(x  2/5) (x  4/3) = 0
This is already a perfectly good answer, but if you like, you can multiply both sides by 15 by multiplying the first factor by 5 and the second by 3 to get:
(5x  2) (3x  4) = 0
Which is the most elegant solution, even though we got here by a roundabout way!
Now we're finished exploring how to find the original numbers if you only know their sum and their product. If the x² coefficient is not one, then we know to find numbers whose sum is the xcoefficient and whose product is the x² coefficient multiplied by the constant.
For "extra credit": did you notice that a small variation on this formula can be used if you know only the Difference and Product of two numbers? For example, suppose ab is 4, and a*b is 21. With a little wizardry, you can convert this to a Sum and Product by thinking of "a" and "b" as your two numbers. The Sum of a and b is 4, and the Product of a and b is 21. Plug them into the formula to see that the two numbers are 7 and 3. In other words, a is 7 and b is 3, so b must be 4. (Or a is 3 and b is 7, so b is 7; both answers are correct.)
Factor the following quadratic equations:
You can check your answers by multiplying out the factors you found. No need to ask me if you're right!
Do you have any questions about this procedure? Send me an email.
The F.O.I.L. method of checking your answer
The webmaster and author of this Math Help site is Graeme McRae.