# How to Factor Quadratics of the Form

## Presentation on theme: "How to Factor Quadratics of the Form"— Presentation transcript:

How to Factor Quadratics of the Form
ax2 + bx + c

The first rule of thumb in factoring is to factor out, if any, the greatest common factor of all the terms, for example: 1. 2x2 + 6x – 8 = 2(x2 + 3x – 4) 2. 5x3 – 20x2 + 15x = 5x(x2 – 4x + 3) On the other hand, what if the expression has no common factors? For example, x2 + 4x + 2 and x2 + 4x + 3. Keep in mind that not all quadratic expression can be factored. Our first example x2 + 4x + 2 is not factorable (in this case, the expression is called a prime). The second example, however, x2 + 4x + 3 can be factored as (x + 1)(x + 3).

There are a couple of ways we can determine whether a quadratic expression is factorable or not. We will show you some of those ways later. But first, we are going to show you how to factor any quadratic expressions if they are indeed factorable. We will divide these quadratic expressions into two classes: 1. x2 + bx + c, that is, a = 1, and 2. ax2 + bx + c, where a  1

Let’s look at some of the examples in the first class and see how to factor them:
1. x2 + 5x x2 – 7x + 10 3. x2 + 2x – x2 – 6x – 16 If they are factorable, each of them must be factored as (x )(x ) since that is the only way x2 can be factored. How about the two numbers that should be also in the parentheses—how do we get them? Those two numbers can be obtained by asking, “What two numbers multiplied will give me the last term?” Let’s repeat this again—what two numbers multiply, not add. There are two pairs of numbers that multiply to be 6, {1, 6} and {2, 3}, and for 10, {1, 10} and {2, 5}, and for 15, {1, 15} and {3, 5}; whereas 16 has three pairs: {1, 16}, {2, 8}, and {4, 4}. Which pair should we use?

1. x2 + 5x x2 – 7x + 10 3. x2 + 2x – x2 – 6x – 16 Actually, it is quite easy to choose the right pair of numbers that multiply to be the last term. This is how: If the last term is positive, like in the first two examples, ask what two numbers multiplied is the last term and their sum is the coefficient of the middle term; If the last term is negative, like in the last two examples, ask what two numbers multiplied is the last term and their difference is the coefficient of the middle term. Here, you can ignore the sign of the coefficient of the middle term. That is, just consider it is positive even if it is negative.

1. x2 + 5x x2 – 7x + 10 3. x2 + 2x – x2 – 6x – 16 If you understood what we said on the previous page, you should have picked {2, 3}, {2, 5}, {3, 5}, and {2, 8} for our four examples, respectively. That is, you should have written something like the following: 1. x2 + 5x + 6 = (x 2)(x 3) 2. x2 – 7x + 10 = (x 2)(x 5) 3. x2 + 2x – 15 = (x 3)(x 5) 4. x2 – 6x – 16 = (x 2)(x 8)

1. x2 + 5x + 6 = (x 2)(x 3) 2. x2 – 7x + 10 = (x 2)(x 5)
How should we determine the sign between the x and the number? If the last term is positive, then both signs should be the same and they follow the sign of the coefficient of the middle term, i.e., if that coefficient is: a) positive, then both signs should be “+,” the plus sign; b) negative, then both signs should be “–,” the minus sign. If the last term is negative and then one sign should be “+” and the other is “–” and if the coefficient of the middle term is: a) positive, then put “+” with the larger number and “–” with the smaller number; b) negative, then put “–” with the larger number and “+” with the smaller number.

If you followed what we said on the previous page, you should have factored the quadratic expressions as follows: 1. x2 + 5x + 6 = (x + 2)(x + 3) 2. x2 – 7x + 10 = (x – 2)(x – 5) 3. x2 + 2x – 15 = (x – 3)(x + 5) 4. x2 – 6x – 16 = (x + 2)(x – 8) If you think the rules on the previous two pages are more than you can handle at this time, no problem—we have an easier way for you to factor quadratic expressions like the ones above. All you need to do is ask: What are the two numbers that if I multiply, they will give me the last term, and if I add or subtract, they will give me the coefficient of the middle term. Again, ignore the signs of the the last term and the coefficient of the middle term (we always can take care of the signs later).

Let’s look at the following example:
x2 + 2x – 24 Recall the last term, 24, is obtained by multiplying. The only two numbers that can add or subtract to be 2 in the middle term are 4 and 6. Therefore, we can forget about the other pairs that also multiply to 24, namely, {1, 24}, {2, 12}, and {3, 8}. You will have at least this: x2 + 2x – 24 = (x 4)(x 6) What about the signs? Since we obtain 24 by multiplying and its sign is minus, so one sign must be “+” (plus) and the other must be “–” (minus). So, just put the signs in. x2 + 2x – 24 = (x – 4)(x + 6) Are you puzzled? What happen if we put the “–” with the 6 and the “+” with the 4—will that work too?

The answer is no, it will not work
The answer is no, it will not work. That’s why we should always check whether we have the signs (if they are different) in the right place or not. To see whether x2 + 2x – 24 is really equal to (x – 4)(x + 6), some text books would suggest to FOIL (x – 4)(x + 6) (i.e., to multiply it out) and see if it is equal to x2 + 2x – 24. (x – 4)(x + 6) = x2 + 6x – 4x – 24 = x2 + 2x – 24 Therefore, it’s the correct factoring, but a waste of time. We obviously know the “F” will give us the first term, x2, and “L” will give us the last term, –24. Therefore, we only need to check “I” (the inner terms) and “O” (the outer terms). F O I L

(x – 4)(x + 6) Combine the inner and outer products by adding, the result is +2x, which is exactly the middle term of x2 + 2x – 24. Therefore, the factoring must be correct. If, unfortunately, you have factored x2 + 2x – 24 as (x + 4)(x – 6), you can still check it by the method mentioned above. (x + 4)(x – 6) Again, combine the inner and outer products, the result is –2x. Although it’s not the +2x we want to have, it is exactly the opposite. In this case, just switch the two signs, i.e., change (x + 4)(x – 6) to (x – 4)(x + 6). It will be correct, with 100% guarantee, you don’t have to check it again. –4x +6x +4x –6x

No! Yes! Let’s look at the following example: x2 + 13x + 30
In this case, we seem to have two pairs of numbers {2, 15} and {3, 10} which will work, since each pair has a product of 30 and while the difference of first pair is 13, the sum of the second pair is also 13. So which pair? Why not write both out and see which is the correct pair? Again, don’t write the signs yet. (x 2)(x 15) (x 3)(x 10) We see 30 has a “+” sign. In order to have the “+” sign by multiplying, either both signs have to be “+” or both be “–.” Since the 13 also has a “+” sign, the signs must both be “+.” Put in the “+” signs, we can see only (x + 3)(x +10) is correct: (x + 2)(x + 15) (x + 3)(x + 10) +2x +15x +17x +3x +10x +13x No! Yes!

If you have been following what we have said so far, it is quite easy to determine whether a quadratic expression is factorable or not. For example, the following quadratic expressions are not factorable: 1. x2 + 6x – 24 2. x2 + 11x – 24 3. x2 + 5x + 24 The first one is not factorable since 6 (the coefficient of the middle term) is neither the sum nor the difference of any desirable pairs of factors of 24. The second one is also not factorable since the last term is –24; we want the difference of the two factors of 24 to be 11, but 11 is not one of such differences. The last one is also not factorable since the last term is +24; we want the sum of the two factors of 24 to be 5, but 5 is not one of such sums. 24 Sum Difference

We have one more trick which we call the “closer factors” trick
We have one more trick which we call the “closer factors” trick. When the last term has more than one desirable pair of factors, usually it is the pair of factors that are close to each other or the pair(s) that are in the times table. For example, if the last term is 42, it is likely the pair (which gives a product of 42) is {6, 7}, more so than {3, 14}, {2, 21} and {1, 42}. If the last term is 24, it is most likely a toss-up between {4, 6} and {3, 8}, rather than a toss-up between {2, 12} and {1, 24}. Although this trick will not be always the case, it is very probable. It’s even more probable for quadratic expressions of the form ax2 + bx + c where a  1. We are going to discuss this type of quadratic expressions next.

Now, let’s look at the following example: 4x2 + 4x – 15.
In order to obtain the first term, 4x2 (by multi- plying), we must use either 2x and 2x or x and 4x. To get the last term, 15 (also by multiplying), we must use either 3 and 5 or 1 and 15. This can be quite a hassle, because it could be one of the six factor forms shown on the right. Which one? We can’t really tell you which one is right, but we can tell you that you should always try the one with the factors that are closer to each other first (recall the “closer factors” trick we have mentioned earlier). Therefore, try (2x 3)(2x 5) first. Check inner and outer products again and see if it really works. The inner product gives 6x while the outer gives 10x. Since the15 of the last term has a “–” sign, so the signs must be one “+” and the other “–”. Therefore, if we place the “+” sign with the 5 and “–” with the 3, we will have +10x and –6x which will combine to +4x. So, 4x2 + 4x – 15 = (2x – 3)(2x + 5). 4x2 + 4x – 15 (2x 3)(2x 5) (2x 1)(2x 15) (x 3)(4x 5) (x 5)(4x 3) (x 1)(4x 15) (x 15)(4x 1) Which one? –6x +10x +4x (2x – 3)(2x + 5)

Let’s look at this example: 6x2 + 23x + 20.
If you are following the “closer factors” trick, you should have something like (2x 5)(3x 4) or (2x 4)(3x 5). Let’s look at the second choice (2x 4)(3x 5), notice that you can factor out a 2 from (2x 4). However, the original expression 6x2 + 23x has no common factor. Therefore, it can’t be (2x 4)(3x 5). We call this the “no common factor” trick. That is, make sure that the numbers appearing in the same parenthesis have no common factor if the given expression has no common factor. Let’s go back to (2x 5)(3x 4) and see if it really works. We see that the inner product gives 15x and the outer gives 8x and we see that the 20 of the last term has a “+” sign and the 23 of the middle term also has a “+” sign, so the signs must be both “+”. So, 6x2 + 23x + 20 = (2x + 5)(3x + 4). 6x2 + 23x + 20 = (2x 4)(3x 5) 2 can be factored out from here Nothing can be factored out from here ?  6x2 + 23x + 20 can’t be factored as (2x 4)(3x 5) Contradiction! +15x +8x +23x (2x + 5)(3x + 4)

6x2 + 7x – 24 (2x 4)(3x 6) (2x 6)(3x 4) (2x 8)(3x 3) (2x 3)(3x 8)
Here is another example: 6x2 + 7x – 24. Using the “closer factors” tactic, you would have (2x )(3x ) for the 6x2, and the closest pair that gives a product of 24 is 4 and 6. In which parenthesis should you put the 4 and the 6? The answer is neither—don’t put either number in. It is be- cause it breaks the “no common factor” rule. The next closest pair that gives a product of 24 is 3 and 8. Again, in which paren- thesis should you put the 3 and the 8? The answer is to put the 3 with 2x and the 8 with 3x, i.e., (2x 3)(3x 8). The inner product of (2x 3)(3x 8) gives 9x while the outer gives 16x. And it’s only possible to obtain the +7x (the mid- dle term) if we assign the “+” sign with 8 and the “–” with 3 (see illustration on the right). Therefore, 6x2 + 7x – 24 = (2x – 3)(3x + 8). 6x2 + 7x – 24 (2x 4)(3x 6) (2x 6)(3x 4) (2x 8)(3x 3) (2x 3)(3x 8) 2 can be factored out from (2x 4) and 3 can be factored out from (3x 6) Can’t be this one! Nothing can be factored out from here 2 can be factored out from (2x 6) Can’t be this one either! 2 can be factored out from (2x 8) and 3 can be factored out from (3x 3) This one is no good either! Nothing can be factored out from here also Hey, maybe it’s this one! –9x +16x +7x (2x – 3)(3x + 8)

By using the “closer factors” and “no common factor” tricks wisely (especially the latter one), it’s quite easy to determine whether the quadratic expression is factorable or not. For example, it’s not difficult to see that 6x2 + 11x is not factorable by using these two tactics. If you have (3x )(2x ), notice that you can’t put in any pairs of factors of 48 into (3x )(2x ), because one way or another they all break the “no common factor” rule. Our last resort is (6x )(x ). We can see that every pair break the rule except the last pair 1 and 48. And we must put 1 with the 6x and 48 with x and see if it works. When we multiply the inner terms and the outer terms, however, we will have 288x and 1x which can never become +11x of the middle term. So, 6x2 + 11x + 48 is not factorable. “No Common Factor” Rule: If an quadratic expression has no common factor in the first place, no matter how you factor it, these two numbers can’t have a common factor, so can’t these two. (x )(x ) 48 6 8 4 12 3 16 2 24 1 48 1x 288x (6x 1)(x 48)

Therefore, 6x2 + 11x + 48 is not factorable.
We can also determine whether a quadratic expression of the form ax2 + bx + c is factorable or not by checking whether b2 – 4ac (1) is a perfect square or not. If the result is a perfect square (2), then the quadratic expression is factorable; otherwise, it is not. Our example 6x2 + 11x + 48 has a = 6, b = 11 and c = 48. In this case, however, b2 – 4ac = 112 – 4(6)(48) is negative—it can’t be a perfect square. Therefore, 6x2 + 11x + 48 is not factorable. Notes: 1. b2 – 4ac is called the discriminant of ax2 + bx + c. 2. A perfect square (including 0) is a number that is the square of some integer. Hence, the perfect squares are 0, 1, 4, 9, 16, 25, etc.

Solving most quadratic equations, Solving rational equations,
Many areas in algebra requires the knowledge of factoring quadratics. These include: Solving most quadratic equations, Solving rational equations, Simplifying rational expressions, and last but not least, Solving trigonometric, logarithmic, exponential and higher-degree equations which are really quadratic equations in disguise. Quadratic Equations Rational Equations x2 + 5x – 6 = 0 2x2 + x – 3 = 0 3x2 + 4x = 7 4x2 – 3 = 4x Rational Expressions Others x4 – 10x2 + 9 = 0 sin2x + 2sin x – 3 = 0 e2x + 7ex + 12 = 0 (log x)2 – 4log x – 5 = 0