to view answer.As with factoring out the greatest common factor, factoring quadratics is simply reversing the process of the distributive property, where we rewrite a sum as a product.
Recall multiplying polynomials when we use the distributive property to multiply:
$$(x-2)(x+1)$$
In this type of problem, we multiply each term by every other term to get the sum of terms:
$$(x-2)(x+1)=x^2+x-2x_2$$
Then, combining like terms, we get:
$$(x-2)(x+1)= x^2+x-2x_2=x^2-x-2$$
So, how do we go in reverse? How do we go from the sum $$x^2-x-2$$ to the product $$(x-2)(x+1)$$? Let’s look at an example:
Factor completely:
$$x^2+6x-16$$
1. The first thing that we should do when we try to factor a polynomial like this is see if we can “factor out” any constants or variables (See Factoring Part 1). In this case, there is no common factor, so there is nothing to factor out.
2. Now, we want to try to find two binomials that, when multiplied together, give us:
$$x^2+6x-16$$
If these binomials exist, we need the first terms of each of them to multiply to be $$x^2$$. So, here, we know that the first term of each binomial should be x. Thus, our factored form will look something like:
$$\left( {x + a} \right)\left( {x + b} \right)$$
in which a and b are constants and may be negative or positive.
3. What happens when we multiply $$\left( {x + a} \right)\left( {x + b} \right)$$? We get: $$x^2 + bx + ax + ab$$. Let’s combine like terms and to get:
$$x^2 + \left( {b + a} \right)x + ab$$. So, we see that whatever constants a and b are, when multiplied together, they will be the constant term (in this case: –16) and when added together, they will be the coefficient of the x term (which in this case is 6).
4. So, how do we find a and b? Well, we know two things about them. They multiply together to be –16 and add together to be 6. So, first, since they multiply to be a negative number, one of them must be negative and the other positive. So, listing the possibilities, we see only 5 choices:
$$\eqalign{
& 1,\;\; - 16 \cr
& 2,\;\; - 8 \cr
& 4,\;\; - 4 \cr
& 8,\;\; - 2 \cr
& 16,\;\; - 1 \cr} $$
5. Now, all we need to do is determine which of these possible values of a and b add to be 6. We can check each one, and we will see that only –2 and 8 add to be 6. So, our factored form must be:
$$\left( {x - 2} \right)\left( {x + 8} \right)$$
6. We can always check our answer by using the distributive property to multiply these binomials together to see that:
$$\left( {x - 2} \right)\left( {x + 8} \right) = x^2 + 8x - 2x - 16 = x^2 + 6x - 16$$
Thus, since our factors multiply to our original sum of terms, the product form must be correct.
Factor completely:
$$12x^2 - 22x + 8$$
This is a slightly more difficult polynomial to factor because the $$x^2 $$
term has a coefficient greater than 1. However, the process is exactly the same.
1. First, let’s check to see if there are any constants or variables that we can factor out immediately, before we try to break this divide the polynomial into binomial factors. We notice that every term is divisible by 2, and thus, we can divide a 2 out of each term:
$$2\left( {6x^2 - 11x + 4} \right)$$
2. Now, we will look at the somewhat less complicated polynomial: $$6x^2 - 11x + 4$$. There are no more constants or variables that can be factored out, so now we will try to rewrite this sum as a product of two binomials. This time, we know that the first terms of those binomials must multiply to be $$6x^2 $$
and the last terms of the binomials must multiply to be 4.
3. So, we know that the first terms in our binomials must either be the pair $$2x$$ & $$3x$$, or $$x$$ & $$6x$$. Then, the last terms in the binomials must be either the pair 1 & 4 or 2 & 2 (or the negatives of those two). Actually, since the middle term is negative, we can simply look at the pairs -1 & -4 or -2 & -2. Let’s list the possibilities, in the binomial form:
$$\eqalign{
& (x - 1)(6x - 4) \cr
& (x - 4)(6x - 1) \cr
& (x - 2)(6x - 2) \cr
& (2x - 1)(3x - 4) \cr
& (2x - 4)(3x - 1) \cr
& (2x - 2)(3x - 2) \cr} $$
In each of these products, the first term is$$6x^2 $$ and the constant term is 4. If the original quadratic sum is factorable, one of the above products will also produce the correct middle term, –11x.
Try each combination and to see if it produces the correct middle term. Suppose that the factored form is:
$$\left( {x - 1} \right)\left( {6x - 4} \right)$$
Well, let’s multiply it out and see what happens:
$$\left( {x - 1} \right)\left( {6x - 4} \right) = 6x^2 - 4x - 6x + 4 = 6x^2 - 10x + 4$$
So, that didn’t work, because the middle term is not correct.
4. Keep trying the other possibilities, until
$$\left( {2x - 1} \right)\left( {3x - 4} \right) = 6x^2 - 8x - 3x + 4 = 6x^2 - 11x + 4$$
And, there is our polynomial, so we have found our factors. So, is this the answer? NO. Remember that we factored out a 2 to begin with, so we need to remember that we were working with only one factor of our original polynomial. Our factored polynomial is:
$$12x^2 - 22x + 8 = 2\left( {2x - 1} \right)\left( {3x - 4} \right)$$
How do I know where to start? You may have noticed that quadratics with coefficients higher than 1 on the $$x^2 $$ term are more difficult to factor than those with an $$x^2 $$ coefficient of 1. So, the first thing that you should do is make a note, as we did above, of all of the possible coefficients of x’s that will give your coefficient of $$x^2 $$. Then, make a list of all possible numbers that can multiply to be your final, constant term. Using these, you can at least guess and check until you get your answer. However, that may take a while with some problems (if your terms each have a lot of factors). One good approach is to try the least “extreme” possibilities first. Try factors that are not as big or as small as they could be. For instance, in Example 2, instead of starting with x and 6x (the biggest/smallest possible factor pair), we should have tried 2x and 3x. Using these terms first will help you to determine if you need a bigger or smaller coefficient on the x terms or on the constant terms. |
Some practice problems to check your skills:
Rewrite the following sums as products by factoring the following polynomials completely:
1. $$x^2 - 2x - 35$$
Think about it for a moment and then access this link to view answer.
2. $$x^2 + 3x - 15$$
Think about it for a moment and then access this link to view answer.
3. $$5x^2 - 35x + 60$$
Think about it for a moment and then access this link to view answer.
4. $$2x^2 - 9x - 5$$
Think about it for a moment and then access this link to view answer.
5. $$30x^2 + 4x - 16$$
Think about it for a moment and then access this link to view answer.