Explain how to factor the following trinomials forms: x² + bx + c and ax² + bx + c.

Question

Is there more than one way to factor this? Show your answer using both words and mathematical notation?

Answer 1

x² + bx + c

example: x² + 5x+6

In this case you need to two factors (x+ a) and (x+ d).  First, you have to look for the factor of c, that c= (a)(d), and b must be a + d.

In this example, c= 6.  Which are the factors of 6?  6*1 and 3*2

Now, you can verify: c= 6*1=6; b= 6+1= 7.  IN this case 6 and 1 could not be the factors of x²+5x+6 because 6+1= 7, and not 5.

In the other hand, c= 3*2=6; b= 3+2=5

Now you can factorize x² + 5x + 6 = (x+3) (x+2).

 

ax² + bx + c

First, you have to look for the factors of a and c to know which one will be in th b value:

 

example: 7x² + 9x + 2

Factor of 7x²= 7x * x; factors of 2: 2 * 1

now, you can do the following:

7x         +         1

x           +         2          and use cross multiplication and sum the results (which can give you the value of bx).

 

(7x)*(2)= 14 x; (x)*1 = x; 14x + x= 15x.  If we notice, b is 9, so, this are not the factors for that polynomial.  If you have no the correct answer, you must switch the factors:

7x      +       2

x        +       1 and make the cross multiplication:

 

(7x)*1= 7x ; (x)*2 = 2x ; 7x + 2x= 9x

Now you have found the correct answer:

7x² + 9x + 2 = (7x+2) (x+1).

 

I hope this can help you.

 

Answer 2

Whenever I look at these problems, I put them into equation form, just because it makes more sense to me. In my example below, I not only factor the trinomial, but also find out the "zeroes" of the trinomial expressions corresponding equation:

For the first part of your question (x²+bx+c), we know we need to have two quantities (represented by a set of parentheses) that have an x in each quantity. (so far that looks like (x + _)*(x + _). We know this so far because when we foil these two quantities, we'll end up with the first term, x². Now to handle the next to terms, I always like to put the terms into words so it makes more sense to me. For example, if the equation were x² + 5x + 4 = 0 I would first look at both the signs of the terms and their coefficients. I'd eventually ask myself "What two factors multiply to four that add to five?" You could use trial and error until you found out that those two factors are 4 and 1. 

Continuing the example above, we'd simplify the equation to (x + 4)*(x+1) = 0 **Use foiling to make sure that this is correct! You should end up with the original equation!**. Now we solve for the equation. What we have right now is two quantities multiplied against each other to equal ZERO. This means we can separate the quantities into two separate equations (x+4) = 0 and (x+1) = 0. Now we use algebra to solve each of those equations, which will give us our answers. x = -4 and x = -1. 

Answer 3

ax² + bx + c=a(x² + bx/a + c/a)=a(x² + bx/a +(b/2a)²-(b/2a)²+c/a)

x² + bx/a +(b/2a)²=(x+(b/2a))²

ax² + bx + c=a((x+b/2a)²+c/a-(b/2a)²)

If c/a-(b/2a)²=0, then ax² + bx + c=a(x+b/2a)²

If c/a-(b/2a)²>0,  the expression cannot be factorized

If c/a-(b/2a)²<0, the expression can be factorized. Let d=sqrt((b/2a)²-c/a)

     ax² + bx + c=a((x+b/2a)²+c/a-(b/2a)²)=a((x+b/2a)²-d²)

     The factorization of (x+b/2a)²-d² is (x+b/2a+d)(x+b/2a-d)

      Finally, we have ax² + bx + c=a(x+b/2a+d)(x+b/2a-d)

example1: x²+2x+1, a=1 , b=2, c=1 

             c/a-(b/2a)²=1/1-(2/2)²=1-1=0, so x²+2x+1=(x+1)²

example2: 2x²+8x+10, a=2, b=8, c=10

                  c/a-(b/2a)²=10/2-(8/(2*2))²=5 - 4=1>0, so we cannot factorize the expression.

example3 3x²+12x-15, a=3, b=12, c=-15

          (b/2a)²-c/a=(12/(2*3))²+15/3=9

             d=sqrt(9)=3

               3x²+12x-15=3(x+2+3)(x+2-3)=3(x+5)(x-1)

 

         

Answer 4

Your question has been recently discussed.

https://www.wyzant.com/resources/answers/12859/explain_how_to_factor_the_following_trinomials_forms_x_bx_c_and_ax_bx_c

I do not want to repeat my arguments here, but I disagree with tutors here about the efficiency of this method of factoring. Look by the link at my answer and my comments to other tutors' responses. Look also at what tutor Natalia said, she essentially gave the same method I did, but in more concise form, omitting all details. Hope it helps more to see the logic behind the method. Also look up Viet's theorem, because this is what other tutors applied here. It may prove useful for factoring other polynomials too.