Help with Factoring 2x^2 + 5x - 12

  • Thread starter gabrielh
  • Start date
  • Tags
    Factoring
In summary: This method is as follows: - Start by finding a number m such that mp=a, and nq=c. - Then find a number p such that mp+np=b. - Finally, find a number q such that nq+mq=c. The product of these two factors will give your original quadratic as being (2x+4)(x-4).
  • #1
gabrielh
79
0
I'm currently taking a Math Analysis class. We started reviewing basic factoring from Algebra 2 before starting new material. I can factor a basic trinomial with the coefficient of x^2 being 1, but if the coefficient is 2 or more, I am not sure how to factor it. I am able to factor these very easily with a graphing calculator, but I'd like to know how to do it by hand.

Homework Statement


Factor: 2x^2 + 5x - 122. The attempt at a solution
I know, by using a graphing calculator, that the trinomial factors into (2x-3)(x+4). I know that if the trinomial were, for instance, x^2 - 2x - 8, one could say the product of 8 and the coefficient of x is 8, and the factors of 8 that when added together result in negative 2 are -4 and +2, therefore the factors of x^2 - 2x - 8 are (x-4)(x+2).

Can trinomials with coefficients greater than one be factored in a similar way?

Using the method described above, I come up with a product of 24. Factors of 24 that result in a sum of 5 when added together are +8 and -3, which by the method I am acquainted with would mean the factors are (x+8)(x-3), but that obviously doesn't equal 2x^2 + 5x - 12 when multiplied. I know I'm missing a step somewhere.

Thanks for the help in advance. I know this is a simple question, but it plagued me all day.
 
Last edited:
Physics news on Phys.org
  • #2


You do it in the same manner

2x2 + 5x - 12 must be factored into something like (2x+a)(x+b)

the last times the last should give -12 or ab=-12, so what products can give -12?
There is -12 and 1, 2 and -6, 3 and -4 (and the others switch around the signs)

expanding (2x+a)(x+b) in your head, you'd see that the coefficient of x is a+2b. So looking at your choices, you want a+2b=5. Right away 12 and -1 (or 1 and -12) is eliminated.

2 and -6 is gone since 2+2(-6)≠5 or -2+2(6)≠5

so you are left with 3 and -4 or -3 and 4

3+2(-4)= -5 .So we are seeing 5 but negative, so we need to change the signs. So the choice is -3 and 4

so it is factored as (2x-3)(x+4)

as you practice, you can quickly do this and eliminate the obvious ones it can't be.

For ax2+b+c

Another way is to compute b2-4ac and find the square root of that. If it is an integer, then using the quadratic equation formula

[tex]x_1,x_2=\frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]

In your example b2-4ac=121, so √(b2-4ac)=11

x1,x2= (-5±11)/2(2)

x1=(-5+11)/4=6/4=3/2

So one root is x=3/2 or 2x+3=0

x2=(-5-11)/4=-4. So the other root is x=-4 or x+4=0

the product of these two factors will give your original quadratic as being (2x+4)(x-4)
 
  • #3


First not all trinomials can be factored with integer coefficients. But trinomials with coefficients of [itex]x^2[/itex] not equal to one are only a little more tedious. To factor [itex]ax^2+ bx+ c[/itex], you need to think about what (mx+ n)(px+ q) would look like: [itex]mx(px+ q)+n(px+q)= mpx^2+ mqx+ npx+ nq[/itex][itex]= mpx^2+ (mq+np)x+ nq[/itex]. So we are looking for a numbers m, n, p, q such that mp= a, nq= c, and mq+np= b. Start by thinking of factors for a and b and look at all the different ways you could form mq+ np.

With you example, [itex]2x^2 + 5x - 12[/itex], there are two ways to factor 2: (1)(2) and (2)(1). But there are many ways to factor 12: (1)(12), (2)(6), (3)(4), (4)(3), (6)(2), and (12)(1). Since there are 2 ways to factor 2 and 6 ways to factor 12, there are (2)(6)= 12 possible combinations. Also, since the last term is negative we know the last term in the two factors must have opposite sign:
(1)(2) and (1)(12): [itex](x+ 1)(2x- 12)= 2x^2- 10x- 12[/itex]
(2)(1) and (1)(12): [itex](2x+ 1)(x- 12)= 2x^2- 23x- 12[/itex]
(1)(2) and (2)(6): [itex](x+ 2)(2x- 6)= 2x^2- 2x- 12[/itex]
(2)(1) and (2)(6): [itex](2x+ 2)(x- 6)= 2x^2- 10x- 12[/itex]
(1)(2) and (3)(4): [itex](x+ 3)(2x- 4)= 2x^2+ 2x- 12[/itex]
(2)(1) and (3)(4): [itex](2x+ 3)(x- 4)= 2x^2- 5x- 12[/itex]
(1)(2) and (4)(3): [itex](x+ 4)(2x- 3)= 2x^2+ 5x- 12[/itex] !
(2)(1) and (4)(3): [itex](2x+ 4)(x- 3)= 2x^2- 2x- 12[/itex]
(1)(2) and (6)(2): [itex](x+ 6)(2x- 2)= 2x^2+ 10x- 12[/itex]
(2)(1) and (6)(2): [itex](2x+ 6)(x- 2)= 2x^2+ 2x- 12[/itex]
(1)(2) and (12)(1): [itex](x+ 12)(2x- 1)= 2x^2+ 23x- 12[/itex]
(2)(1) and (12)(1): [itex](2x+12)(x- 1)= 2x^2+ 10x+ 12[/itex]
Whew! Of course, we could have stopped when we got "(1)(2) and (4)(3): [itex](x+ 4)(2x- 3)= 2x^2_ 5x- 12[/itex]" but I wanted to show what they all looked like.
 
  • #4


Thank you both for your in depth replies. I'm able to do this now. Again, thanks a lot :)
 
  • #5


There is a clever factorization method (I believe developed by Viete) for non-monic quadratic trinomials with integer coefficients. (i.e. of the form [itex]ax^2 + bx + c \text{ for }a,b,c \text{ integers where } a \neq 1 \text{ or } 0[/itex]).

Step 0.
Factor out any common divisors of a, b, and c and work with the cofactor.

Step 1.
Multiply a and c and call the product d. This is the key number to unlocking the factorization.

Step 2.
Determine if there is a pair of factors of d whose sum is b. If such a factor pair exists, then the polynomial is factorable, otherwise it is not factorable (and you'd stop here).

If a factor pair exists, call them m and n.

Step 3.
Rewrite [itex]ax^2 + bx + c \text{ as }(ax^2 + mx) + (nx + c)[/itex] and factor the latter expression by grouping. Be careful when n is negative.

Step 4.
You should have your factorization.

The advantage to this method is that it cuts down the "try-this-then-try-this-other-way"-ness of other methods.



Example: Factor [itex]12x^2 -31x-30[/itex].

Step 0. No common divisors exist so I'm working with it as is.

Step 1. The product of 12 and -30 gives a key number of -360.

Step 2. The factos of -360 the add to -31 are -40 and 9.

Step 3.

[itex]12x^2-31x-30=(12x^2-40x)+(9x-30).[/itex]

[itex]=4x(3x-10)+3(3x-10)[/itex]

[itex]=(3x-10)(4x+3).[/itex]​

Done.

--Elucidus
 

1. How do I factor the expression 2x^2 + 5x - 12?

The first step in factoring any expression is to check for a common factor. In this case, we can factor out a 2, giving us 2(x^2 + 2.5x - 6). Then, we need to find two numbers that multiply to -6 and add to 2. These numbers are 3 and -2. So, our final factored expression is 2(x + 3)(x - 2).

2. What is the difference between factoring and simplifying?

Factoring is the process of breaking down an expression into smaller, simpler expressions. This is done by identifying common factors and using algebraic techniques to rewrite the expression. Simplifying, on the other hand, involves simplifying an expression to its most basic form by combining like terms and using the order of operations. Factoring and simplifying are two different ways of manipulating an expression, but they can often be used together to solve a problem.

3. Can I use the quadratic formula to factor this expression?

No, the quadratic formula is used to solve quadratic equations, not factor expressions. To factor an expression, we need to find two numbers that multiply to the constant term and add to the coefficient of the x-term. Then, we can use these numbers to rewrite the expression as a product of two binomials.

4. What is the purpose of factoring an expression?

Factoring is a useful tool in algebra that allows us to simplify and solve equations. It can help us identify patterns and make solving equations easier. Factoring is also useful in real-world applications, such as finding the roots of a polynomial function or simplifying complex fractions.

5. Can I use the FOIL method to factor this expression?

No, the FOIL method is used to multiply two binomials, not factor them. In factoring, we are essentially doing the opposite of FOIL - breaking down a polynomial into smaller binomials. However, the process of factoring can sometimes involve using the distributive property, which is similar to the FOIL method.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
6
Views
544
  • Precalculus Mathematics Homework Help
Replies
1
Views
1K
  • Precalculus Mathematics Homework Help
Replies
6
Views
906
  • Precalculus Mathematics Homework Help
Replies
4
Views
561
  • Precalculus Mathematics Homework Help
Replies
10
Views
2K
  • Precalculus Mathematics Homework Help
Replies
8
Views
173
  • Precalculus Mathematics Homework Help
Replies
11
Views
318
  • Precalculus Mathematics Homework Help
Replies
2
Views
856
  • Precalculus Mathematics Homework Help
Replies
10
Views
506
  • Precalculus Mathematics Homework Help
Replies
3
Views
1K
Back
Top