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I am factoring quardratic trinomials. The problem I'm working with is the following: 12x2 + -46x + 42. How do I factor it?

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4 Answers

 12 X2 - 46X + 42 = 2 ( 6X2 - 23X + 21) 
 
       two factor : Must be 2 numbers a,b such that    a . b = 6 ( 21) = 126    a+b =-23
 
          first we write 126 as a product of its prime factors:
 
         126 = 2 . 3 . 3 . 7 = 2 * 63 = 6 * 21= 18 * 7 = 14 * 9
 
             We find that:  -14 - 9 = -23
 
            break up -23 into -14 - 9 = -23
 
             2( 6 X2 - 9X - 14 X  + 21) = 2 [ 3X( 2X -3) - 7( 2X -3) ] = 2( 2X -3 ) ( 3X -7 )
 
          This is the methods by which step by step can be followed to get the result, be able to factor.  
Factor a 2 out of the equation and you get 2(6x^2-23x+21).  Then factor the polynomial 2(3x-7)(2x-3).
 
Good Luck!

Comments

Thank you! That was what I had been getting. But what I wasn't sure about was how to check it. Once you have 2(3x-7)(2x-3) then how do you get it to come out with 12x2 + -46x + 42? If I just take (3x-7)(2x-3) and use the FOIL method then it comes out with 6x2-23x+21. That of course is just half of 12x2 + -46x + 42. So obviously, if I used 2(3x-7)(2x-3) then it should come out with the right answer. I'm sure I'm making some stupid mistake while doing this, but apparently there's something I'm just not getting. To get 2(3x-7)(2x-3) to come back with 12x2 + -46x + 42, what do I do? What I was doing was starting out with 2 times 3x and 2 times 2x. Then multiply those together. But then that comes out with 24x2 when I need 12x2. So what do I do? Use the FOIL method and then multiply everything by 2?
Yes, follow PEDMAS and multiply out the terms in the parentheses first.  You end up with 2[6x2 - 23x + 21] and then you just multiply each term in the parentheses by 2 to get 12x2 - 46x + 42.
The first thing I would do is to see if I could factor out a constant. Notice that all the coefficients are even; you could at least factor out 2.
2(6x^2 - 23x + 21)
 
Let's list the possible factor pairs for 6 and 21
6= 2*3 = 6*1
21= 3*7 = 21*1
 
Notice that the middle term is negative and the final term is positive. This means that the form of the factors will be (  -  )(  -  ) because two negatives added is a negative while two negatives multiplied is a positive. Now we can try different combinations of the factors of 6 and 21 and test using FOIL.
 
(2x-7)(3x-3)= 6x^2-27x+21 Not equal to 6x^2 - 23x + 21
(2x-3)(3x-7)=6x^2 - 23x + 21 This is it!
 
So the the factors are 2(2x-3)(3x-7).
 
Let us look at this trinomial:
 
12x2-46x+42;
 
The best way to factor it is to use a quadratic formula. If you know the roots of the equation 12x2-46x+42=0, then it can be factored as follows:
12x2-46x+42=12(x-x1)(x-x2), where x1 and x2 are the roots.
 
Quadratic formula gives us:
 
x1,2=[46±√(462-12*42*4)]/24;
 
x1,2=(46±√100)/24=(46±10)/24
x1=3/2;
x2=7/3;
Thus, we have:
12x2-46x+42=12(x-3/2)(x-7/3)=2*(2x-3)(3x-7)

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