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7x^2+17x+6 how do you factor the trinomial of the form ax^2+bx+c

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3 Answers

(ax+b)(cx+d) = (ac)x^2 + (ad+bc)x + (bd)
 
So you need to find two numbers, (ad) & (bc), that are both factors of the product of the leading coefficient and constant terms, (ac)(bd).
 
So, as Praviz indicates, multiply the leading coefficient and constant terms together, then find factors that add up to the coefficient of x.
 
You can find all the integer factors and test them all. If you can't find a pair of factors that work, then the trinomial is Not Factorizable Over The Integers.
You basically have to guess and check numbers
 
7(a) has factor pairs of 1,7 only
6(c) has factor pairs of 1,6 or 2,3
 
We need to guess and check combinations to get 17(b)
Combine pairs of number in a with pairs of numbers from c
This is the hard part, but if you practice it, you will learn how to guess the correct numbers.
7*6 +1*1 = 43...no
7*1+1*6 = 13... no
7*2 + 1*3 = 17... yes
 
So, 7 and 1 are the terms for x... (7x   ) (x   )
+2 and +3 are the other terms, but you have to place them on the opposite sides(so that they multiply correctly with the x terms)
 
Answer is
(7x + 3)(x+2)  
 
When you multiply them together...
7x^2 +   (2*7x + 3*x) + 6
 

Comments

If it  is  12 X^2 + 17X +5 , there is too many guesses for braking up 12 ( 3*4, 4*3, 6*2, 2*6,12*1, 1*12) for 12 , and other numbers , that gives the constant of 5.
 
   It is best just choose 12*5 = 60 , 12 +5 =17   
 
 7 X^2 + 17X + 6 :
****
  In your scratch paper :
 
     6*7 = 42 = 21* 2 = 7 * 6 = 14 *3  / look for 2 factors whose sum are +17 ( b) , you see that
     
 14 + 3 = 17
 
    Break up 17x = 14X + 3X
 
******
     7 X^2 + 14X + 3X + 6 =
 
      ( X + 2 ) + 3 ( X +2 ) =             / Perform factoring by grouping  
 
     ( X +2 ) ( 7x + 3 )  /  is the final answer.
   
    To find its zeros , let each factor equal zero and evaluate the zeros.                                                                                 

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