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How do use use factoring to solve quadratic equations?

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2 Answers

Hi Harry,
 
ax2 + bx + c = 0 is a quadratic equation where a, b and c are all numbers.
 
When factoring a quadratic equation you want to find two numbers (z,y) that satisfy the following if a = 1:
  1. (z)(y) = c
  2. z + y = b 
ax2 + bx + c = 0 can then be written as (x + z)(x + y) = 0. Now we can solve for x since 
x + z = 0 or
x + y = 0
 
Example:
Solve x+ 6x + 8 = 0. 
(2)(4) = 8 and 2+4 = 6. Therefore the factored form is (x + 2)(x + 4) = 0. 
=> x + 2 = 0 => x = -2 OR
x + 4 = 0 => x = -4.
 
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If a is not equal to 1 then you want to find four numbers (e,d,z and y) that satisfy the following:
  1. (e)(d) = a
  2. (z)(y) = c
  3. (e)(y) + (d)(z) = b
Therefore ax2 + x + c = 0 where a is not equal to 1 then it can also be written as (ex + z)(dx + y) = 0.
ex + z = 0 => ex = -z and x = -z/e OR
dx + y = 0 => 
Example:
Solve 6x2 + 11x + 3 = 0
(2)(3) = 6, (3)(1) = 3 and (2)(1) + (3)(3) = 11, therefore (2x + 3)(3x +1) = 0
2x + 3 = 0 => x = -3/2 OR
(3x + 1) = 0 => x = -1/3
 
Hope this helps!
( X + a) ( X + b) = X2 + ( a +b) X + ab (1) ( X + a) 2 = X2 + 2aX +a2 (2)
 
( X +a ) ( X -a )= X2 -a2  (3)

The 3 above identities are key to factoring quadratic, and finding the roots
 
In general every quadratic, trinomial  aX2  + bX + c is  generated  by multiplication of 2 binomial (linear) expression in the form of 3 identity given above, and subsequently any given Ttrinomial ( quadratic), can
 be factored to binomial factors i.e.( X + a) ( X + b) = X2 + ( a +b) X + ab, that is used in evaluation of roots of the quadratic.
,  .  
 In other words identities (1), (2) , ( 3) can work both ways.
given a quadratic like : X2 + 7X + 10

here we see that ( a + b) = 7 ab =10
a = 2 b =5 is the answer, therefore X2 + 7X + 10 = ( X +2) ( X + 5 )
equation (2) is a special case of (1) where a=b , a+b = 2a , ab=a2
equation (3) is a special case of ( 1) where b = -a, a + ( -a) = 0 , ab = a2

These 3 identities are used in factoring a quadratic.

Equation (1) is factorable if there exists 2 whole number whose sum is ( a+ b), and product =ab.

If the answer of the system of equation is not a whole number, then have to do factoring by competing the square, and come up with a factors of irrational and complex numbers, yielding to Irrational and complex roots .