How do use use factoring to solve quadratic equations?

Question

What property do you need to use? Can you demonstrate an example of utilizing factoring to solve equations?
Don't forget to include a quadratic equation

Answer 1

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 x² + 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!

Answer 2

( X + a) ( X + b) = X² + ( a +b) X + ab (1) ( X + a) 2 = X² + 2aX +a² (2)

 

( X +a ) ( X -a )= X^{2 -}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 .