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Q? Factoring quadratic eqn.

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by vk_vinayak » Mon Aug 27, 2012 4:01 am
When solving a quadratic equation: x2 + 2x - 8 = 0

We know that sum -2 and + 4 equal to middle part and their product is equal to -8. We can safely write the factors as (x + a)(x + b) =0 ie (x-2)(x+4)=0

But, if the equation is 3x^2 + 5x -12 =0 , we come to know that a and b as +9 and -4. But we can't write it directly (x + 9)(x -4)=0 because of the quotient 3. We then need to divide 9 and -4 by 3. SO, we get the equation: (x+3)(x-4/3)=0.

Just wanted to confirm if this is a general rule (to divide the roots by co-ef of x^2) and works all the time.
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by Jim@StratusPrep » Mon Aug 27, 2012 7:42 am
Your middle term will not work with this. Here is a good way to approach:

5x^2 + 11x + 2

Step 1: List the factors of the product of the coefficient and the last term 5(2) = 10 -- Factors are 1, 2, 5, and 10.
Step 2: Find the 2 factors that add up to the coefficient of the middle term 10 + 1 = 11
Step 3: Rewrite the middle term, splitting it with the factors your found in step 2 5x^2 + 10x + x + 2
Step 4: Factor grouped terms
5x(x+2)+1(x+2)= (5x+1)(x+2)
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by vk_vinayak » Mon Aug 27, 2012 8:33 am
Jim@StratusPrep wrote:Your middle term will not work with this. Here is a good way to approach:

5x^2 + 11x + 2

Step 1: List the factors of the product of the coefficient and the last term 5(2) = 10 -- Factors are 1, 2, 5, and 10.
Step 2: Find the 2 factors that add up to the coefficient of the middle term 10 + 1 = 11
Step 3: Rewrite the middle term, splitting it with the factors your found in step 2 5x^2 + 10x + x + 2
Step 4: Factor grouped terms
5x(x+2)+1(x+2)= (5x+1)(x+2)
Jim, Thanks. I know the calculations, but I am looking for a shortcut.

After you figure out the factors as 10 and 1 (your step 2), I want to directly (without going through step 3 and 4) write the equation in the form of (x+a)(x+b)=0. It works perfectly when co-ef of x^2 is 1, but I was having problem when co-ef of x^2 as something else (eg: In your equation co-ef of x^2 is 5)

Here a=+10 and b=+1, but as the co-ef of x^2 is 5, I divided these a and b by 5, getting 2 and 1/5, so that I could write (x+2)(x+ 1/5)=0. I wanted confirm this method.
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