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1. 
   kjcdb8er Teacher
   

   eNotes Editor

   To factor x² - 6x - 16 = 0, you can complete the square or use the quadratic formula.

   The quadratic formula is for equations, like yours, that have the form ax² + bx + c = 0. The quadratic formula states that x = (-b +/- sqrt(b² -4ac))/2a

   In your case, a = 1, b = -6, and c = -16. So,

   x = (-(-6) +/- sqrt ( (-6)² -4*(1)(-16) ))/(2*1) = (6 +/- sqrt(36 + 64))/2

   and x = (6 +/- 10)/2 = 8 or -2

   To complete the square of the quadratic equation x² + bx + c = 0, we want to put the equation into this form: (x + M)(x + N) = 0. So:

   0 = x² + bx + c = (x + M)(x + N) = x² + (M+N)x + M*N

   Comparing the right and left sides, we can see that b = M + N and c = M*N. So, in your problem, -6 = M + N and -16 = M*N. You can solve this quickly on paper, or we can guess. Lets guess:

   What numbers multiplied together give -16 and added together give -6?

   -1*16 or -2*8 or -4*4 = -16 Can you see one pair that added together gives 6 or -6?

   8 - 2 = 6, but we need -6. So lets make that 2 - 8 = -6, and -8*2 = -16. So M = 2 and N = -8. So,

   (x + M)(x + n) = (x + 2)(x - 8) = 0

   We can now solve for x, since we know that only zero * anything = 0. So either (x + 2) = 0 OR (x - 8) = 0

   Thus, x = -2 or x = 8

    

   Sources:

   http://www.enotes.com/math/q-and-a/solve-for-x-4-3x-7-7-8...

   http://www.enotes.com/math/q-and-a/x-1-x-4-2x-1-x-6-92557

   http://www.enotes.com/math/q-and-a/x-4-lt-7-92161

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   Posted by kjcdb8er on Saturday August 1, 2009 at 1:53 PM
   
   

2. 
   cburr Teacher
   Middle School

   eNotes Editor

   Wow!  Here's a less complicated answer.

   When you have an equation like this, what you want to do is find the factors.  Since it takes the form x2 - ___x - ___=0, you can assume that the factors will be:

   (x + __) (x - __)

   When you multiply these factors together, you will get x^2, you will get two elements that include x, and a final element that has no x, but is the product of the two numbers.

   NOTE:  I have used one '+' and one '-'.  This is because the only way to get a negative number for the last element is to multiply a plus and a minus (two + or two - would give you a positive.

   The next step is to figure out what numbers to use.  First look at the factors of 16:  1 x 16, 4 x 4, and 8 x 2.  You need to choose the pair that will get you 6 when you subtract one from the other.  Clearly, 8 and 2 is the correct pair.

   The last step is to figure out which number goes with the - and which goes with the +.  If you did

   (x + 8) (x - 2), you would end up with 8x - 2x or 6x.

   This doesn't work, because we need -6x.  So, the correct answer is

   (x + 2) (x - 8)

   x^2 + 2x - 8x - 16

   x^2 -6x - 16

    

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   Posted by cburr on Saturday August 1, 2009 at 3:29 PM
   
   

3. 
   neela Teacher
   Graduate School

   eNotes Editor

   x^2-6x-16=0.

   It seems your main problem is to factorise the left side.

   The expression on the left is a quadratic expression which  can be factorised by splitting the middle term, -6x in such a  way that  the product of the split expressions is equal to the  product of the end terms:x^2 and (-16) are the end terms here.So, the end term product  is -16x^2. That is like doing as below:

   -6x = (-8x)+(2x) and

   (-8x)(2x) = -16x^2.

   Note that this is the most important part of facrorisation of quadratic expression.
   

   Therefore , now the arragement of middle term -6x in the  left side of the  equation becomes  :

   x^2 -8x+2x-16=0.   Here -6x is replaced by -8x + 2x

   x(x-8)+2(x-8)=0. Take the common factor out.

   Therefore the factors of the quadratic x^2-6x-16 are (x-8)  and (x+2) or x^2-6x-16  is identically equal to  the factorised form (x-8)(x+2).

   Now let us go for the solution part of  the equation:

   (x-8)(x+2)=0.Please remember that if a product pq = 0,then either p=0 or q=0.

   Therefore, x-8=0 or x+2=0

   x-8=0 gives the solution, x=8

   x+2=0 gives x=-2.

   Second method:

   Any quadratic expression can be reduced to the form X^2-n^2 ,where X = ax+b , n and  a,b are some numbers and then it can be factorised like (X+n)(X-n):

   x^2-6x -16 canbe written like:

   (x-3)^2-9 - 16 --->

   (x-3)^2     - 25---->

   (x-3)^2 -  5^2

   This is in X^2-n^2 form.

   [(x-3)+5][(x-3)-5]

   (x+2)(x-8)

   Hope this helps.

    

    

    

    

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   Posted by neela on Saturday August 1, 2009 at 7:46 PM