Equation problem

Anthonyk2013

Junior Member
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Sep 15, 2013
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Hi, I was doing calculus turning points of a curve in evening class last night and this happened in the middle of a problem.

the teacher described the method as the big X.
could anyone explain it for me.

Things were moving very fast last night and meant to ask after class. Think its simple enough but would like an explanation.
 

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That's not Calculus, but fairly basic algebra- for factoring a quadratic. You probably learned it long ago but in another form.

(x+ a)(x- b)= x^2+ (a- b)x- ab. So to factor something like x^2- 3x= 6, you look for "a" and "b"- the factors of 6. 6 happens to be particularly easy, it factors only as (6)(1) or (3)(2). Trying (x- 6)(x- 1) or (x+ 6)(x- 1) we see that "a+ b" is 6- 1= 5 or -6+ 1= -5, not -3. Trying (x- 3)(x+ 2) or (x+ 3)(x- 2) we see that "a+ b" is -3+ 2= -1 or 3- 2= 1. The first of those is correct: x^2- x- 6= (x- 3)(x+ 2).

The "big X" is just a way of visualising the middle term. At each end of one arm of the X are "x" and "-3": -3x. At each end of the other arm are "x" and "2": 2x. The sum of the numbers on the right, -3+ 2= -1, is the coefficient of x.
 
That's not Calculus, but fairly basic algebra- for factoring a quadratic. You probably learned it long ago but in another form.

(x+ a)(x- b)= x^2+ (a- b)x- ab. So to factor something like x^2- 3x= 6, you look for "a" and "b"- the factors of 6. 6 happens to be particularly easy, it factors only as (6)(1) or (3)(2). Trying (x- 6)(x- 1) or (x+ 6)(x- 1) we see that "a+ b" is 6- 1= 5 or -6+ 1= -5, not -3. Trying (x- 3)(x+ 2) or (x+ 3)(x- 2) we see that "a+ b" is -3+ 2= -1 or 3- 2= 1. The first of those is correct: x^2- x- 6= (x- 3)(x+ 2).

The "big X" is just a way of visualising the middle term. At each end of one arm of the X are "x" and "-3": -3x. At each end of the other arm are "x" and "2": 2x. The sum of the numbers on the right, -3+ 2= -1, is the coefficient of x.

Sorry I should have stated what we were doing, it came up in a calculus problem while we were finding the turning point of curves. Thanks
 
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