I need some help
- Thread starter GlennaN
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mmm4444bot
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You need to find two factors of -36 that sum to 5.
Can you do that? If not, then let us know.
Otherwise, once you find these two numbers, the factorization looks like:
(y + ?)(y + ?)
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You need to find two factors of -36 that sum to 5.
Can you do that? If not, then let us know.
Otherwise, once you find these two numbers, the factorization looks like:
(y + ?)(y + ?)
?
The most common way to find the factors of these functions is to calculate the delta
With a function f(x) = a*x^2 + b*x +c (a,b,c are constants)
delta = b^2 - 4*a*c
then the two factors of the function are x1 = (-b - delta^(1/2))/(2*a) and x2 = (-b + delta^(1/2))/(2*a)
If delta < 0 then you can not factorize the function, or the there is no x for f(x) = 0.
If delta > 0, you can factorize the function f(x) = (x-x1)*(x-x2) with x1 and x2 calculated as shown above.
If delta = 0, x1 = x2 and f(x) = (x-x1)^2
__________
For example, f(y) = y^2 + 5y - 36
we have a = 1, b = 5 and c = -36, thus delta = 5^2 - 4*1*(-36) = 169 = 13^2
delta > 0, we have two different factor x1 and x2. (delta)^(1/2) = 13. Therefore,
x1 = (-5 - 13)/2 = -9
x2 = (-5 +13)/2 = 4
Thus, f(y) = y^2 +5y - 36 can be factorized as f(y) = (y +9)*(y-4).
With a function f(x) = a*x^2 + b*x +c (a,b,c are constants)
delta = b^2 - 4*a*c
then the two factors of the function are x1 = (-b - delta^(1/2))/(2*a) and x2 = (-b + delta^(1/2))/(2*a)
If delta < 0 then you can not factorize the function, or the there is no x for f(x) = 0.
If delta > 0, you can factorize the function f(x) = (x-x1)*(x-x2) with x1 and x2 calculated as shown above.
If delta = 0, x1 = x2 and f(x) = (x-x1)^2
__________
For example, f(y) = y^2 + 5y - 36
we have a = 1, b = 5 and c = -36, thus delta = 5^2 - 4*1*(-36) = 169 = 13^2
delta > 0, we have two different factor x1 and x2. (delta)^(1/2) = 13. Therefore,
x1 = (-5 - 13)/2 = -9
x2 = (-5 +13)/2 = 4
Thus, f(y) = y^2 +5y - 36 can be factorized as f(y) = (y +9)*(y-4).
mmm4444bot
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I do not have enough knowledge to declare what is "most common", but it seems to me that second-degree polynomials which factor over the Integers, whose leading coefficient is one and whose remaining coefficients have absolute value less than 145, are easily factored mentally.
(Of course, this will not be the case for people who have not memorized the multiplication table.)
I mean, it takes all of ten seconds to realize that 36 factors as 9 times 4, and that these two factors differ by 5. Another five seconds of consideration leads to the proper signs.
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toidayma said:
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I do not have enough knowledge to declare what is "most common", but it seems to me that second-degree polynomials which factor over the Integers, whose leading coefficient is one and whose remaining coefficients have absolute value less than 145, are easily factored mentally.
(Of course, this will not be the case for people who have not memorized the multiplication table.)
I mean, it takes all of ten seconds to realize that 36 factors as 9 times 4, and that these two factors differ by 5. Another five seconds of consideration leads to the proper signs.
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Totally agree!Another way to mentally factorize a second-degree function whose factors are integers is the consequence of the formal method. i.e. for the function f(x) = ax^2 + bx + c, we have
x1 + x2 = -b/a (1)
x1 * x2 = c/a (2)
In the case f(x) = x^2 + 5y - 36, we have a = 1, b = 5, c = -36. So
x1 + x2 = -5
x1*x2 = -36
From this point, we can quickly mentally the two factors are -9 and 4.
x1 + x2 = -b/a (1)
x1 * x2 = c/a (2)
In the case f(x) = x^2 + 5y - 36, we have a = 1, b = 5, c = -36. So
x1 + x2 = -5
x1*x2 = -36
From this point, we can quickly mentally the two factors are -9 and 4.