Find the limt x-> -1; (x^3 - 4x^2 + X + 6) / (X + 1)?

Question

Limits. I've tried factoring it a million ways and it doesn't seem to work. Please show all work as I already know the answer but can't work the problem out. Thx :)

Answer 1

factor.
[(x-3)(x-2)(x+1)]/(x+1)
x+1's become 1
(x-3)(x-2)
plug in -1 for x
(-1-3)(-1-2)
(-4)(-3)
lim as x->-1 = 12

Answer 2

You can do a polynomial division:
(x+1) goes into x^3, x^2 times. substract x^2(x+1) and you get: -5x^2 + x + 6.
x+1 goes into -5x^2, -5x times, subtract -5x(x+1) and you get: 6x + 6.
x+1 goes into 6x+6 six times and there is no remainder. Therefore your fraction can be written: x^2 - 5x +6, the limit of which is 12 when x-> -1

If this is too complicated you could also change the variable and write Y = X+1 (which means that X = Y-1), when X -> -1, Y -> 0. Plug in the Y into your fraction, you will get a polynome and then calculate its value when Y=0

Answer 3

x^3 - 4x^2 + x + 6 = x^3 + x^2 - 5x^2 + x + 6 =
= x^3 + x^2 - 5x^2 -5x + 6x + 6 =
= (x + 1)x^2 - 5x(x + 1) + 6(x + 1) =
= (x + 1)(x^2 - 5x + 6)

limt x-> -1; (x^3 - 4x^2 + X + 6) / (X + 1) =
= limt x-> -1; (x + 1)(x^2 - 5x + 6)/(x + 1) =
= limt x-> -1; x^2 - 5x + 6 = (-1)^2 - 5(-1) + 6 = 1 + 5 + 6 = 12

Answer 4

x^3 - 4x^2 + x + 6 = 0 for x = -1
Then x^3 - 4x^2 + x + 6 = (x + 1)(ax² + bx +c)
Expand (x + 1)(ax² + bx +c) and identify
a = 1
a + b = -4
b + c = 1
c = 6
x^3 - 4x^2 + x + 6 = (x + 1)(x² - 5x + 6)

(x^3 - 4x^2 + x + 6) / (x + 1) = (x + 1)(x² - 5x + 6) / (x + 1)
= x² - 5x + 6 if x different of -1

Then limt x-> -1; (x^3 - 4x^2 + X + 6) / (X + 1) is (-1)² -5*(-1) + 6 = 12

Answer 5

Without differentiating the answer u will get 0....Differentiating the equation once u will get
(3x^2-8x+1)/1....Substitute x= -1 here...The answer is 12...