Pre-Calculus/Calculus Problem?
This problem asks to solve for x, I can solve it out fine, but I get a polynomial with 5 pieces, which I have no idea what to do with, heres the problem:
(x+1)^3(4x-9)-(16x+9)(X+1)^2=0
And what I get:
4x^4-13x^3-56x^2-57x-18=0
Comments
x + 1)*(4x - 9) - (16x + 9) = 0
4x^2 + 5x - 9- 16x - 9 = 0
4x^2 - 11x - 18 = 0
This is a simple quadratic, solve for two values of x
You can get some of the answer from inspection. For example, plug x = -1 into the equation, and the (x + 1) terms go to zero, making the whole equation zero. So x = -1 is one solution. Exclude this x value, and divide through by (x + 1)^2 to get
(x + 1)*(4x - 9) - (16x + 9) = 0
4x^2 - 5x - 9- 16x - 9 = 0
4x^2 - 21x - 18 = 0
This is a simple quadratic, solve for two values of x. x = 6, x = -3/4
From the original equation remove (x+1)^2 as a common factor and try solving it. I got x = 6 or x= -3/4.
just make it y=4x^4-13x^3-56x^2-57x-18, and plug it into a handy calculator, and see where it crosses x axis. works best for big ones for me.
make it a good day
(x+1)^3(4x-9)=(16x+9)(X+1)^2
*assuming x doesn't equal -1 and that we can divide by (x+1)^2
(x+1)(4x-9)=16x+9
4x^2-5x-9=16x+9
4x^2-21x-18=0
The roots of the polynomial are the roots of this expression, and also x=-1 (if you check, you can see it is a root).
-1 and 6 are two solutions by trial and error.
The other two roots may be complex.