how to solve y=2x^4-6x^3+5x^2-18

Question

paki sagot please di ko na talaga kaya yung assignment namen e..

Answer 1

y(-1)=2+6+5-18=-5; y(-1.5)=23.625;

y(2.5)=-2.375; y(3)=162-162+45-18=27.

Let zeroes be x=a, x=b, x=c, x=d and -1.5<a<-1, 2.5<b<3.

Let f(x)=y, then f'(x)=df/dx=dy/dx=8x³-18x²+10x.

Newton's Method for n=0, 1, 2, 3, ...

a₀=-1;

a_n+1=a_n-f(a_n)/f'(a_n); 

b₀=2.5;

b_n+1=b_n-f(b_n)/f'(b_n).

We can find approximations for a and b which are increasingly accurate using this method.

We arrive at a=-1.123353 approx and b=2.559903 approx.

We can express these zeroes as a factorised quadratic:

(x-a)(x-b)=(x+1.123353)(x-2.559903).

This expands to the quadratic: x²-x(a+b)+ab=x²-1.436550x-2.875675 approx.

If we divide this quadratic into the original quartic we have:

                                         2x²-3.126899x+6.259401                                           

x²-1.436550x-2.875675 ) 2x⁴-6.000000x³ +5.000000x²+0.000000x-18.000000

                                        2x⁴-2.873101x³  -5.751349x²

                                                       -3.126899x³+10.751349x²+0.000000x

                                            -3.126899x³   +4.491948x²+8.991945x

                                                                    6.259401x²-...

We need divide no further because we already have a quadratic quotient.

So the quotient is quadratic 2x²-3.126899x+6.259401 which has no real solutions (but it has two conjugate complex solutions).

Therefore the only real solutions are x=-1.123353 or 2.559903, rounded to 6 decimal places.

To find the complex solutions, you can use the quadratic formula on 2x²-3.126899x+6.259401.

The complicated nature of this solution may indicate that the question contains an error (perhaps a missing x term).