f(x)=x^4+x^3=16-8x-6x^2

Question

how do i solve f(x)=x^4+x+^3=16-8x-6x^2 polynominal equation

Answer 1

f(x)=x⁴+x³=16-8x-6x² means f(x)=x⁴+x³ and f(x)=16-8x-6x², therefore x⁴+x³=16-8x-6x².

So x⁴+x³-16+8x+6x²=0,

x⁴+x³+6x²+8x-16=0.

Note that if x=1: 1+1+6+8-16=0, therefore x=1 is a zero of the polynomial.

Use synthetic division to divide by this zero:

1 | 1 1 6  8  -16

     1 1 2  8 | 16

     1 2 8 16 | 0 = x³+2x²+8x+16.

There may be another rational zero. If it exists, it will be a factor of 16 and it will be negative because this cubic can only factorise with positive factors (all signs are plus) of the type x+a, where a is positive, which means that the zeroes must be such that x+a=0, that is, x=-a. So let's start by plugging in x=-2:

-8+8-16+16=0, making x=-2 another zero. So we divide by this zero:

-2 | 1  2 8    16

      1 -2 0 | -16

      1  0 8 |    0 = x²+8 which has no real zeroes (only complex factors ±2i√2).

Assuming only real zeroes apply in this question, f(x)=x⁴+x³. Plug in the two zeroes f(x)=1+1=2, or f(x)=16-8=8.

The same zeroes can be plugged into f(x)=16-8x-6x²=16-8-6=2, or f(x)=16+16-24=8.