Question

Claire Judice

Answer 1

Answer:

The values of 'x' are -1.2, 0, 0, $-4i$ or $4i$.

Step-by-step explanation:

Given:

The equation to solve is given as:

$5x^5+6x^4+80x^3+96x^2=0$

Factoring $x^2$ from all the terms, we get:

$x^2(5x^3+6x^2+80x+96)=0$

Now, rearranging the terms, we get:

$x^2(5x^3+80x+6x^2+96)=0$

Now, factoring $5x$ from the first two terms and 6 from the last two terms, we get:

$x^2(5x(x^2+16)+6(x^2+16))=0\\x^2(x^2+16)(5x+6)=0$

Now, equating each factor to 0 and solving for 'x', we get:

$x^2=0\\x=0\ and\ 0\\\\5x+6=0\\x=\frac{-6}{5}=1.2\\\\x^2+16=0\\x^2=-16\\x=\sqrt{-16}=\pm 4i$

There are 3 real values and 2 imaginary values. The value of 'x' as 0 is repeated twice.

Therefore, the values of 'x' are -1.2, 0, 0, $-4i$ or $4i$.