Question

Consider this polynomial equation.

Answer 1

Answer:

x = 3, -1, 2, -2

Step-by-step explanation:$x^{2}$

6($x$-3)($x^{2}$+4)($x$+1) = 0

6($x$-3)($x$+2)($x$-2)($x$+1) = 0

Using the fact that ($x^{2}$+4) = ($x$-2)($x$+2) - known as the difference of two squares rule

Thus solve by inserting values of $x$ that will result in the equation equating to 0.

ie: if $x$ = 3,

6(3-3)(3+2)(3-2)(3+1) = 0

0=0

RHS = LHS therefore true

Answer 2

Answer:

The equation has 5 solutions and its real soluations are x=3, +sqrt2, -sqrt2, and -1

Step-by-step explanation: