Question

factorize ((x) ^6)+9((x) ^2)+8

Answer 1

Your question might be x^6+8x²+9

x^6+8x²+9

Let x²=a

(x²)³+2x²+9

=a³+2a+9

=a³+a²-a²-a+9a+9

=a²(a+1)-a(a+1)+9(a+1)

=(a+1)(a²-a+9)

Put a=x²

=(x²+1)[(x²)²-x²+9

=(x²+1)(x⁴-x²+9)

Let's check now according to your question,

x^6+9x²+8=0

x²(x⁴+9)=-8

If you take any value of x in LHS,the value would be positive,but if you see RHS it is negative.
So,positive term is not equal to negative term.
So,the question is x^6+8x²+9

Answer 2

I think you took typing mistake x^6 + 9x^2 + 8 not factorise in real factors . I think the question is

x^6 + 8x^2 + 9

let x² = P

(x²)³ + 8(x²) + 9

= P³ + 8P + 9
=P³ + P² -P² -P + 9P + 9
=P²(P + 1) -P( P + 1) + 9(P + 1)
=(P + 1)( P² - P + 9 )

now, put P = x²

=(x² + 1)(x⁴ - x² + 9)

hence, x^6 + 8x² + 9 have two factors (x² + 1) and ( x⁴ - x² + 9) .


but you say your question is correct then,
x^6 + 9x² + 8 = 0
x²(x⁴ + 9) = - 8

you are seeing in LHS
x²(x⁴ + 9) always positive term ,
but in RHS = -8 is a negative term ,

how negative term = positive term .
this is not possible in real . but here using complex we can find . but I think you are 9-10th class student . so, tough to understand complex no .