How do you factor 3x to the 6 power plus 54x to the 4 power plus 243x to the second power?

Answer 1

3x5+ 15x3+ 12x

All of these terms have a common factor of 3x, so we can factor it out to get:

3x(x4+ 5x2+ 4)

Our 'bracketed' term now has the layout of a quadratic equation, except the powers of x have been doubled - x2+ 5x + 4x0(x0= 1, x0*2= x0)

This means we can factorise it just like a quadratic, except making sure the x's are squared in the final brackets.

x2+ 5x + 4 factorises to (x+1)(x+4), so x4+ 5x2+ 4 will factorise to give (x2+1)(x2+4).

Put this all together to get:

3x(x2+1)(x2+4)

This could be factorised further by treating each of these new bracketed terms as quadratics, but only if complex numbers are allowed - so for simplicity's sake, this is the final product.

Otherwise, you could turn it into:

3x(x+i)(x-i)(x+2i)(x-2i)

Answer 2

All terms include 3x2. So we can factor the expression as :-

3x6 + 54x4 + 243x2 = 3x2(x4 + 18x2 + 81)

And (x4 + 18x2 + 81) = (x2 + 9)2

Putting the two parts together gives, 3x2(x2 + 9)2.