What substitution should be used to rewrite 16(x3 + 1)2 – 22(x3 + 1) – 3 = 0 as a quadratic equation?

  z = x^3 +1

Step-by-step explanation:

Noting the squared term, it makes sense to substitute for that term:

  z = x^3 +1

gives ...

  16z^2 -22z -3 = 0 . . . . the quadratic you want

Solutions derived from that substitution

Factoring gives ...

  16z^2 -24z +2z -3 = 0

  8z(2z -3) +1(2z -3) = 0

  (8z +1)(2z -3) = 0

  z = -1/8  or  3/2

Then we can find x:

  x^3 +1 = -1/8

  x^3 = -9/8 . . . . . subtract 1

  x = (-1/2)∛9 . . . . . one of the real solutions

__

  x^3 +1 = 3/2

  x^3 = 1/2 = 4/8 . . . . . . subtract 1

  x = (1/2)∛4 . . . . . . the other real solution

The complex solutions will be the two complex cube roots of -9/8 and the two complex cube roots of 1/2.

step-by-step explanation:

first, eliminate the parenthesis:

3

x

+

2

−

4

x

second, group like terms:

3

x

−

4

x

+

2

third, combine like terms:

(

3

−

4

)

x

+

2

→

−

1

x

+

2

→

−

x

+

2