Let a = x^2 + 4. Rewrite the following equation in terms of a and set it equal to zero. (x^2+4) ^2+32=12x^2+48 what are the solutions for x

Just replace x^2+4 with a

factor out on left side

(x^2+4) ^2+32=12 (x^2+4)

replace x^2+4 with a

a^2+32=12a

minus 12a both sides

a^2-12a+32=0

solve

factor

what 2 numbers mulitply to get 32 and add to get - 12

-4 and - 8

(a-4) (a-8) = 0

set each to zero

a-4=0

a=4

a-8=0

a=8

remember

a=x^2+4

so

4=x^2+4

0=x^2

0=x

8=x^2+4

4=x^2

+/-2=x

x=-2, 0, 2