A=x^2+4 (x^2+4)^2+32=12x^2+48 12x^2+48=12(x^2+4) (x^2+4)^2+32=12(x^2+4) a^2+32=12a subtract 12a from both sides a^2-12a+32=0 factor (a-4)(a-8)=0 set each to zero a-4=0 a=4 a-8=0 a=8 a=4 or 8 a=x^2+4
8=x^2+4 and 4=x^2+4 subtract 4 4=x^2 0=x^2 square root +/-2=x 0=x x=-2,0,2 ax^2+bx+c=0 coefficient of the a term is 1 constant=32
A = x² + 4 & (x² + 4) +32 = 12x² + 48. You can factorize the left side of this equation
So 12x² + 48 = 12(x² + 4)==> (x² + 4) +32 = 12(x² + 4) Now replace (x² +4) by a ==> a +32 = 12a transpose all on the left side: a+32 -12a = 0==> -11a + 32 = 0 If you are asked to calculate then -11a=-32 & a=32/11
