Let #a=x^2+4#. How do you rewrite #(x^2+4)^2 +32 = 12x^2+48# in terms of #a# and set it equal to zero.? Algebra Polynomials and Factoring Polynomials in Standard Form 1 Answer Shwetank Mauria Feb 22, 2017 #a^2-12a+32=0# Explanation: #(x^2+4)^2+32=12x^2+48# is equivalent to (factorizing RHS) #(x^2+4)^2+32=12(x^2+4)# But as #a=x^2+4#, this can be written as #a^2+32=12xxa# or #a^2+32=12a# or #a^2-12a+32=0# Answer link Related questions What is a Polynomial? How do you rewrite a polynomial in standard form? How do you determine the degree of a polynomial? What is a coefficient of a term? Is #x^2+3x^{\frac{1}{2}}# a polynomial? How do you express #-16+5f^8-7f^3# in standard form? What is the degree of #16x^2y^3-3xy^5-2x^3y^2+2xy-7x^2y^3+2x^3y^2#? What is the degree of the polynomial #x^4-3x^3y^2+8x-12#? What is the difference between a monomial, binomial and polynomial? How do you write #y = 2/3x + 5# in standard form? See all questions in Polynomials in Standard Form Impact of this question 4880 views around the world You can reuse this answer Creative Commons License