Question

представьте в виде многочлена:
(4+а)^2=
(2х-1)^2=
(2а+3b)^2=
(х^3-3)^2=

Answer 1

(4+а)^2=4^2+(2*4*a)+a^2=16+8a+a^2

(2х-1)^2=(2x)^2-(2*2x*1)+1^2=4x^2-4x+1

(2а+3b)^2=(2a)^2+(2*2a*3b)+(3b)^2=4a^2+12ab+9b^2

(х^3-3)^2=(x^3)^2-(2*x^3*3)+3^2=x^6-6x^3+9

Answer 2

${(4 + a)}^{2} = {a}^{2} + 8a + 16$
${(2x - 1)}^{2} = {4x}^{2} - 4x + 1$
${(2a + 3b)}^{2} = {4a}^{2} + 12ab + {9b}^{2}$
${( {x}^{3} - 3) }^{2} = {x}^{6} - {6x}^{3} + 9$