Come consideri # x ^ 2 + 16x + 64 #?
Risposta:
#(x+8)^2#
Spiegazione:
#x^2+16x+64" is a "color(blue)"perfect square"#
#"that is "(x+a)^2=x^2+2ax+a^2#
#"compare the coefficients of the x-term"#
#rArr2a=16rArra=8#
#rarrx^2+16a+64=(x+8)^2#
#(x+8)^2#
#x^2+16x+64" is a "color(blue)"perfect square"#
#"that is "(x+a)^2=x^2+2ax+a^2#
#"compare the coefficients of the x-term"#
#rArr2a=16rArra=8#
#rarrx^2+16a+64=(x+8)^2#