Question

factorise a³+ 4a²+3a+1​

Answer 1

Answer:

[√{a(a+1)(a+3)}+i]×[√{a(a+1)(a+3)}-i]

Step-by-step explanation:

well, other factors can beade but look at this method.

a(a²+4a+3) +

a(a²+3a+a+3)+1

a[a(a+3)+1(a+3)]+1

a[ (a+1)(a+3)] +1

√[a(a+1)(a+3)]² +1²

also, a²+b² = (a+bi)(a-bi) where, i=√-1

now put a(left term) and b(right term=1)

[√{a(a+1)(a+3)}+i]×[√{a(a+1)(a+3)}-i]

Answer 2

Answer:

A:B= 3:4, then what is the value of the expression (3A^2+4B/3A-4B^2)? ... If 2a/b =1/2, what is the value of b²/a²? ... If a+1/a=-2 then how do you find the value of a⁴+a³+a²+a+1?