Question

Factorise. a³-3a²-9a-5

Answer 1

Step-by-step explanation:

The given question is solved in the uploaded photos

Additional information-

For the given polynomial,

$p(x) = a {x}^{3} + b {x}^{2} + cx + d$

(where,

a,b and c are rational constants and x is the

variable)

$find \: thefactors \: of \: \frac{d}{a} \\ and \: then \: put \: each \: of \: the \\ factors \: in \: place \: of \: x \: in \: p(x).$

Then find the zero of p(x) ,among the factors,

by using trial and error method,solve the rest(shown in my photos)

Answer 2

p(a)=a³-3a²-9a-5

p(1)=1³-3*1²-9*1-5

=1-3-9-5

=-16 , which is not equal to 0

therefore,

p(-1)=(-1)³-3*(-1)²-9*(-1)-5

=-1-3+9-5

=-4+4

=0

therefore

a+1 is a factor of a³-3a²-9a-5

now,

a³-3a²-9a-5/a+1

quotient=a²-4a-5

remainder=0

now,

a³-3a²-9a-5=(a+1)(a²-4a-5)

=.(a+1)(a²+a-5a-5)

=(a+1){a(a+1)-5(a+1)}

=(a+1)(a+1)(a-5)

hope it will helpfull