How do you factor #a^3 + 3a^2 - a - 3#?

1 Answer
Mar 27, 2018

Answer:

a=1, -1 and 3
see below

Explanation:

we have #a^3+3a^2-a-3#
first find a factor of ur expression
in this case
lets take a=1 (subs in expression)
#1^3+3(1)^2_1-3=0#

now lets assume a=x

#therefore# we can say that x=1 => (x-1) is a factor of the expression
now
#(x-1)(ax^2+bx+c)#
expand it
#ax^3+bx^2+cx-ax^2-bx-c#

collect the coefficients of #x^3, x^2 and x#
#ax^3+bx^2-ax^2+cx-bx-c#
now

#ax^3+(b-a)x^2+(c-b)x-c# <------|
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now compare the coefficients of this and original expression

we get a=1
-c=-3
c=3
b-a=3
b-1=3
b=3+1
b=4
now
#(x-1)(x^2+4x+3)#
#(a-1)(a^2+4a+3)#
now simplify the quadratic expression
#(a-1)(a^2+3a+a+3)#
#(a-1)[a(a+3)+1(a+3)]#
#(a-1)(a+1)(a+3)#
#therefore a=1, -1 and 3#
we can say that 1, -1 and 3 are all factors of #a^3+3a^2-a-3#