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

Mar 27, 2018

a=1, -1 and 3
see below

#### Explanation:

we have ${a}^{3} + 3 {a}^{2} - a - 3$
first find a factor of ur expression
in this case
lets take a=1 (subs in expression)
${1}^{3} + 3 {\left(1\right)}^{2} _ 1 - 3 = 0$

now lets assume a=x

$\therefore$ we can say that x=1 => (x-1) is a factor of the expression
now
$\left(x - 1\right) \left(a {x}^{2} + b x + c\right)$
expand it
$a {x}^{3} + b {x}^{2} + c x - a {x}^{2} - b x - c$

collect the coefficients of ${x}^{3} , {x}^{2} \mathmr{and} x$
$a {x}^{3} + b {x}^{2} - a {x}^{2} + c x - b x - c$
now

$a {x}^{3} + \left(b - a\right) {x}^{2} + \left(c - b\right) x - c$ <------|
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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
$\left(x - 1\right) \left({x}^{2} + 4 x + 3\right)$
$\left(a - 1\right) \left({a}^{2} + 4 a + 3\right)$
$\left(a - 1\right) \left({a}^{2} + 3 a + a + 3\right)$
$\left(a - 1\right) \left[a \left(a + 3\right) + 1 \left(a + 3\right)\right]$
$\left(a - 1\right) \left(a + 1\right) \left(a + 3\right)$
$\therefore a = 1 , - 1 \mathmr{and} 3$
we can say that 1, -1 and 3 are all factors of ${a}^{3} + 3 {a}^{2} - a - 3$