Question

3.
Factorise x3 - 18x2 + 99x - 162, using factor
theorem.​

Answer 1

Answer:

$x^3-18x^2+99x-162=(x-3)(x-6)(x-9)$

Step-by-step explanation:

Factor theorem:

(x-a) is a factor of f(x) iff f(a) =0

Let,

$f(x)=x^3-18x^2+99x-162$

put x=3,

$f(3)=3^3-18(3)^2+99(3)-162$

$f(3)=27-162+297-162$

$f(3)=324-324$

$f(3)=0$

By factor theorem, (x-3) is a factor

$x^3-18x^2+99x-162=(x-3)(x^2+px+54)$

Equating coefficients of x on both sides, we get

$99=-3p+54$

$-3p=45$

$p=-15$

$\implies\:$other factor is $x^2-15x+54$

$=(x-6)(x-9)$

$x^3-18x^2+99x-162=(x-3)(x-6)(x-9)$