Question

Given, $$9x^2 - 4$$ is a factor of $$9x^3 - mx^2 - nx + 8$$. 

1. find the values of $$m$$ and $$n$$ using the remainder and factor theorem 
2. factorise the given polynomial completely.

Answer 1

i. Find the values of m and n

Let \[f(x) = 9x^3 - mx^2 - nx + 8.\]

The given factor can be written as \[9x^2 - 4 = (3x - 2)(3x + 2).\]

At the first zero:

\[f\left(\frac{2}{3}\right) = 9\left(\frac{2}{3}\right)^3 - m\left(\frac{2}{3}\right)^2 - n\left(\frac{2}{3}\right) + 8\]

\[= \frac{8}{3} - \frac{4m}{9} - \frac{2n}{3} + 8\]

\[= 0\]

Multiplying by 9 gives

\[24 - 4m - 6n + 72 = 0\]

\[\Rightarrow 4m + 6n = 96\]

\[\Rightarrow 2m + 3n = 48 \quad(1)\]

At the second zero:

\[f\left(-\frac{2}{3}\right) = 9\left(-\frac{2}{3}\right)^3 - m\left(-\frac{2}{3}\right)^2 - n\left(-\frac{2}{3}\right) + 8\]

\[= -\frac{8}{3} - \frac{4m}{9} + \frac{2n}{3} + 8\]

\[= 0\]

Multiplying by 9 gives

\[-24 - 4m + 6n + 72 = 0\]

\[\Rightarrow 4m-6n=48\]

\[\Rightarrow 2m - 3n = 24 \quad(2)\]

Adding the two equations,

\[(2m + 3n) + (2m - 3n) = 48 + 24\]

\[\Rightarrow 4m = 72\]

\[\Rightarrow m=18\]

Substituting into the first equation,

\[2(18)+3n=48\]

\[\Rightarrow 3n = 12\]

\[\Rightarrow n = 4\]

ii. Factorise the polynomial completely

With the obtained values,

\[9x^3 - 18x^2 - 4x + 8 = (9x^2 - 4)(x - 2)\]

\[= (3x - 2)(3x + 2)(x - 2)\]

Thus, \[m = 18, n = 4.\]