Question

Given, 9x² – 4 is a factor of 9x³ – mx² + nx + 8: 

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

Answer 1

Given:

9x² – 4 is a factor of 9x³ – mx² + nx + 8.

So, (3x – 2)(3x + 2) are factors; by the factor/remainder theorem, `f(2/3) = 0` and `f(-2/3) = 0`.

Step-wise calculation:

1. Let f(x) = 9x³ – mx² + nx + 8.

2. Evaluate at `x = 2/3`: 

`f(2/3) = 9(2/3)^3 - m(2/3)^2 + n(2/3) + 8`

= `8/3 - (4m)/9 + (2n)/3 + 8`

= 0 

Multiply by 9: 96 – 4m + 6n = 0

Simplify: 48 – 2m + 3n = 0   ...(Equation 1)

3. Evaluate at `x = -2/3`:

`f(-2/3) = 9(-2/3)^3 - m(-2/3)^2 + n(-2/3) + 8`

= `-8/3 - (4m)/9 - (2n)/3 + 8`

= 0

Multiply by 9: 48 – 4m – 6n = 0

Simplify: 24 – 2m – 3n = 0   ...(Equation 2)

4. Solve the system:

Equation 1: –2m + 3n + 48 = 0

Equation 2: –2m – 3n + 24 = 0

Subtract equation 1 from equation 2:

(–2m – 3n) – (–2m + 3n) = –24 – (–48)

⇒ –6n = 24 

⇒ n = –4 

Substitute n = –4 into equation 2:

–2m – 3(–4) + 24 = 0

⇒ –2m + 12 + 24 = 0

⇒ –2m + 36 = 0

⇒ m = 18

5. With m = 18 and n = –4

f(x) = 9x³ – 18x² – 4x + 8

Since 9x² – 4 is a known factor, divide or match:

(9x² – 4)(x – 2) = 9x³ – 18x² – 4x + 8 

So, the full factorisation is (9x² – 4)(x – 2) = (3x – 2)(3x + 2)(x – 2).