Question

Using factor theorem, show that (x – 3) is a factor of \[(x^{3} - 7x^{2} + 15x - 9)\]. Hence, factorise the given expression completely.

Answer 1

Given: $$f(x) = x^3 - 7x^2 + 15x - 9$$.

To prove and find: $$(x - 3)$$ is a factor and the complete factorisation of $$f(x)$$.

Proof:

1. By the Factor Theorem: $$(x - \alpha)$$ is a factor of $$f(x) \Leftrightarrow f(\alpha) = 0$$.
2. $$f(3) = 27 - 63 + 45 - 9 = 0$$.
3. Therefore, $$(x - 3)$$ is a factor of $$f(x)$$.
4. On dividing $$f(x)$$ by $$(x - 3)$$, the quotient is $$x^2 - 4x + 3$$.
5. $$x^2 - 4x + 3 = (x - 1)(x - 3)$$.
6. Hence, $$x^3 - 7x^2 + 15x - 9 = (x - 3)(x - 1)(x - 3)$$.
7. Therefore, $$x^3 - 7x^2 + 15x - 9 = (x - 3)^2(x - 1)$$.

Hence proved.