Question

Show that $$(x - 3)$$ is a factor of $$(2x^{3} - 3x^{2} - 11x + 6)$$ and hence factorise $$(2x^{3} - 3x^{2} - 11x + 6)$$.

Answer 1

Let $$f(x) = 2x^{3} - 3x^{2} - 11x + 6.$$ 

By the Factor Theorem, $$(x - 3)$$ is a factor if $$f(3) = 0.$$ 

Now, $$f(3) = 2(3)^{3} - 3(3)^{2} - 11(3) + 6$$ 

$$= 54 - 27 - 33 + 6$$ 

$$= 0$$ 

Therefore, $$(x - 3)$$ is a factor. 

On dividing the polynomial by $$(x - 3)$$, we obtain $$2x^{3} - 3x^{2} - 11x + 6 = (x - 3)(2x^{2} + 3x - 2).$$ 

Now, $$2x^{2} + 3x - 2 = 2x^{2} + 4x - x - 2$$

$$= 2x(x + 2) - (x + 2)$$

$$= (2x - 1)(x + 2)$$

Hence, the complete factorisation is $${2x^{3} - 3x^{2} - 11x + 6 = (x - 3)(2x - 1)(x + 2)}.$$