Question

Using the factor theorem, show that (x – 2) is a factor of \[x^{3} + x^{2} - 4x - 4\]. Hence factorise the polynomial completely.

Answer 1

Given: $$f(x) = x^3 + x^2 - 4x - 4$$.

To prove and find: $$(x - 2)$$ 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. Let $$f(x) = x^3 + x^2 - 4x - 4$$.
3. The constant term in $$f(x)$$ is – 4.
4. Its factors are 1, –1, 2, –2, 4 and – 4.
5. Now, $$f(2) = 2^3 + 2^2 - 4 \times 2 - 4 = 0$$.
6. Therefore, $$(x - 2)$$ is a factor of $$f(x) = x^3 + x^2 - 4x - 4$$.
7. On dividing $$f(x)$$ by $$(x - 2)$$, we get quotient $$(x^2 + 3x + 2)$$.
8. Therefore, $$(x^3 + x^2 - 4x - 4) = (x - 2)(x^2 + 3x + 2)$$.
9. $$(x^3 + x^2 - 4x - 4) = (x - 2)(x^2 + 2x + x + 2)$$.
10. $$(x^3 + x^2 - 4x - 4) = (x - 2)[x(x + 2) + (x + 2))$$.
11. Hence, $$x^3 + x^2 - 4x - 4 = (x - 2)(x + 1)(x + 2)$$.

Hence proved.