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Question
Using factor theorem, show that (x – 4) is a factor of \[(2x^{3} + x^{2} - 26x - 40)\] and hence factorise \[(2x^{3} + x^{2} - 26x - 40)\].
Theorem
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Solution
Given: $$f(x) = 2x^3 + x^2 - 26x - 40$$.
To prove: $$(x - 4)$$ is a factor and the complete factorisation of $$f(x)$$.
Proof:
- By the Factor Theorem: $$(x - \alpha)$$ is a factor of $$f(x) \Leftrightarrow f(\alpha) = 0$$.
- $$f(4) = 2(4)^3 + 4^2 - 26(4) - 40 = 128 + 16 - 104 - 40 = 0$$.
- Therefore, $$(x - 4)$$ is a factor of $$f(x)$$.
- On dividing $$f(x)$$ by $$(x - 4)$$, the quotient is $$2x^2 + 9x + 10$$.
- $$2x^2 + 9x + 10 = (2x + 5)(x + 2)$$.
- Hence, $$2x^3 + x^2 - 26x - 40 = (x - 4)(2x + 5)(x + 2)$$.
Hence proved.
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