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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)].

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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:

  1. By the Factor Theorem: $$(x - \alpha)$$ is a factor of $$f(x) \Leftrightarrow f(\alpha) = 0$$.
  2. $$f(4) = 2(4)^3 + 4^2 - 26(4) - 40 = 128 + 16 - 104 - 40 = 0$$.
  3. Therefore, $$(x - 4)$$ is a factor of $$f(x)$$.
  4. On dividing $$f(x)$$ by $$(x - 4)$$, the quotient is $$2x^2 + 9x + 10$$.
  5. $$2x^2 + 9x + 10 = (2x + 5)(x + 2)$$.
  6. Hence, $$2x^3 + x^2 - 26x - 40 = (x - 4)(2x + 5)(x + 2)$$.

Hence proved.

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Chapter 8: Remainder Theorem and Factor Theorem - EXERCISE 8B [Page 127]

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R.S. Aggarwal Mathematics [English] Class 10 ICSE
Chapter 8 Remainder Theorem and Factor Theorem
EXERCISE 8B | Q 6. | Page 127
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