Show that p(x) is not a multiple of g(x), when p(x) = x3 + x – 1 g(x) = 3x – Related: Very Short Answers - Polynomials?

Question

Show that p(x) is not a multiple of g(x), when p(x) = x3 + x – 1 g(x) = 3x – Related:  Very Short Answers - Polynomials?

Answer

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Answer

Understanding Polynomial Division

To determine if $ p(x) = x^3 + x - 1 $ is a multiple of $ g(x) = 3x $, we can use polynomial division or check the remainder when dividing $ p(x) $ by $ g(x) $.

Step 1: Polynomial Division

- When dividing polynomials, if $ p(x) $ is a multiple of $ g(x) $, the remainder must be zero.

Step 2: Dividing $ p(x) $ by $ g(x) $

- Divide $ p(x) $ by $ g(x) $:
$$
\frac{x^3 + x - 1}{3x}
$$
- The first term of the division gives $ \frac{1}{3} x^2 $.
- Multiply $ g(x) $ by $ \frac{1}{3} x^2 $:
$$
\frac{1}{3} x^2 (3x) = x^3
$$
- Subtract this from $ p(x) $:
$$
(x^3 + x - 1) - (x^3) = x - 1
$$

Step 3: Continuing the Division

- Now, divide $ x - 1 $ by $ 3x $:
$$
\frac{x - 1}{3x}
$$
- The first term gives $ \frac{1}{3} $.
- Multiply $ g(x) $ by $ \frac{1}{3} $:
$$
\frac{1}{3} (3x) = x
$$
- Subtract:
$$
(x - 1) - (x) = -1
$$

Final Conclusion

- The remainder is $ -1 $, which is not zero.
- Hence, $ p(x) $ is not a multiple of $ g(x) $.

Key Point

- If the remainder is not zero when dividing $ p(x) $ by $ g(x) $, then $ p(x) $ cannot be expressed as $ k \cdot g(x) $ for any polynomial $ k(x) $. Thus, $ p(x) $ is not a multiple of $ g(x) $.

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