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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
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Show that p(x) is not a multiple of g(x), when p(x) = x3 + x – 1 g(x) ...
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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Show that p(x) is not a multiple of g(x), when p(x) = x3 + x – 1 g(x) ...
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