UPSKILL MATH PLUS
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If \(p(x)\) is a polynomial of degree \(n \geq 1\)and ‘\(a\)’ is any real number then
(i) \(p(a) = 0\) implies \((x - a)\) is a factor of \(p(x)\).
(ii) \((x-a)\) is a factor of \(p(x)\) implies \(p(a) = 0\).
(i) \(p(a) = 0\) implies \((x - a)\) is a factor of \(p(x)\).
(ii) \((x-a)\) is a factor of \(p(x)\) implies \(p(a) = 0\).
Proof:
Let \(p(x)\) be the dividend and \((x-a)\) be the divisor.
Division algorithm
Given any two integers \(a\) and \(b\) with \(a > 0\), there exists a unique interger \(q\) and \(r\) such that \(b = qa + r\) with \(0 \leq r < a\).
By division algorithm we have, \(p(x) = q(x)(x-a) + p(a)\) where \(q(x)\) is the quotient and \(p(a)\) is the remainder.
(i) Assume \(p(a) = 0\).
To prove: \((x - a)\) is a factor of \(p(x)\).
Substitute \(p(a) = 0\) in \(p(x) = q(x)(x-a) + p(a)\).
Then, \(p(x) = q(x)(x-a)\)
This implies, \((x - a)\) is a factor of \(p(x)\).
(ii) Assume \((x-a)\) is a factor of \(p(x)\).
To prove: \(p(a) = 0\).
By assumption, we have \(p(x) = q(x)(x-a)\), where \(q(x)\) is some polynomial.
Substitute \((x = a)\) in \(p(x) = q(x)(x-a)\).
Then, \(p(a) = q(a)(a-a)\)
\(= q(a)(0)\)
\(= 0\)
Hence, the proof.
Example:
Check whether \(x - 5\) is the factor of \(x^{3} + x^{2} + 2x - 3\).
Given:
The polynomial \(p(x) = x^{3} + x^{2} + 2x - 3\).
To check:
If \(x - 5\) is the factor of \(p(x)\).
Solution:
Step 1: Find the zero of the polynomial \(x = a\).
Equate \(x - 5\) to zero and solve for \(x\).
\(x\) \(-\) \(5\) \(=\) \(0\)
\(x\) \(=\) \(5\)
Step 2: Check if \(p(5) = 0\).
By the factor theorem, for \(x - 5\) to be a factor of \(p(x)\) it must satisfy \(p(5) = 0\).
Substitute \(x = 5\) in \(p(x) = x^{3} + x^{2} + 2x - 3\).
\(p(5) = 5^{3} + 5^{2} + 2(5) - 3\)
\(= 125 + 25 + 10 - 3\)
\(= 157\)
Here, \(p(5) \neq 0\).
This implies that, \(x - 5\) is not the factor of \(x^{3} + x^{2} + 2x - 3\).
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