15.3 Practice Problems
Problem 15.1. The polynomial \(P(x)=3x^{3}+2x^{2}-px+q\) is divisible by \(x+1\) and leaves a remainder of \(10\) when divided by \(x-1\). Find \(p\) and \(q\).
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Solution. By the factor theorem, divisibility by \(x+1\) means \(P(-1)=0\): \[3(-1)^{3}+2(-1)^{2}-p(-1)+q=-3+2+p+q=0 \implies p+q=1 .\] By the remainder theorem, the remainder on division by \(x-1\) is \(P(1)\): \[3+2-p+q=10 \implies -p+q=5 .\] Adding the two equations gives \(2q=6\), so \(q=3\) and then \(p=-2\).
Check: with \(p=-2\) and \(q=3\) the polynomial is \(3x^{3}+2x^{2}+2x+3\). Then \(P(-1)=-3+2-2+3=0\) and \(P(1)=3+2+2+3=10\), as required.
Note 15.18. Note the sign care needed at \(-px\) with \(p\) negative: \(-px\) becomes \(+2x\), not \(-2x\). Substituting the answers back into the original polynomial, as above, is what catches this.
Note also how little work the two theorems did compared with dividing twice. Each condition became a single equation in \(p\) and \(q\), and two equations in two unknowns finish it.
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