When P(x)=(ax+b)Q(x)+R, with a 0, the remainder R is constant. Set x=-b/a. The product containing Q becomes zero, so R=P(-b/a). This is why substitution replaces division when only the remainder is wanted. Method: set the divisor equal to zero; substitute that value into every term of P; simplify exactly. Example: for P(x)=2x3+x+1, division by 2x+1 leaves P(- 12)=- 14- 12+1= 14. Check: 2x+1=0 gives - 12, not -1. Dividing P(- 12) by 2 again is incorrect.

When \(\mathrm{P}(x)=(ax+b)\mathrm{Q}(x)+R\), with \(a\ne0\), the remainder \(R\) is constant.

Set \(x=-\dfrac{b}{a}\). The product containing \(Q\) becomes zero, so \(R=\mathrm{P}\left(-\dfrac{b}{a}\right)\). This is why substitution replaces division when only the remainder is wanted.

Method: set the divisor equal to zero; substitute that value into every term of \(P\); simplify exactly.

Example: for \(\mathrm{P}(x)=2x^3+x+1\), division by \(2x+1\) leaves \(\mathrm{P}\left(-\dfrac12\right)=-\dfrac14-\dfrac12+1=\dfrac14\).

Check: \(2x+1=0\) gives \(-\dfrac12\), not \(-1\). Dividing \(\mathrm{P}\left(-\dfrac12\right)\) by 2 again is incorrect.

Remainders from linear divisors

Remainders from linear divisors