Proof. Dividing gives a quotient \(Q(x)\) and a remainder \(R\) with \[P(x)=(x-a)Q(x)+R .\] The remainder is a constant, because dividing by a linear factor must leave something of lower degree. Substituting \(x=a\), \[P(a)=(a-a)Q(a)+R=0\cdot Q(a)+R=R .\] □

Solution.

\(f(2)=2^3-7(2)^2+6(2)-2=8-28+12-2 =-10\)

Solution.

\begin {align*} x+3=0 & \implies x=-3\\\\ R & = P(-3) = 2(-3)^3+2(-3)+31=-54-6+31\\ & = -60 + 31\\ & = -29\\ \end {align*}

Solution.

Let \(P(x)=2x^3+ax^2+bx+31\)

1.

The remainder when \(P(x)\) is divided by \(x+2\) is given by \begin {align*} P(-2) & = 2(-2)^3+a(-2)^2+b(-2)+31\\ & = -16+4a-2b+31\\ & = 4a-2b+15 \end {align*}

but this remainder is 11 \begin {align*} 4a-2b+15 & = 11 \end {align*}

that is \begin {equation} \label {eq:polyone} 4a-2b=-4 . \end {equation}

2.

The remainder when \(P(x)\) is divided by \(x+3\) is \begin {align*} P(-3) & = 2(-3)^3+a(-3)^2+b(-3)+31\\ & = -54+9a-3b+31\\ & = 9a-3b-23 \end {align*}

But the remainder is \(-29\) \begin {align*} 9a-3b-23 & = -29 \end {align*}

that is \begin {equation} \label {eq:polytwo} 9a-3b=-6 . \end {equation}

Solving simultaneously: multiplying (2) by \(3\) and (3) by \(2\) makes the coefficients of \(b\) agree, \begin {align*} 12a-6b & = -12\\ 18a-6b & = -12 . \end {align*}

Subtracting the first from the second gives \(6a=0\), so \(a=0\). Substituting \(a=0\) into (2), \[-2b=-4 \qquad \implies \qquad b=2 .\] Hence \(a=0\) and \(b=2\).