Remainder Theorem and Factor Theorem
If a polynomial \(F(x)\) divided by \(p(x)\), the quotient is \(H(x)\) and the remainder is \(S(x)\), then the polynomial can be written as:
\(F(x) = p(x) \:.\: H(x) + S(x)\)
Example 01
Find the quotient and the remainder of \(2x^3 + 5x^2 - 6x + 3\) divided by \((x + 2)\)
\begin{array}{r} \bbox[5px, border: 2px solid red] {2x^2 + x - 8}\\ x + 2\enclose{longdiv}{2x^3 + 5x^2 - 6x + 3}\\ \underline{2x^3 + 4x^2}\hspace{4em}\\ x^2 - 6x + 3\hspace{.33em}\\ \underline{x^2 + 2x}\hspace{2em}\\ -8x + 3\hspace{.33em}\\ \underline{-8x - 16}\\ \bbox[5px, border: 2px solid blue] {19} \end{array}
Quotient: \({\color {red} 2x^2 + x - 8}\)
Remainder: \({\color {blue} 19}\)
\(2x^3 + 5x^2 - 6x + 3 = (x + 2) \:.\: (2x^2 + x - 8) + 19\)
Remainder Theorem
If a polynomial \(F(x)\) divided by \((x - a)\), the remainder is \(F(a)\).
Factor Theorem
If \((x - a)\) is a factor of a polynomial \(F(x)\), then \(F(a) = 0\).
Example 02
Find the remainder of \(F(x) = x^3 + 4x^2 - 6x + 1\) divided by \((x - 2)\)
\(F(x) = x^3 + 4x^2 - 6x + 1\)
\(F(2) = 2^3 + 4 \:.\: 2^2 - 6 \:.\: 2 + 1\)
\(F(2) = 13\)
We can write the polynomial as
\(x^3 + 4x^2 - 6x + 1 = (x - 2) \:.\: H(x) + 13\)
Example 03
Prove that \((x + 3)\) is a factor of \(F(x) = x^3 - x^2 - 10x + 6\)
\(F(-3) = (-3)^3 - (-3)^2 - 10(-3) + 6 = 0\)
Since \(F(-3) = 0\), then \((x + 3)\) is a factor of \(F(x)\).
Exercise