More about polynomials Examples

ExamplesSimplify \(\frac{{{x^2} + 2x  3}}{{{x^2} + 7x + 12}} \times \frac{{{x^2}  2x  15}}{{{x^2}  2x  8}} \div \frac{{{x^2}  2x  15}}{{{x^2} + 6x + 8}}\).

Solutions\(\begin{array}{1}{\frac{{{x^2} + 2x  3}}{{{x^2} + 7x + 12}} \times \frac{{{x^2}  2x  15}}{{{x^2}  2x  8}} \div \frac{{{x^2}  2x  15}}{{{x^2} + 6x + 8}}}
= {\frac{{{x^2} + 2x  3}}{{{x^2} + 7x + 12}} \times \frac{{{x^2}  2x  15}}{{{x^2}  2x  8}} \times \frac{{{x^2} + 6x + 8}}{{{x^2}  2x  15}}}\\
= {\frac{{(x  1)(x + 3)(x + 3)(x  5)(x + 2)(x + 4)}}{{(x + 3)(x + 4)(x + 2)(x  4)(x + 3)(x  5)}}^{}}\\
= {\underline{\underline {\frac{{x  1}}{{x  4}}}} ^{}}\end{array}\)

ExamplesLet A and B be constants. If \(\frac{{3x + 5}}{{(x + 1)(2x + 3)}} \equiv \frac{A}{{x + 1}} + \frac{B}{{2x + 3}}\), find the values of A and B.

Solutions\(\begin{array}{1}R.H.S. = \frac{A}{{x + 1}} + \frac{B}{{2x + 3}}\\ = {\frac{{A(2x + 3) + B(x + 1)}}{{(x + 1)(2x + 3)}}^{}}\\ = {\frac{{(2A + B)x + 3A + B}}{{(x + 1)(2x + 3)}}^{}}\end{array}\)
By comparing the like terms on the L.H.S. and R.H.S., we have
\(\left\{ \begin{array}{l}2A + B = 3\;{\kern 1pt} .....................\;(1)\\3A + B = 5\;{\kern 1pt} .....................\;{(2)^{}}\end{array} \right.\)
(2)  (1), \(A = 2\)
Substitute \(A = 2\) into (1),
\(\begin{array}{1}2(2) + B = 3\\B =  {1^{}}\end{array}\)
∴ \(\underline{\underline {A = 2{\rm{ }},{\rm{ }}B =  1}} \)

ExamplesIt is given that \(f(x) = \frac{1}{{{x^2} + 2x  8}}\) is a rational function.
Find another rational function \(g(x)\) such that the denominator of \(f(x) \times g(x)\) is \({x^2}  x  2\). 
SolutionsSuggested answer:
\(\begin{array}{1} f(x) = \frac{1}{{{x^2} + 2x  8}}\\ = \frac{1}{{(x  2)(x + 4)}}\end{array}\)
Denominator of
\(\begin{array}{1} f(x)\times g(x) = {x^2}  x  2\\ = (x  2)(x + 1)\end{array}\)
Let \(f(x) \times g(x) = \frac{{P(x)}}{{(x  2)(x + 1)}}\), where \(P(x)\) is a polynomial.
\(\begin{array}{1}\frac{1}{{(x  2)(x + 4)}} \times g(x) = \frac{{P(x)}}{{(x  2)(x + 1)}}\\g(x) = \frac{{P(x)(x  2)(x + 4)}}{{(x  2)(x + 1)}}\\ = \frac{{P(x)(x + 4)}}{{x + 1}}\end{array}\)
∴ One possible \(g(x)\) is \(\frac{{x + 4}}{{x + 1}}\).