### More about polynomials Examples

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

• Let 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.
• $$\begin{array}{1}R.H.S. = \frac{A}{{x + 1}} + \frac{B}{{2x + 3}}\\ = {\frac{{A(2x + 3) + B(x + 1)}}{{(x + 1)(2x + 3)}}^{}}\\ = {\frac{{(2A + B)x + 3A + B}}{{(x + 1)(2x + 3)}}^{}}\end{array}$$
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),
$$\begin{array}{1}2(2) + B = 3\\B = - {1^{}}\end{array}$$
∴ $$\underline{\underline {A = 2{\rm{ }},{\rm{ }}B = - 1}}$$

• It 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 $${x^2} - x - 2$$.
• Suggested 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 $$\frac{{x + 4}}{{x + 1}}$$.