2.12 Rational expressions of
polynomials:
We have seen the following properties of
polynomials:
If p(x) is a polynomial then:
1.
Sum/difference/Product of two polynomials is a polynomial
2.
p(x)+0 = p(x)
3.
p(x)-p(x) = 0
4.
p(x)*1 =p(x)
Thus, operations of polynomials are similar to the
operations we carry out on integers.
Also if m, n and p are integers, we have learnt
that (m/n)+(p/n) = (m+p)/n and (m/n)+(p/q) = (mq+np)/nq.
Integers and polynomials are algebraically similar.
An expression of the form p(x)/q(x) where p(x) and
q(x) are polynomials and q(x) 
0 is called a ‘rational expression
of polynomial’.
Note: A rational expression is always represented
in its lowest term.(Numerator and denominator should not have any common
factors. If they have, then divide them by their HCF).
Notes:
If p(x)/q(x) and g(x)/h(x) are two rational
expressions, then:
1.
{p(x)/q(x)}*{g(x)/h(x)} = {p(x)*g(x)}/{q(x)*h(x)}
2.
{p(x)/q(x)}/{g(x)/h(x)} = {p(x)*h(x)}/{q(x)*g(x)}
3.
{p(x)/q(x)} 
{g(x)/h(x)} = {p(x)*h(x) {q(x)*g(x)}/{q(x)*h(x)}
4.
p(x)/q(x) + r(x)/s(x) = {p(x)*s(x)+q(x)r(x)}/{q(x)s(x)}
1.
p(x)/q(x) - p(x)/q(x) = 0 (-p(x)/q(x) is called ‘additive
inverse’ of p(x)/q(x))
2.
{p(x)/q(x)}/{p(x)/q(x)} = 1 (q(x)/p(x) is called ‘multiplicative
inverse’ of p(x)/q(x))
2.12 Problem 1: What should be subtracted from (x2+2)/(x-1)
to get (x-3)/(x+1) ?
Solution:
Let p(x)/q(x) be the rational expression to be
subtracted
(x2+2)/(x-1)
- p(x)/q(x) = (x-3)/(x+1)
(x2+2)/(x-1)
- (x-3)/(x+1) =p(x)/q(x)
LHS = {(x2+2)(x+1) –(x-1)(x-3)}/ {(x+1)(x-1)}
= {(x3+x2+2x+2) - (x2-3x-x+3)}/{(x2-1)}
= {(x3+x2+2x+2) - x2+4x-3}/{(x2-1)}
= {(x3+6x-1}/{(x2-1)}
Thus the rational expression to be subtracted is {(x3+6x-1}/{(x2-1)}
2.12 Problem 2. Simplify 1/(x+a) + 1/(x+b)+1/(x+c) +ax/(x3+ax2) +bx/(x3+bx2) + cx/(x3+cx2)
Solution:
The given equation =
1/(x+a) + ax/(x3+ax2)
+1/(x+b)+ bx/(x3+bx2)
+1/(x+c) + cx/(x3+cx2)(
Rearrange terms)
= {1/(x+a)} + {ax/x2(x+a)}
+ {1/(x+b)} + {bx/x2(x+b)} +{1/(x+c)} + {cx/x2(x+c)}
(Take common factor out)
= {1/(x+a)} + {a/x(x+a)} + {1/(x+b)} + {b/x(x+b)} + {1/(x+c)} + {c/x(x+c)}( Cancel same terms)
= {(x+a)/x(x+a)} + {(x+b)/x(x+b)} + {(x+c)/x(x+c)} ( Simplify)
= 1/x+1/x+1/x = 3/x
2.12 Problem 3. If P = (x3+y3)/{(x-y)2+3xy},
Q = {(x+y)2-3xy}/ (x3-y3) and R = xy/(x2-y2)
Find (P/Q)*R
Solution:
P = (x3+y3)/
{(x-y)2+3xy}
= (x+y) (x2-xy+y2)/
{(x2-2xy+y2)+3xy)}
(Apply formula for (x3+y3) and (x-y)2)
=(x+y) (x2-xy+y2)/
(x2+xy+y2)
Q = {(x+y)2-3xy}/ (x3-y3)
= {(x2+2xy+y2)-3xy)}/ {(x-y) (x2+xy+y2)}
(Apply formula for (x3-y3) and (x+y)2)
= {(x2-xy+y2)}/ {(x-y) (x2+xy+y2)}
R = xy/(x2-y2)
=xy/{(x+y)(x-y)}
P/Q = {(x+y) (x2-xy+y2)/ (x2+xy+y2)}/ [{(x2-xy+y2)} /{ (x-y) (x2+xy+y2)}]
= {(x+y) (x2-xy+y2)/
(x2+xy+y2)}*[{(x-y) (x2+xy+y2)}/ {(x2-xy+y2)}] (Reciprocal)
= (x+y)(x-y)
P/Q*R = (x+y)(x-y) * {xy/{(x+y)(x-y)}
= xy
2.12 Summary of
learning
|
No |
Points to remember |
|
1 |
Rules
of mathematical operations (+,-,*, / ) on rational expressions of polynomials
follow the same rules that are applicable to rational numbers |