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 (x^{2}+2)/(x-1)
to get (x-3)/(x+1) ?
Solution:
Let p(x)/q(x) be the rational expression to be
subtracted
_{} (x^{2}+2)/(x-1)
- p(x)/q(x) = (x-3)/(x+1)
_{} (x^{2}+2)/(x-1)
- (x-3)/(x+1) =p(x)/q(x)
LHS = {(x^{2}+2)(x+1) –(x-1)(x-3)}/ {(x+1)(x-1)}
= {(x^{3}+x^{2}+2x+2) - (x^{2}-3x-x+3)}/{(x^{2}-1)}
= {(x^{3}+x^{2}+2x+2) - x^{2}+4x-3}/{(x^{2}-1)}
= {(x^{3}+6x-1}/{(x^{2}-1)}
Thus the rational expression to be subtracted is {(x^{3}+6x-1}/{(x^{2}-1)}
2.12 Problem 2. Simplify 1/(x+a) + 1/(x+b)+1/(x+c) +ax/(x^{3}+ax^{2}) +bx/(x^{3}+bx^{2}) + cx/(x^{3}+cx^{2})
Solution:
The given equation =
1/(x+a) + ax/(x^{3}+ax^{2})
+1/(x+b)+ bx/(x^{3}+bx^{2})
+1/(x+c) + cx/(x^{3}+cx^{2})(
Rearrange terms)
= {1/(x+a)} + {ax/x^{2}(x+a)}
+ {1/(x+b)} + {bx/x^{2}(x+b)} +{1/(x+c)} + {cx/x^{2}(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 = (x^{3}+y^{3})/{(x-y)^{2}+3xy},
Q = {(x+y)^{2}-3xy}/ (x^{3}-y^{3}) and R = xy/(x^{2}-y^{2})
Find (P/Q)*R
Solution:
P = (x^{3}+y^{3})/
{(x-y)^{2}+3xy}
= (x+y) (x^{2}-xy+y^{2})/
{(x^{2}-2xy+y^{2})+3xy)}
(Apply formula for (x^{3}+y^{3}) and (x-y)^{2})
=(x+y) (x^{2}-xy+y^{2})/
(x^{2}+xy+y^{2})
Q = {(x+y)^{2}-3xy}/ (x^{3}-y^{3})
= {(x^{2}+2xy+y^{2})-3xy)}/ {(x-y) (x^{2}+xy+y^{2})}
(Apply formula for (x^{3}-y^{3}) and (x+y)^{2})
= {(x^{2}-xy+y^{2})}/ {(x-y) (x^{2}+xy+y^{2})}
R = xy/(x^{2}-y^{2})
=xy/{(x+y)(x-y)}
_{}P/Q = {(x+y) (x^{2}-xy+y^{2})/ (x^{2}+xy+y^{2})}/ [{(x^{2}-xy+y^{2})} /{ (x-y) (x^{2}+xy+y^{2})}]
= {(x+y) (x^{2}-xy+y^{2})/
(x^{2}+xy+y^{2})}*[{(x-y) (x^{2}+xy+y^{2})}/ {(x^{2}-xy+y^{2})}] (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 |