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${{{x^2} + 13x + 15} \over {(2x + 3)\,{{(x + 3)}^2}}} = $
${1 \over {x + 3}} - {1 \over {2x + 3}} + {5 \over {{{(x + 3)}^2}}}$
${1 \over {2x + 3}} - {1 \over {x + 3}} + {5 \over {{{(x + 3)}^2}}}$
${1 \over {2x + 3}} + {1 \over {x + 3}} - {5 \over {{{(x + 3)}^2}}}$
${1 \over {2x + 3}} - {1 \over {x + 3}} - {5 \over {{{(x + 3)}^2}}}$
Solution
(a) ${{{x^2} + 13x + 15} \over {(2x + 3)\,{{(x + 3)}^2}}} = {A \over {2x + 3}} + {B \over {x + 3}} + {C \over {{{(x + 3)}^2}}}$
==> ${x^2} + 13x + 15 = A{(x + 3)^2} + B(2x + 3)\,(x + 3) + C(2x + 3)$
For $x = – 3,\,C = 5$ and for $x = – {3 \over 2};\,A = – 1$
Equating coefficient of ${x^2}$
$1 = A + 2B \Rightarrow B = {{1 – A} \over 2} = 1$
$\therefore $ Given expression = ${1 \over {x + 3}} – {1 \over {2x + 3}} + {5 \over {{{(x + 3)}^2}}}$.
