Theory of Intelligence
11TH•Mathematics•Partial Fractions•Question 8

Solution

Resolve into partial fractions: $\dfrac{(x-1)(x-2)(x-3)}{(x-4)(x-5)(x-6)}$

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Solution

(x−1)(x−2)(x−3)(x−4)(x−5)(x−6)\frac{(x-1)(x-2)(x-3)}{(x-4)(x-5)(x-6)}

Since degrees are equal, divide first.

(x−1)(x−2)(x−3)(x−4)(x−5)(x−6)=1+R(x)(x−4)(x−5)(x−6)\frac{(x-1)(x-2)(x-3)}{(x-4)(x-5)(x-6)}=1+\frac{R(x)}{(x-4)(x-5)(x-6)}

Now let:

(x−1)(x−2)(x−3)(x−4)(x−5)(x−6)=1+Ax−4+Bx−5+Cx−6\frac{(x-1)(x-2)(x-3)}{(x-4)(x-5)(x-6)}=1+\frac{A}{x-4}+\frac{B}{x-5}+\frac{C}{x-6}

Solving gives A=3A=3, B=−24B=-24, C=30C=30.

(x−1)(x−2)(x−3)(x−4)(x−5)(x−6)=1+3x−4−24x−5+30x−6\boxed{\frac{(x-1)(x-2)(x-3)}{(x-4)(x-5)(x-6)}=1+\frac{3}{x-4}-\frac{24}{x-5}+\frac{30}{x-6}}

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