I'm supposed to show that: \(\displaystyle{y}={\frac{{{5}{\left({x}-{1}\right)}{\left({x}+{2}\right)}}}{{{\left({x}-{2}\right)}{\left({x}+{3}\right)}}}}={P}+{\frac{{{Q}}}{{{\left({x}-{2}\right)}}}}+{\frac{{{R}}}{{{\left({x}+{3}\right)}}}}\)

Trent Fuller

Trent Fuller

Answered question

2022-04-06

I'm supposed to show that:
y=5(x1)(x+2)(x2)(x+3)=P+Q(x2)+R(x+3)

Answer & Explanation

star04iks7

star04iks7

Beginner2022-04-07Added 14 answers

y=5(x1)(x+2)(x2)(x+3)=P+Q(x2)+R(x+3)
gives
5(x1)(x+2)=P(x2)(x+3)+Q(x+3)+R(x2) (1)
Now put x=2,x=3,x=0 respectively on both sides of (1) to get, Q=4,R=4,P=5 respectively. Hence you can reach the desired result.
posciad48o

posciad48o

Beginner2022-04-08Added 15 answers

You are given the wrong expression. After some calculation I figure out that in order for these two expressions to be equivalent,
5(x1)(x2)(x2)(x+3)=5+4(x2)+4(x+3)
instead of having 5(x1)(x2)(x2)(x+3), it should be
5(x1)(x+2)(x2)(x+3)
Now solve it again it would work well.
My approach to solving it is
5(x1)(x+2)(x2)(x+3)=5x2+510(x2)(x+3)=5x2+510(x2)(x+3)
5x2+510(x2)(x+3)=5x+15+40(x2)(x+3)
5x+15+40(x2)(x+3)=5+40(x2)(x+3)
5+40(x2)(x+3)=5+4(x2)+4(x+3)

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