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How do I solve `(3p^2-3p-60)/(p^2-2p-8)-:(4p^2-12p-40)/(4p^2-24p+20)` ?

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monique06 | Valedictorian

Posted May 17, 2013 at 1:54 PM via web

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How do I solve

`(3p^2-3p-60)/(p^2-2p-8)-:(4p^2-12p-40)/(4p^2-24p+20)` ?

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mjripalda | High School Teacher | (Level 3) Educator

Posted May 17, 2013 at 2:56 PM (Answer #1)

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`(3p^2-3p-60)/(p^2-2p-8)-:(4p^2-12p-40)/(4p^2-24p+20)`

First, factor the polynomials present.

`=(3 (p-5)(p+4))/((p-4)(p+2))-:(4 (p-5)(p+2))/(4 (p-5)(p-1))`

` `Then, flip the second rational expression and change the operation from division to multiplication.

`=(3 (p-5)(p+4))/((p-4)(p+2)) xx (4 (p-5)(p-1))/(4 (p-5)(p+2))`

` `Cancel common factors between numerators and denominators. 

 `=(3 (p-5)(p+4))/((p-4)(p+2))xx((p-1))/((p+2))`

` ` And multiply straight across.

`=(3 (p-5)(p+4)(p-1))/((p-4)(p+2)^2)`

` `Hence,`(3p^2-3p-60)/(p^2-2p-8)-:(4p^2-12p-40)/(4p^2-24p+20)=(3 (p-5)(p+4)(p-1))/((p-4)(p+2)^2).`

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oldnick | Valedictorian

Posted May 18, 2013 at 3:23 AM (Answer #2)

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`(3p^2-3p-60)/(p^2-2p-8) : (4p^2-12p-40)/(4p^2-24p+20)=` `(3(p^2-p-20))/(p^2-2p-8) xx (4(p^2-6p+5))/(4(p^2-3p-10))=`

`=(3(p^2-p-20))/(p^2-2p-8) xx (p^2-6p+5)/(p^2-3p-10)=` `(3(p^2-5p+4p-20))/(p^2-4p+2p-8)xx (p^2-5p-p+5)/(p^2-5p+2p-10)=`

`=(3[p(p-5)+4(p-5)])/[p(p-4)+2(p-4)] xx [p(p-5)-(p-5)]/[p(p-5)+2(p-5)]` `=[3(p-5)(p+4)]/[(p+2)(p-4)] xx [(p-1)(p-5)]/[(p+2)(p-5)]=` `[3(p-5)(p+4)(p-1)]/[(p+2)^2(p-4)]`

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