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# how do you solve r^2-4r-5 / 5r-25?

*confused*

r^2 - 4r - 5

_____________

5r - 25

### 5 Answers

- 1 decade agoFavorite Answer
(r^2 - 4r - 5)/[5(r-5)] = (r-5)(r+1)/5(r-5)

simplify by r-5:

(r+1)/5

that's your result :)

if you're not sure how to find roots of r^2 - 4r - 5, here's how you do it.

1. Find the delta (Δ)

Δ = b^2 - 4*a*c.

here:

a=1

b=-4

c=-5

so, Δ = (-4)^2 - 4*1*(-5) = 16 - (-20) = 16+20 = 36

then, the roots of the polynomial are:

x1 = [-b-sqrt(Δ)]/2 and x2 = [-b+sqrt(Δ)]/2

(sqrt(Δ) - square root of delta)

so, let's calculate it now

x1 = [4-sqrt(36)]/2 and x2 = [4+sqrt(36)]/2

x1 = [4-6]/2 and x2 = [4+6]/2

x1 = -2/2 and x2 = 10/2

x1 = -1 and x2 = 5

so as the polynomial is: a(x-x1)(x-x2), it looks like that:

1*(x+1)(x-5) = (x+1)(x-5)

yeah, hope it helps ;)

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- digbyLv 44 years ago
[r + 5/(r^2) + 5r - 14] / (r^2 + 4r - 21/r -2] [6r - 14 + 5/(r^2)] / (r^2 + 4r - 2 - 21/r) Multiply suitable & backside by capacity of r^2 (6r^3 - 14r^2 + 5) / (r^4 + 4r^3 - 2r^2 - 21) I did this long branch and have been given r/6 + 19/18 plus a the rest, yet i think of you probably did no longer write the undertaking wisely. [Edit] Oops ... I see what you probably did. You ignored the parentheses. the suited Saul have been given it for you. basically substitute i could make in his answer is to place (r+7)^2 interior the denominator.

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- An ESL LearnerLv 71 decade ago
(r^2 - 4r - 5)/(5r - 25)

= (r^2 + r - 5r - 5)/(5r - 25)

= (r + 1)(r - 5)/5(r - 5)

= (r + 1)/5

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- 1 decade ago
r^2 - 4r - 5

---------------

5r - 25

= r^2 + r - 5r - 5

--------------------

5 ( r - 5)

= r(r + 1) -5 (r + 1)

-------------------------

5 ( r - 5)

= (r + 1) (r - 5)

------------------

5 ( r - 5)

= (r + 1)

---------

5

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- Math HLv 61 decade ago
[r^2 -5r + r - 5]/5(r - 5)

[r(r - 5) +1(r - 5)]/5(r - 5)

[(r - 5)(r + 1)]/5(r - 5)

= (r+1)/5

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