# Divide the polynomial n^2 + 10n + 18 by n+5.

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### 3 Answers

We have to divide n^2 + 10n + 18 by n+5

(n^2 + 10n + 18) / (n+5)

Now, using the remainder theorem we find the remainder when (n^2 + 10n + 18) is divided by (n+5) as (-5)^2 - 10*5 + 18 = 25 - 50 + 18= -7

Now n^2 + 10n + 18 + 7 = n^2 + 10n + 25 = (n + 5)^2.

(n + 5)^2 divided by (n + 5) gives (n + 5)

**Therefore n^2 + 10n + 18 divided by n+5 gives a quotient of (n+5) and a remainder of -7.**

We'll divide the polynomial n^2+10n+18 by n+5.

We'll re-write n^2+10n+18, by completing the square:

n^2+10n+_

We'll add and subtract 25 and we'll get:

(n^2+10n+ 25) - 25 + 18

We notice that we can write (n^2+10n+ 25) = (n + 5)^2

We'll re-write the numerator n^2+10n+18:

n^2+10n+18 = (n + 5)^2 - 25 + 18

n^2+10n+18 = (n + 5)^2 - 7

We'll re-write the division:

( n^2+10n+18)/(n+5) = (n + 5)^2/(n + 5) - 7/(n + 5)

**We'll simplify and we'll get:**

**( n^2+10n+18)/(n+5) = n + 5 - [7/(n + 5)]**

To divide the polynomial n^2+10n+5 by n+5.

(n^2+10n+5) /(n+5) .

n+5) n^2+10n+18(n

n^2+5n

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

n+5) 5n+18 (n+5.

5n+ 25

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

-7.

Therefore n^2+10n+18/(n+5) = (n+5) is the quotient and -7 is the remainder.