Question

X3-3x2-9x-5/x+1 Divide or factorise that question of polynomials

Answer 1

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Answer 2

It can be seen that the remainder is equal to 0 and its quotient is $(x^2-4x-5)$.

Step-by-step explanation:

We need to divide $(x^3-3x^2-9x-5)$ by $(x+1)$. We can do it by long division method such that,

$\begin{array}{r|l}&x^2-4x-5\\\cline{2-2}\\x+1&x^3-3x^2-9x-5\\&x^3+x^2\\\cline{2-2}&\;\;\;\;\;-4x^2-9x\\&\;\;\;\;\;-4x^2-4x\\\cline{2-2}&\;\;\;\;\;\;\;\;\;\;\;\;-5x-5\\&\;\;\;\;\;\;\;\;\;\;\;\;-5x-5\\\cline{2-2}&\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;0\\\cline{2-2}\end{array}$

It can be seen that the remainder is equal to 0 and its quotient is $(x^2-4x-5)$.

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