How do you find the remainder for $(x^{3} + 11 x^{2} + 24 x + 9) \div (x + 2)$ $(x^3+11x^2+24x+9)div(x+2)$ ?

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Answer 1 · 2016-09-13T04:18:00

$The Remainder = - 3$ $" The Remainder"=-3$ .

We will use the Remainder Theorem :

$A poly. P (x), when divided by (p x + q) leaves remainder P (- \frac{q}{p})$ $"A poly. "P(x)," when divided by "(px+q)" leaves remainder "P(-q/p)$ .

Let $P (x) = x^{3} + 11 x^{2} + 24 x + 9$ $P(x)=x^3+11x^2+24x+9$ .

Comparing $(x + 2)$ $(x+2)$ with $(p x + q), p = 1, q = 2$ $(px+q), p=1, q=2$ .

Hence by the Remainder Theorem,

The Remainder $= P (- \frac{q}{p}) = P (- 2)$ $=P(-q/p)=P(-2)$

$= {(- 2)}^{3} + 11 {(- 2)}^{2} + 24 (- 2) + 9$ $=(-2)^3+11(-2)^2+24(-2)+9$

$= - 8 + 44 - 48 + 9$ $=-8+44-48+9$ .

$:." The Remainder"=-3$ .

Enjoy Maths.!

How do you find the remainder for (x^3+11x^2+24x+9)div(x+2)(x3+11x2+24x+9)÷(x+2)(x^3+11x^2+24x+9)div(x+2)?