Question

show that x+3 is a factor of 69+11x-x²+x³.
step by step explanation
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Answer 1

Answer:

x^3-x^2+11x+69 is perfectly divisable by x+3

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

We need to show that (x+3) is a factor of 69+11x-x^2+x^369+11x−x

2

+x

3

. If it is the factor of (x+3), then f(-3) should be equal to 0.

Let f(x)=69+11x-x^2+x^3f(x)=69+11x−x

2

+x

3

Put x = -3 in above function

\begin{gathered}f(-3)=69+11\left(-3\right)-\left(-3\right)^{2}+\left(-3\right)^{3}\\\\f(-3)=0\end{gathered}

f(−3)=69+11(−3)−(−3)

2

+(−3)

3

f(−3)=0

It means that (x+3) is a factor of 69+11x-x^2+x^369+11x−x

2

+x

3

.